EE105LectureNotes(6)

Lecture Notes

Jay A. Farrell, ProfessorCollege of Engineering

University of California, Riverside

February 10, 2014

Contents

1 System & Model Basics: Lecture 1 1

2 Review – Basic Matrix & Linear Algebra Concepts 5

3 Types of Models: Lectures 2 – 3 73.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Advanced Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4 Frequency and Step Responses: Lecture 4 164.1 Frequency response. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Simulation: Lecture 5 19

6 Energy Domains & Power Connections: Lecture 6 – 7 216.1 Energy Domains (Karnopp Section 2.1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216.2 Power Connections (Karnopp Section 2.1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.3 Word Bond Graphs (Karnopp Section 2.2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

7 Bond Graphs: Lecture 8 29

8 Multiports: Lecture 9 328.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

9 Causality Propagation and State Space Models: Lecture 10 – 11 369.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

10 Electrical Systems & Bond Graphs: Lectures 12 46

11 Bond Graph modeling (Circuits, Fluids, Thermal): Lects. 12 – 14 4711.1 Modeling Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4711.2 Electrical Circuit Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4811.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

12 Exam 1 55

13 Bond Graph modeling – Two Ports: Lecture 15 5613.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5813.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

i

CONTENTS February 10, 2014

14 Bond Graph modeling of Networks: Lecture 16 6314.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

15 Fluids: Lectures 17-18 6515.1 Fluid Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6615.2 Additional Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6815.3 Fluids Bond Graph Modeling Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7115.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7115.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

16 Thermal Systems: Lecture 19-20 7616.1 Steady State (static) Thermal Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7716.2 Dynamic Thermal Systems: Lecture 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8216.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

17 Exam 2 88

18 Causality Problems: Lecture 21 8918.1 Derivative Causality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8918.2 Algebraic Loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9118.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

19 Linear Dependence, Determinants, Matrix Inverse: Lecture 22 9719.1 Linear Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9719.2 Determinants, Minors, and Cofactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9819.3 Matrix Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9919.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

20 Eigenvalues and Eigenvectors: Lecture 23 10420.1 Matrix Diagonalization: Extra material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10620.2 Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10620.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

21 State Space to Transfer Function: Lecture 23 10821.1 Equivalent State Space Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11021.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

22 Mechanical Systems: Lecture 24 11222.1 Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11222.2 Mechanical: Translation Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11322.3 Mechanical: Rotation Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11522.4 Right hand rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

23 Bond Graph modeling–Mechanical Systems: Lecture 25 11623.1 Mechanical Modeling Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

24 Bond Graph modeling – Translation Examples: Lecture 26-27 11624.1 Sign convention examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11624.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

25 Bond Graph modeling – Rotation Examples: Lecture 27-28 12225.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

26 Mechanical Coupling – Slip or No slip: Lecture 29 12526.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Copyright ©2000-2014, J. A. Farrell. All Rights reserved. p. ii


27 Simple Mixed Systems: Lecture 29 12827.1 Wheels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12827.2 Electric Water Pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12927.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

28 State Space to Convolution: Interesting Lecture 133

29 System Design and Analysis: Lecture 30 13429.1 Stability Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13429.2 Steady State & Transient Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13529.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

30 Extra Lecture - Diagram Models (w/ Simulink) 142

A Vector & Matrix Review 146A.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146A.2 Matrix and Vector Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146A.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Copyright ©2000-2014, J. A. Farrell. All Rights reserved. p. iii


Preface

This document contains my lecture notes for a one quarter course on dynamic system modeling and simulation.The objective is to train students:

1. to develop state space and transfer function models for dynamic systems;

2. to understand how to transform between state space and transfer functions models;

3. to simulate the dynamic system models; and

4. to perform basic analysis of the system performance.

These notes are not a text book. I make the notes available to students in my class, because I know that there maybe errors either when I write on the board or when students write their notes. Distributing the notes allows students:

� to fix both types of errors,

� to work through examples at a slower pace,

� to try additional examples that are in these notes that I do not get to in lecture, and

� to double check these notes while working through the homework.

The notes will contain errors. Please help improve the notes by reporting errors to me, by making constructivesuggestions, and by asking questions during class.

Please do not think that, because I distribute my lecture notes, it is not necessary to come to class. These notesare only a guide to the lecture material. The notes are note a textbook. I will discuss many issues in class that arenot in these notes. Also, questions ask by students will raise important clarifications. If you skip class you will missmany important points, as well as problem solving suggestions that will save you time both on homework, on exams,and later in your career.

The main references for this set of notes are [2, 3]. The textbook [1] is available on-line as a pdf; however, itsnotation is slightly different from the standard notation established in [2] that we use in this class.

The course will introduce many new concepts. Most will be mathematically expressed. You may not be com-fortable with the concepts or mathematics when we start, but like anything else in life, we improve with practice,practice, and more practice.

Jay A. FarrellDepartment of Electrical EngineeringUniversity of California, Riverside

Copyright ©2000-2014, J. A. Farrell. All Rights reserved. p. iv

1 SYSTEM & MODEL BASICS: LECTURE 1 February 10, 2014

1 System & Model Basics: Lecture 1

Course Organization.

1. Syllabus

2. Curve

3. ILearn

4. Lab handout distribution

Course Objectives. Modeling and Simulation of Dynamics Systems

1. Dynamic Systems – a system whose status changes with time (i.e., position of a vehicle)

2. Modeling – development of mathematical representations to enable quantitative analysis and design

3. Simulation – numeric analysis of system models

Course Focus. State space modeling for analysis and simulation.

1. State space models allow convenient and efficient analysis via numeric methods.

Inputs: External variables that affect the system behavior.

States: Internal memory of the system. Initial values of the state variables are required to solved thesystem equations.

Outputs: Algebraic functions of the input and state variables that are of interest.

2. Main Tools:

(a) Matlab: Originally written by engineers for engineer to analyze models for the design of dynamicphysical systems.

(b) Bond graphs: Organize the variables and equation necessary for model derivation.

i. emphasize power interactions between subsystems

ii. emphasize analogies between different energy domains (Electrical, translational, rotational, fluids,thermal)

iii. provide a systematic approach to the development of state space models

(c) Intellect: Ultimately, the designer is responsible for ensuring the accuracy and reasonableness of themodel and the analysis.

Copyright ©2000-2014, J. A. Farrell. All Rights reserved. p. 1

February 10, 2014

System Variables. [3] The mathematical model of a system may contain a variety of quantities that can be cate-gorized as follows:

� Constants – do not vary as a function of time

– System parameters – Any constant that is defined by the system and not changeable by the designer.

– Design parameters – Any constant that the design can choose to affect the cost or behavior of thesystem.

� Signals – may vary as a function of time

– External variable – Any signal that can affect the behavior of the system, but is not (significantly)affected by the system.

* Inputs – An external signal that the designer/user can choose. Inputs are denoted by u(t).

* Disturbances – An external signal that the designer/user cannot choose. Disturbances are denotedby d(t) or w(t).

– Outputs – A variable that is of primary interest to the user/designer of the system. The output isdenoted by y(t).

– Internal variables (i.e., states)– Additional signals required to completely represent the systems be-havior that are distinct from the output and external variables.

Example 1.1 A bank checking account has a balance that changes as a function of time based on the net depositsand fees. A model is

bk = bk−1 + dk − fk

where bk is the checking account balance at the end of month k, i is the monthly interest rate, dk is the total netdeposits during month k, and fk is the total amount of fees charged during month k.

A bank savings account has a balance that changes as a function of time based on the net deposits and interest.A model is

sk = i sk−1 + wk

where sk is the savings account balance at the end of month k, i is the monthly interest rate and wk is the total netdeposits to this account during month k.

For this system, the variables are categorized as follows:

Inputs dk, wk

Disturbance fkInternal variables bk, skSystem parameter i

Definition of the output requires a little more discussion, as it is dependent on the purpose of the model. If the maininterest of the model is to specify total networth, then

yk = bk + sk;

however, some users may have other interests and therefore specify other outputs. For example, if the balance ineither account gets too low, it may trigger fees. To allow us to prevent this, we may want to monitor the two accountbalances by defining the vector output

yk =

[bksk

].


1 SYSTEM & MODEL BASICS: LECTURE 1 February 10, 2014

Example 1.2 Corresponding to Figure 1, the system variables are categorized as follows:

Output vL(t)Input u(t)Disturbance iL(t)Internal variables v1(t), v2(t), v3(t)Design parameters R1, R2, R3, R4 C1, C2, C3, C4

+_ Loadu C1 C2 C3 C4 vL

v1 v2 v3 R1 R2 R3 R4

+_

iL

Figure 1: Electrical example.

Discuss why iL is the disturbance while vL is the output.

Example 1.3 1 Repeat above with an output voltage.

Example 1.4 2 Rectifier with inductor determining output current.

Example 1.5 3 Room temperature.

1Unfinished.2Unfinished.3Unfinished.


February 10, 2014

Model Characteristics. [3] The models for systems can be sorted according to their various characteristics:

Static vs. Dynamic: A system is static if there is a direct and instantaneous relationship between theinputs and the outputs. A resistor is an example of static system where the voltage and current areinstantaneously related via Ohm’s law: v(t) = Ri(t). Static systems have not memory of past inputs.A system is dynamic if the output at time t depends on some past history of the the input. Inductorsand capacitors are both examples of dynamic systems. For example, with a capacitor, the voltage isdependent on the entire past history of the current through it: vc(t) =

1C

∫ t

−∞ i(τ)dτ . Another exampleis the velocity of a mass m which is dependent on the entire past history of the applied force according tov(t) = 1

m

∫ t

−∞ f(τ)dτ .

Continuous-time vs. Discrete-time: Models that represent the change of the system variables as a functionof the variable t ∈ ℜ are continuous-time models. Models that represent the changing system variablesas a function of k ∈ I are discrete-time models. Often discrete-time models result from the samplingof continuous-time systems at discrete-time instants. Most physical systems evolve in continuous-time.

Lumped vs. Distributed Parameter: Many physical systems are described by partial differential equa-tions. For example, the change of temperature u as a function of location and time is described by

∂u

∂t= k

(∂2u

∂2x+

∂2u

∂2y+

∂2u

∂2z

).

This represents the fact that the temperature change is distributed as a function position. The solutionu(x, y, z, t) is infinite dimensional. Many systems can be accurately represented by lumped models wherethe changing system status is represents by a finite number of lumped variables.

Linear vs. Nonlinear: A system model is linear if it includes only linear algebraic and differential equations.Otherwise, the model is nonlinear.

Deterministic vs. Stochastic: A model is stochastic when it involves random variables, processes, or ef-fects as representation of uncertainty. Otherwise, the model is deterministic.

This course will construct and analyze of continuous-time, lumped, deterministic, linear models for dynamic systems.Nonlinear models and their linearization may be discussed briefly. Conversion of distributed parameter models tolumped parameter models will also be considered briefly in the discussion of thermal systems.


2 REVIEW – BASIC MATRIX & LINEAR ALGEBRA CONCEPTS February 10, 2014

2 Review – Basic Matrix & Linear Algebra Concepts

This section is intened as a review. We will build on these concepts for the rest of the quarter. Please make surethat you have them now. If you are not understanding, please read the appendix and ask questions.

Matrices:

� The notation A ∈ ℜn×m means that A is a n row by m column tabular arrangement of real numbers.

Example 2.1 — An example for n = 3 and m = 4 is

A =

0.20 0.00 −3.14 0.00−1.71 0.00 2.00 0.001.00 1.00 1.00 0.00

∈ ℜ3×4.

△

� The notation A = [aij ] indicates that the element in row i and column j will be named aij .

Example 2.2 — In Example 2.1, a2,3 = 2.00 and a1,1 = 0.20. △

There are a few important matrices with special symbols.

– The identity matrix A = In ∈ ℜn×n has

ai,j =

{1 for i = j0 otherwise.

Example 2.3 — The identity matrix ℜ3×3 is I3 =

1.00 0.00 0.000.00 1.00 0.000.00 0.00 1.00

. △

– The null matrix is Z = 0n,m ∈ ℜn×m has zi,j = 0 for all i ∈ [1, n] and j ∈ [1,m].

Example 2.4 — The zero matrix ℜ2×3 is 02,3 =

[0.00 0.00 0.000.00 0.00 0.00

]. △

Matrix equality:For matrices A, B ∈ ℜn×m the notation A = B means that ai,j = bi,j for all i ∈ [1, n] and j ∈ [1,m].

Matrix Addition and Subtraction:For matrices A, B, C ∈ ℜn×m:

� The notation C = A+B means that ci,j = ai,j + bi,j for all i ∈ [1, n] and j ∈ [1,m].

� The notation C = A−B means that ci,j = ai,j − bi,j for all i ∈ [1, n] and j ∈ [1,m].

For example, if

A =

[0.20 0.00

−1.71 0.01

]and B =

[0.00 3.001.71 2.00

], then C = A+B =

[0.20 3.000.00 2.01

].

Scalar multiplication:When α ∈ ℜ and A ∈ ℜn×m, then the notation C = αA means that ci,j = αai,j for all i ∈ [1, n] and j ∈ [1,m].For example, with A as defined above,

D = 2.0A =

[0.40 0.00

−3.42 0.02

].


February 10, 2014

Matrix multiplication:

� For matrices A ∈ ℜn×p and B ∈ ℜp×m, the matrix C = A B ∈ ℜn×m is defined as

ci,j =

p∑k=1

ai,kbk,j

for all i ∈ [1, n] and j ∈ [1,m].

� Note that matrix multiplication is only defined when the ‘inner dimensions’ (number of columns of thematrix A and number of rows in B) are compatible. This is referred to as A and B being conformable.

� For example, for

A =

[0.20 0.001.00 3.00

]and B =

[1.00 2.00 3.00

−1.00 4.00 1.00

], then C = A B =

[0.20 0.40 0.60

−2.00 14.00 6.00

].

� The identity matrix I has the property that for any situation where A and I are conformable, AI = A.

� The order of matrix multiplications is important. For conformable A, B, and C,

– (AB)C = A(BC)

– AB = BA

– The product AB can exist when BA is not even conformable. Consider above example.

� Matrix multiplication is very useful for organizing (and solving linear systems of equations). For example,

1 = 2x1 − 4x2 − 8x3

2 = 4x2 + 15x1 + 16x3

3 = 3x2 − 3x3

is equivalent to 123

=

2 −4 −815 4 160 3 −3

x1

x2

x3

.

Linear Systems of Equations:

� The general form of such equations isb = Ax

where b ∈ ℜn×1 is known, A ∈ ℜn×m is known, and x ∈ ℜm is the unknown vector variable.

� This equation would be easy to solve if we new a matrix B such that B A = I, then

b = Ax

B b = B A x

B b = I x = x

� The matrix B is referred to as the left inverse of A. The inverse of A is denoted as A−1. This is not 1A ,

which is not even defined. We will be very interested in matrix inverses:

– When does such a matrix exist?

– How can the inverse be calculated?

The matrix inverse is discussed in Section 19.

For additional matrix review, see Appendix A.


3 TYPES OF MODELS: LECTURES 2 – 3 February 10, 2014

3 Types of Models: Lectures 2 – 3

We are interested in mathematical models that relate the input, disturbances, and constants to the time evolutionof the outputs and internal variables.

3.1 Fundamental Concepts

There are various types of equation-based models.

Systems of Ordinary Differential Equations: Consists of one ODE for each output. The ODE relates theoutput and its derivatives to the inputs, other outputs and their derivatives, the disturbances and the constants.

� A general n-th order SISO model could have the form

g(y(n)(t), y(n−1)(t), . . . , y(t), u(n−1)(t), u(n−2)(t), . . . , u(t)

)= 0

where y(k)(t) = dk

dtky(t) and g is a (possibly) nonlinear function.

� A nonlinear 3-rd order SISO model is

y(3)(t) + d1|y(2)(t)|+ d2y(1)(t) + d3 sin (y(t)) = n1u

(2)(t) + n2

((u(1)(t)

)2

+(u(t)

)2)

� A linear 3-rd order SISO model is

y(3)(t) + d1y(2)(t) + d2y

(1)(t) + d3y(t) = n1u(2)(t) + n2u

(1)(t) + n3u(t) (1)

� A set of coupled ODE’s

z + z = y + u (2)

y = −z (3)

Often, the direct application of physical laws will yield a set of higher order ODE’s.

Transfer Functions: For a linear system with all initial conditions set to zero, it is possible to find a transferfunction from each input to each output. The transfer function is defined as the ratio of the Laplace transformof the output divided by the Laplace transform of the input. For a linear system without pure delay, thetransfer function is the ratio of two polynomials in the Laplace variable s. Setting the numerator of the TFto zero, N(s) = 0, allows determination of the TF zeros. Setting the denominator of the TF to zero, D(s) = 0,allows determination of the TF poles.

� Corresponding to eqn. (1), the transfer function is4

L(y(3)(t) + d1y

(2)(t) + d2y(1)(t) + d3y(t)

)= L

(n1u

(2)(t) + n2u(1)(t) + n3u(t)

)s3Y (s) + +d1s

2Y (s) + d2sY (s) + d3Y (s) = n1s2U(s) + n2sU(s) + n3U(s)(

s3 ++d1s2 + d2s+ d3

)Y (s) =

(n1s

2 + n2s+ n3

)U(s)

Y (s)

U(s)=

n1s2 + n2s+ n3

s3 ++d1s2 + d2s+ d3=

n1

∏2i=1(s+ zi)∏3

i=1(s+ pi)=

N(s)

D(s)(4)

This third order, strictly proper system has three poles and two finite zeros.

� Example 3.1 — Corresponding to eqns. (2–3) the transfer functions from the input u to the outputs yand z are

L (z + z) = L (y + u) (5)

s2Z(s) + sZ(s) = Y (s) + U(s) (6)

4Note that L(f)= s2F (s)− sf(0)− f(0).


3.1 Fundamental Concepts February 10, 2014

L (y) = L (−z) (7)

s2Y (s) = −Z(s) (8)

s2Z(s) + sZ(s) = − 1

s2Z(s) + U(s) (9)

s4Z(s) + s3Z(s) + Z(s) = s2U(s) (10)

Z(s)

U(s)=

s2

s4 + s3 + 1(11)

Y (s)

U(s)=

Y (s)

Z(s)

Z(s)

U(s)=

−1

s2s2

s4 + s3 + 1(12)

=−1

s4 + s3 + 1(13)

△

State Space: Earlier discussion mentioned that a dynamic system has memory in the sense that the present valueof its output y(t) is dependent on the entire past history of the input u(t). Therefore, to determine y(t) for allt ≥ t0 it is not enough to know u(t) for all t ≥ t0. In addition, we require information about that completelycharacterizes the status of the system at t0.

Example. To determine the monthly ending balance of a bank account, it is not enough to know the dailyinflows and outflows for the given month, we also must know the initial account balance for the month.

Concept of state: The state of a system at time t0 is the information required such that knowledge of thestate at time t0 and of u(t) for all t ≥ t0 allows determination of the output y(t) for all t ≥ t0.

� For finite dimensional systems, the state information can be organized into a vector x = [x1, . . . , xn]⊤ ∈

ℜn.

� The definition of the state vector is not unique. Consider a capacitor with an input current andthe output being the capacitor voltage. Knowledge of the initial charge of the capacitor and allsubsequent currents through the capacitor are sufficient to determine the voltage across the capacitorat any future time. Similarly, knowledge of the initial capacitor voltage and all future currents, issufficient to allow computation of the capacitor output voltage at all future times. Either the initialcharge or the initial voltage could be selected as the system state.

� Initial conditions for an ODE are one example of a state definition.

Order of a System: The order of a system is minimum number of independent pieces of information requiredto specify the state.

� Variables are dependent if they are algebraically related.

� In the capacitor example, the capacitor charge and voltage are algebraically related by q(t) = Cv(t).

� The order of the capacitor system is n = 1. Either the capacitor charge or the capacitor voltage cambe selected as the state variable.

State space models: A state space model for an n-th order system with m inputs, p outputs includes n firstorder ODE’s and and p algebraic equations: x ∈ ℜn, u ∈ ℜm, and y ∈ ℜp.

� There is one first order ODE describing the time derivative of each state as a function only of thestate variables and the inputs:

xi(t) = fi(x(t), u(t))i = 1, . . . , n

}⇐⇒

x1(t) = f1(x(t), u(t))

...xn(t) = fn(x(t), u(t))

� There is one algebraic equation relating each output only to the state variables and the inputs:

yj(t) = hj(x(t), u(t))j = 1, . . . , p

}⇐⇒

y1(t) = h1(x(t), u(t))

...yp(t) = hp(x(t), u(t))



The general notation is

x(t) = f(x(t), u(t)) (14)

y(t) = h(x(t), u(t)) (15)

where the vector functions f and h can be nonlinear.

� Linear – A state space model is linear if the (n+p) functions f(x, u) and h(x, u) are all linear functionsof x and u. In this case,

f(x, u) = Ax+Bu

h(x, u) = Cx+Du

where A ∈ ℜn×n, B ∈ ℜn×m, C ∈ ℜp×n, and D ∈ ℜp×m. In this case, the linear state space model is

x = Ax+Bu (16)

y = Cx+Du. (17)

Consult Linear Algebra review.

State Variable Selection:

� One method to define the state variables is to select the set of initial conditions that would be requiredto solve the set of differential equations.

Example 3.2 — The ODE describing a rigid pendulum is

Jθ + bθ +mg sin(θ) = 0.

To solve this 2nd order ODE, we require knowledge of the initial angle and initial angular rate. If weselect the output and state vector as y = θ and x = [θ, θ]⊤, then the state space model is

x1 = x2 (18)

x2 = −mg

Jsin(x1)−

d

Jx2 (19)

y = x1. (20)

This model is in the form of eqns. (14–15) where

f(x, u) =

[x2

−mgJ sin(x1)− d

J x2

]and h(x, u) = x1.

This is a nonlinear model due to the presence of the sin(x1) term. △

Example 3.3 — For the system of eqns. (2–3), define

x1 = z

x2 = z

x3 = y

x4 = y.

Take the derivative of each of these equations, and eliminate all variables except the states and theinput. For x1,

x1 = z

= x2.

For x2,

x2 = z

= y + u− z

= x3 + u− x2.


3.1 Fundamental Concepts February 10, 2014

For x3,

x3 = y

= x4.

For x4,

x4 = y

= −z

= −x1.

The state space model is

x1 = x2

x2 = −x2 + x3 + u

x3 = x4

x4 = −x1

with outputs defined by

y = x3

z = x1.

Because this system is linear, we can write it in the form of eqns. (16–17) with

A =

0 1 0 00 −1 1 00 0 0 1

−1 0 0 0

B =

0100

C =

[0 0 1 01 0 0 0

]D =

[00

]△

� In future lectures, we will see how to read the state variable definitions and the state space modeldirectly from a bond graph.

� When a linear ODE contains derivatives of the input, it is sometimes difficult to proceed. One methodis exemplified below.

Example 3.4 — Corresponding to eqns. (1), the transfer function of eqn. (4) is

Y (s)

U(s)=

n1s2 + n2s+ n3

s3 ++d1s2 + d2s+ d3.

To simplify the process, we introduce the intermediate signal r(t):

u(t)−→ 1s3+d1s2+d2s2+d3

r(t)−→ n1s2+n2s+n3

1

y(t)−→

such that Y (s)U(s) = Y (s)

R(s)R(s)U(s) where

Y (s)

R(s)=

n1s2 + n2s+ n3

1, and

R(s)

U(s)=

1

s3 + d1s2 + d2s2 + d3;

therefore, we have

y(t) = n1r(2)(t) + n2r

(1)(t) + n3r(t) (21)



and

r(3)(t) + d1r(2)(t) + d2r

(1)(t) + d3r(t) = u(t). (22)

By defining the state vector x = [r, r, r]⊤, eqn. (21) can be rewritten as

y(t) = n1x3(t) + n2x2(t) + n3x1(t) (23)

and eqn. (22) can be written as

r(3)(t) + d1x3(t) + d2x2(t) + d3x1(t) = u(t). (24)

Now we are in a position to simplify and find the state space model

r(3)(t) = −d1x3(t)− d2x2(t)− d3x1(t) + u(t)

x3(t) = −d1x3(t)− d2x2(t)− d3x1(t) + u(t).

It is also straightforward to show that x1 = x2 and x2 = x3. Summarizing these equations yields thestate space model:

x1 = x2

x2 = x3

x3 = u(t)− d1x3 − d2x2 − d3x1

andy = n1x3 + n2x2 + n3x1.

△

3.2 Causality

Simple devices are configured to construct systems to perform desired objectives. Some simple devices have a causalitythat is natural or preferred. Device causality has three types: algebraic, integral, and derivative.

To explain the concept of causality, we will follow the convention of writing equations with the inputs (i.e.,independent) variables on the right of the equals sign and the output (i.e., dependent or determined) values on theleft hand side. The equation y = f(u) should be read as y is the output of the system f when the input is u.

Consider a resistor with resistance R connected across the terminals of a voltage source. In this case the voltagesource is the input v(t), the resulting current i(t) = 1

Rv(t) is the output of the resistor in response to the appliedinput. This is an example of algebraic causality because the equation relating the input v(t) to the output i(t) isalgebraic. It is algebraic because the equation does not involve any integrals or derivatives. If the resistor wereconnected to a current source input, then the input-output relationship would be v(t) = Ri(t), where this is stillan algebraic expression. Now the input is i(t) and the output is v(t). Many simple devices have algebraic causality:dampers, friction, valves, etc.

Alternatively, consider a capacitor with capacitance C connected across the terminals of a voltage source. Theresulting current i(t) = C d

dtv(t) is the output of the capacitor in response to the applied input voltage. Because theoutput is determined by taking the derivative of the input, this is an example of derivative causality. If the capacitorwere connected to a current source as its input, then the input-output relationship would be v(t) = 1

C

∫ t

−∞ i(τ)dτ .Because the output v(t) is a function of the integral of the input i(t), this is an example of integral causality. Notethat the manner in which the capacitor is connected determines it causality. Many simple devices will have eitherintegral or derivative causality depending on the manner in which they are connected: inductors, masses, inertias,springs, fluid tanks and pipes, etc.

Causality is a critical issue, as nature prefers integral causality and abhors derivative causality. Models showingdevices with derivative causality are missing critical assumptions or have unmodeled effects and should be carefullyevaluated. Systems connected with devices having derivative causality may have extra unneeded components ordesign flaws that may lead to short system life spans. In your circuits course, you should have learned not to connectan inductor in series with a current source or a capacitor in parallel with a voltage source. Both of these are simpleexamples of devices connected with derivative causality. In more complex systems, devices with derivative causalityare more difficult to detect. The bond graph method will detect instances of derivative causality.


3.3 Advanced Concepts February 10, 2014

3.3 Advanced Concepts

Following is a brief discussion of important concepts that are slightly more advanced.

Stationary solutions: If u(t) = u0 (i.e., constant) and x0 is a solution to f(x0, u0) = 0, then x(t;x0, t0, u0) = x0

is a solution to the initial value problemx(t) = f(x(t)), u0)

with initial conditionx(t0) = x0.

The solution (x0, u0) is a stationary point, singular point, or equilibrium of the system. All stationarypoints of a system must satisfy

f(x0, u0) = 0.

Example. A fluid tank with inflow u is described by the ODE

h(t) = −√h(t) + u(t)

with output flow rate described byq(t) =

√h(t).

The variable h is the height of the fluid in the tank and must be positive. This is a nonlinear first order statespace model with x = [h], f(x, u) = −

√x+u, and h(x, u) =

√x. For a constant input u(t) = u0, the equilibria

are defined by the solutions √x0 = uo or x0 = u2

0.

Example. A model of two species where species one preys N1 on species two is

N1 = (λ1 − γ1)N1 + α1N1N2

N2 = (λ2 − γ2)N2 − α2N1N2

where Ni ≥ 0 is the number of individuals of species i, λi is the reproduction rates of species i, γi is thedeath rate of species i, and αi represent predation rates. This is a second order state space model with noinputs. A valid definition of the state is x = [N1, N2]

⊤. The system has two equilibria: x0 = (0, 0) and

x0 =(

λ1−γ1

α1, λ2−γ2

α2

). If the number of species were to start out at either of these values, they would remain

constant forever. Otherwise, the number of each species will change.

Stability: This refers to the tendency for an equilibrium to attract solutions. An equilibrium is

Locally stable – If initial conditions sufficiently near to the equilibrium do not diverge away from the equi-librium.

Locally asymptotically stable – If it is locally stable and initial conditions sufficiently near to the equilib-rium eventually converge to the equilibrium.

Globally asymptotically stable – If all initial conditions eventually converge to the equilibrium.

Unstable – If there exist initial conditions arbitrarily near the equilibrium that diverge from the equilibrium.

As engineers, we are usually interested in designing systems with (at least) locally asymptotically stable equi-libria.

Linearization: Engineers are often interested in analyzing solutions of a nonlinear system with initial conditionsnear an equilibrium (x0, u0). We define the deviation variables:

δx = x− x0

δu = u− u0

δy = y − h(x0, u0).

Then, by Taylor’s Theorem,

δx = x− 0

δx = f(x0, u0) +∂f

∂x(x, u)

∣∣∣∣(x0,u0)

δx+∂f

∂u(x, u)

∣∣∣∣(x0,u0)

δu+ h.o.t′s

δx ≈ Aδx+Bδu



and

δy = h(x, u)− h(x0, u0)

δy = h(x0, u0) +∂h

∂x(x, u)

∣∣∣∣(x0,u0)

δx+∂h

∂u(x, u)

∣∣∣∣(x0,u0)

δu+ h.o.t′s− h(x0, u0)

δy ≈ Cδx+Dδu

which is a linear model that is valid for ∥δx∥ and ∥δu∥ small, where A = ∂f∂x (x, u)

∣∣∣(x0,u0)

, B = ∂f∂u (x, u)

∣∣∣(x0,u0)

,

C = ∂h∂x (x, u)

∣∣(x0,u0)

, and D = ∂h∂u (x, u)

∣∣(x0,u0)

.

The stability of an equilibrium of a nonlinear system can sometimes be determine by analysis of the linearizedsystem valid at that equilibrium location.

Example 3.5 — The system of eqn. (18–20) has equilibria at x = [nπ, 0]⊤ for integer values of n, which correspondto two physical equilibria.

If we wish to find a linearized model valid for x near [π, 0]⊤, then we define

δx = x− (π, 0)⊤

δy = y − π.

Then, because ∂ sin(x1)∂x1

∣∣∣π= cos(x1)|π = −1,

δx1 = δx2

δx2 =mg

Jδx1 −

d

Jδx2

δy = δx1.

In matrix form this is

δx =

[0 1mgJ − d

J

]x

y =[

1 0]x.

△

3.4 Exercises

Exercise 3.1 – For each of the following:

� Determine if the system is linear. If not, clearly state why not?

� If the system is linear, find the transfer function from input to output.

In each subproblem, the inputs are either u(t) or v(t). The outputs are y(t) and z(t).

1. y(5)(t) = u(t)

2. y5(t)− y(1)(t) = u(t). Note that y5 = y ∗ y ∗ y ∗ y ∗ y and y(5)(t) = d5

dt5 y are not the same!

3. y(t) + 15y(t) + sin(y(t)) = u(t)

4. y(t)− y(t) + 3y(t) = u(t) + u(t)

5. y(t) = u(t) and z(t) = u(t)− z(t)− y(t)

Exercise 3.2 – For Example 3.4:


3.4 Exercises February 10, 2014

� Define the functions f(x, u) and h(x, u) for the state space model.

� If the systems is linear, specify the matrices A, B, C, and D.

Exercise 3.3 – For each of the following I/O ODE’s:

� Clearly state the order of the system.

� Clearly define a state vector of that order.

� Find the state space model. Organize the state space model in the format

˙x(t) = f(x(t), u(t))

y(t) = h(x(t), u(t))

and clearly define the functions f and h.

� If the systems is linear, specify the matrices A, B, C, and D and write the system in matrix vector form.

Treat each numbered item as a separate system with inputs u and v; and, outputs y and z.

1. y(3)(t)− 2y(2)(t) + y(t) = u(t)

2. z(3)(t)− 2|z(2)(t)|+ z(t) = u(t)

3. z(3)(t)− 2z(2)(t) + z(t) =(u(t)

)24. z(t)− 2z(t) = u(t) + 2z(t)

5. z(t) = u(t) + 2y(t)− z(t) and y(t) = u(t)− z(t)

6. y(3)(t) = u(t) + z(t) and z(2)(t) = u(t)− y(t)− y(t)

7. y(t) = u(t)− v(t)

8. y(4)(t)− 2y(3)(t) = 7u(t)

9. y(t) + 17|y(t)|+ 10y(t) = u(t)

10. y(t) = 4u(t) and z(t) = 3y(t) + z(t)

11. z(t) + 2z(t) + z(t) = u(t)− u(t) (See Example 3.4.)

12. y(3)(t)− 2y(2)(t) + y(t) = u(2)(t) + 2u(1)(t) + u(t) (See Example 3.4.)

Exercise 3.4 – For each of the ODE’s in Exercise 3.1, find a state space model.

Exercise 3.5 – For the linear system

v − w = 0

w − v = u

z = v

where the output vector is y =

vwz

.

1. Define a valid state vector.



2. Specify the order of the system.

3. Find the matrices A, B, C, and D so that the state space model can be written as

x = Ax+Bu

y = Cx+Du

Exercise 3.6 – An inductor with a current source as an input has derivative causality. This is because the inputcurrent would be differentiated by the inductor to determine the output voltage across the inductor. The sameinductor with a voltage source input would have integral causality, because the voltage input would be integrated bythe inductor to determine the current through the inductor.

Questions:

1. Consider a capacitor connected to either a voltage source or a current source. Which interconnection yieldsintegral causality and which yields derivative causality for the capacitor? Draw the simple schematics. Labelthe voltages and currents. Write down the voltage-current relationships for the capacitor for each of the twoschematics

2. For the inductor with derivative causality

vL(t) = LdiLdt

.

If the input iL is a step function at time t = 0 that turns off at t = 1, what is vL(t)? What is the powersupplied by the power source? Is this physically reasonable?

3. For the inductor with integral causality

iL(t) =1

L

∫ t

−∞vL(τ)dτ.

If the input vL is a step function at time t = 0 that turns off at t = 1, what is iL(t)? What is the powersupplied by the power source? Is this physically reasonable?

4. Why is it reasonable to claim that nature prefers integral causality?

Exercise 3.7 –

For the nonlinear system

x1 = 1− |x2|x2 = x1,

find all of the equilibria (i.e., stationary points) of the system.

Exercise 3.8 – For the system of eqns. (18–20), find the linearized model around the equilibrium point at x =[0, 0]⊤.


February 10, 2014

4 Frequency and Step Responses: Lecture 4

4.1 Frequency response.

Let a strictly stable5 linear system have a transfer function indicated by G(s):

Y (s)

U(s)= G(s).

If the input is u(t) = A sin(ωt), then the output after transients have decayed away is

y(t) = |G(jω)|A sin(ωt+ ∠G(jω)).

Note that a complex number z, can be written in either rectangular z = a+ jb or polar form z = Mejθ = M∠θ.It is necessary to know how to convert between the two forms. Using Euler’s identity,

z = Mejθ = M (cos(θ) + j sin(θ)) = M cos(θ) + jM sin(θ);

therefore, given the polar form we can compute the rectangular form as a = M cos(θ) and b = M sin(θ). Given therectangular form, M = |z| =

√zz∗ =

√a2 + b2 and θ = ∠z = atan2(b, a). The function atan2 is s four quadrant

arctangent. If you use atan(ba

), you need to ensure that you manually adjust the result to be in the correct quadrant.

Example 4.1 — What is the steady state response of the linear system with transfer function G(s) = 2(s+2)(s−2)

to the input u(t) = 4 sin(2πt)?This system has poles at s = ±2. The pole as s = 2 is in the right half plane. The natural response of the

system will contain terms proportional to e2t and e−2t. The system is not stable. No steady state response will exist.Therefore, the frequency response approach cannot be used. △

Example 4.2 — What is the steady state response of the linear system with transfer function H(s) = s(s+1) to the

input u(t) = sin(ωt)?Evaluating H(s) at s = jω we can compute its magnitude and phase. Its magnitude is

|H(jω)|2 = H(jω)H(jω)∗ = H(jω)H(−jω)

=(jω)(−jω)

(1 + jω)(1− jω)

=ω2

1 + ω2

|H(jω)| =|ω|√1 + ω2

.

Its phase is computed as follows

H(s)|s=jw =jω

1 + jω

=

(jω

1 + jω

)(1− jω

1− jω

)=

ω2 + jω

1 + ω2

∠H(jω) = = atan

(1

ω

)= 90− atan(ω).

Therefore, for a fixed value of ω the steady state output is

y(t) =|ω|√1 + ω2

sin (ωt+ 90− atan(ω)) .

For this example, the time constant is τ = 1 second. Therefore, steady-state would be achieved in approximately4τ = 4 seconds. △

5All the system poles are in the left-half of the complex plane.


4 FREQUENCY AND STEP RESPONSES: LECTURE 4 February 10, 2014

Example 4.3 — If the system with transfer function G(s) = s2+1s2+3s+2 were excited by the signal u(t) = 4 sin(ωt).

Find the steady state output signal for ω = 0, 1, 2.

ω G(jω) |G(jω)| ∠G(jω) y0 0.5 0.5 0 21 0 0 DNM 02 −3

−2+6j 0.4743 71.56◦ 1.897 sin(2t+ 71.56◦)

For this example, the system has two poles, one at s1 = −1 and the other at s2 = −2. The transient response willhave the form a1e

−1t + a2e−2t. The corresponding time constants are τ1 = 1 sec and τ2 = 0.5 sec, respectively. The

term that decays the slowest is a1e−1t. Therefore, the dominant time constant is τ1 = 1 second and steady-state

would be achieved in approximately 4τ1 = 4 seconds. △

Proof of Freq. Resp. Result: In the following, the denominator D(s) is factored such that

G(s) =N(s)∏n

i=1(s+ pi)

where each pi > 0. In addition, since we are interested in steady state for a stable system, WOLG, we let the initialconditions be zero.

Y (s) = G(s)U(s) =N(s)∏n

i=1(s+ pi)

Aw

s2 + ω2

=

n∑i=1

Di

s+ pi+

C

s− jω+

C∗

s+ jω

G(s)Aw

(s+ jω)(s− jω)=

n∑i=1

Di

s+ pi+

C

s− jω+

C∗

s+ jω

where C = G(s) As+jw

∣∣∣s=jw

= G(jω) A2jω . Therefore,

Y (t) = L−1

{n∑

i=1

Di

s+ pi+

C

s− jω+

C∗

s+ jω

}

=n∑

i=1

Die−pit +G(jω)

A

2jejωt −G∗(jω)

A

2je−jωt

=n∑

i=1

Die−pit + |G(jω)|A 1

2j

(ej(ωt+∠G(jω)) − e−j(ωt+∠G(jω)

)=

n∑i=1

Die−pit + |G(jω)|A sin (ωt+ ∠G(jω)) .

The term |G(jω)|A sin (ωt+ ∠G(jω)) is the steady state response because for pi > 0 i = 1, . . . , n, the term∑ni=1 Die

−pit approaches zero as t → ∞.The summation

∑ni=1 Die

−pit contains n terms each of which decays away to zero. The rate at which each termdecays is determined by the value of pi. The slowest rate of decay will be due to the smallest value of pi. A quickestimate of the time required for the system to reach steady state of within 2% is Tss = 4mini(pi). Some books willsay Tss = 3mini(pi), which is within 5% if the final value. In either case, the idea is the same.

The shape of the transient response as a function of the pole locations in the complex plane is further discussedin Section 29.



4.2 Exercises

Exercise 4.1 – For each of the following systems and inputs:

1. Check whether the system is stable.

2. If the system is stable, use frequency response to determine the steady state response?

3. For a stable system, the time to reach steady state can be estimated as four times the time constant of theslowest pole. When a system is stable, all poles are in the left half plane (LHP). The slowest pole is the onewith the largest (least negative) real part. For each stable system, approximately how long will it take for thesteady state condition to be achieved. The time constant is the reciprocal of the absolute value of the real partof the pole location.

Problem System Input

1 H(s) = s2+101(s2+2s+1) u(t) = sin(1t)

2 H(s) = ss+10 u(t) = 10 sin(10t)

3 H(s) = 1s+5 u(t) = cos(5t)

4 H(s) = 10 ss+4 u(t) = sin(3t)

5 H(s) = 2s−17 u(t) = sin(ωt)

6 H(s) = s−1(s+10)(s+1) u(t) = cos(t)

7 H(s) = 10 ss+10 u(t) = sin(ωt)

8 H(s) = ss+10 u(t) = 10 sin(ωt)

9 H(s) = 1s+3 u(t) = cos(3t)

10 H(s) = s2+100(s2+200s+100) u(t) = sin(10t)

11 H(s) = 2s−10 u(t) = sin(ωt)

12 H(s) = s−1(s+10)(s+1) u(t) = sin(t)

13 H(s) = s2+4(s+10)(s+1) u(t) = cos(2t)

Exercise 4.2 –

1. For the I/O ODE,y(3)(t) + 5y(2)(t) + 7y(1)(t) + 3y(t) = u(t) + 2u(t)

find the transfer function from the input u(t) to the output y(t).

2. For the input u(t) = cos(2t), find

(a) the steady state output; and

(b) the approximate time to reach steady state.

Feel free to use Matlab to help with the evaluation of the transfer function. Useful functions are ‘polyval’ and‘root’.

3. Find a linear state space model. (See Example 3.4.)

4. Clearly define the matrices A, B, C, and D of the matrix form of the state space model:

x(t) = Ax(t) +Bu(t)

y(t) = Cx(t) +Du(t)


5 SIMULATION: LECTURE 5 February 10, 2014

5 Simulation: Lecture 5

It is very difficult and tedious to find analytic solutions for systems that are either nonlinear or higher order.

Simulation: The creation of one system which acts like another.

� May be motivated by the cost or risk of operating the original system.

� To be useful for design and analysis, the simulation must be an accurate representation of the originalsystem.

– Check variable units and interpret them correctly.

– Keep track of and check assumptions.

Numeric simulation: Solution of ODE’s by numeric methods.

Advantages: � Flexibility – model characteristics are easily changed

� Accuracy – completely controllable by the analyst

� Simplicity – high dimensions and nonlinearities are not problems.

Disadvantages: Results are only as accurate as the model and simulation. Model and simulate carefully.

Euler’s Method: Consider the system

x = f(x, u)

y = h(x, u)

with initial condition x(t0) = x0. By the definition of a derivative

dx

dt≈ x(t+ b)− x(t)

bfor small b;

therefore,

x(t0 + b)− x(t0)

b≈ f(x(t0), u(t0))

x(t0 + b) = x(t0) + bf(x(t0), u(t0)).

If x(0) = x0 is known, u(t) is known for all t ≥ 0, and b is small, then

x(b) = x(0) + bf(x(0), u(0)) (25)

x(2b) = x(b) + bf(x(b), u(b)) (26)

... =... (27)

x((k + 1)b) = x(kb) + df(x(kb), u(kb)) (28)

for k = 1, 2, 3, . . ..

Show simulation: instruction sim for x(t) = −2x(t)+?.

Comments:

� Euler’s method assumes that the slope is constant over time step.

� No solution is known between time steps. Matlab simply connects the dots. Interpret with caution.

� Euler’s method is first order.

– It requires one function eval per step

– Error grows proportional to d. Smaller d means smaller error, but more computation.

Higher order methods:

� An n-th order method has error proportional to dn, but typically requires n function evals per time step.


February 10, 2014

� Predictor-Corrector:

xp(k + 1) = x(k) + b f(x(k), u(k))

xc(k + 1) = x(k) + b f(xp(k + 1), u(k))

x(k + 1) = 0.5 (xc(k + 1) + xp(k + 1))

– This is a second order method.

– It estimates the slope at the beginning and end of each step.

Show simulation: instruction sim leture

Show simulation: instruction sim pred corr

� Matlab has very well designed simulation algorithms (up to order 6) with adaptive step-size formulas, etc.Still analyze with care.


6 ENERGY DOMAINS & POWER CONNECTIONS: LECTURE 6 – 7 February 10, 2014

6 Energy Domains & Power Connections: Lecture 6 – 7

This lecture introduces the effort, flow, momentum, and displacement variables in different energy domains (SeeSection 2.1 in Karnopp [2] or Ch. 5 in Ljung [3]).

The basic idea is that by recognizing parallels between the variables and physical relations in different energydomains, we can develop a general modeling procedure applicable to all the energy domains.

This lecture will work through one energy domain at a time while filling in the entries in Table 1.

Domain Effort Momentum Flow Displacement Power, N-m/secGeneric e p f q P , Joules/sec.Electrical e, Volt λ, Volt-s i, Amp Q, Coul. e(t)i(t)Fluid P , Newton/m2 pP , Newton-s/m

2 QV , m3/sec V , m3 P (t)QV (t)

Thermal T , ◦Kelvin ****** QH , Joules/sec EH , Joules T (t)QH(t)Translational F , Newton p, Newton-sec. v, m/s X, m F (t)v(t)Rotational τ , Newton-m pτ , Newton-m-s ω, rad/sec θ, rad τ(t)ω(t)

Table 1: Physical analogies between variables in different energy domains.

6.1 Energy Domains (Karnopp Section 2.1)

6.1.1 Generic

In each energy domain, we will define the following:

� An effort variable e(t) and its ANSI units

� A momentum variable p(t) =∫e(t)dt and its ANSI units

� A flow variable f(t) and its ANSI units

� A displacement variable q(t) =∫f(t)dt and its ANSI units

� The product of effort and flow P (t) = e(t)i(t) has units of power (Watts=Joules/sec).

6.1.2 Electrical Systems

� The effort variable is voltage e(t) measured in Volts (Joules/Coulomb).

� The momentum variable λ(t) =∫ t

0e(t)dt is the flux linkage. Its units are Volt-seconds. This variable is useful

for the modeling of inductors, transformers, and other devices storing magnetic energy.

� The flow variable is current i(t) measured in Amps (Coulombs/sec)

� The displacement variable is Q(t) =∫ t

0i(t)dt measured in Coulombs.

� The product of effort and flow P (t) = e(t)i(t) has units of power (Joules/sec=Watts).

6.1.3 Fluid Systems

� The effort variable is pressure P (t) measured in Newton/(sq. m) (kg 1/(ms2)).

� The pressure momentum variable is pP (t) =∫ t

0P (t)dt. Its units are Newton-m-seconds (kg 1/(m s)).

� The flow variable is the volume flow rate QV (t) measured in m3/s

� The displacement variable is V (t) =∫ t

0QV (t)dt measured in m3.

� The product of effort and flow p(t) = P (t)QV (t) has units of power((kg 1/(ms2))(m3/s) =

(kg(m/s)2

)/s=Joules/sec=Watts).


6.2 Power Connections (Karnopp Section 2.1) February 10, 2014

6.1.4 Thermal Systems

See discussion of pseudo-bond graphs in Karnopp [2].

� The effort variable is temperature T (t) which is dimensionless, but measured using the ◦Kelvin scale.

� The momentum variable is not used in thermal analysis.

� The flow variable is heat flow rate QH(t) measured in Watts.

� The displacement variable is EH(t) =∫ t

0QH(t)dt measured in Joules.

� The units of QH(t) are already power. The product of effort and flow does not defined power for thermalsystems. This is one reason why for thermal systems we have pseudo bond graphs.

6.1.5 Mechanical: Translation Systems

� The effort variable is force F (t) measured in Newtons (kg m/s2).

� The linear momentum variable is p(t) =∫ t

0f(t)dt. Its units are Newton-seconds (kg m/s).

� The flow variable is velocity v(t) measured in m/s

� A displacement variable X(t) =∫ t

0v(t)dt measured in meters, m.

� The product of effort and flow p(t) = F (t)v(t) has units of power((kg m/s2)(m/s) =

(kg(m/s)2


6.1.6 Mechanical: Rotation Systems

� The effort variable is torque τ(t) measured in Newton-meters (kg m2/s2).

� The angular momentum variable is pτ (t) =∫ t

0τ(t)dt. Its units are Newton-m-seconds (kg m2/s).

� The flow variable is angular velocity ω(t) measured in rad/s

� A displacement variable θ(t) =∫ t

0ω(t)dt measured in radians, rad.

� The product of effort and flow p(t) = τ(t)ω(t) has units of power((kg m2/s2)(rad/s) =

(kg(m/s)2


6.2 Power Connections (Karnopp Section 2.1)

� Energy is transferred between interconnected systems.

� Power is the rate of energy transfer and power is the product of effort times flow.

� The term port refers to the point at which a subsystem may interconnect with another subsystem. A subsystemmay have one or multiple ports. One ports have a single port (e.g., battery). Two ports have two ports (e.g.,pump). ...

� Associated with any port is a pair of effort and flow variables in the same energy domain. The energy domainfor different ports can be different.

– Draw DC motor and pump schematics

– Label power variables at ports

* Field and armature current supplying power to motor.

* Motor supplying power to load. Torque applied to motor opposing rotation.

� When subsystems are connected physically, the complementary power variables of the variable pair at theinterconnected ports are forced to have the same values. This is referred to as a bond.

– Draw interconnected battery,DC motor, and pump



– Discuss variables at ports

� Each bond connects exactly two ports and has exactly two power variables – an effort and a flow in the sameenergy domain. At any bond, one of the ports will determine the effort and the other port will determine theflow. The variable that the port determines is its output. The other variable is the input of the port. Thedetermination depends on the connected subsystems.

Examples.

– Consider the circuit shown in Fig. 2. In the left portion of the figure, we see two disconnectedsubsystems. The voltage source has one set of effort and flow variables, vB and iB and therefore hasone port. The resistor also has one set of effort and flow variables, vR and iR and therefore has oneport. While disconnected, the effort and flow variables of these ports are independent. The rightportion of the figure shows the connected system. Once connected, the connection (or bond) forcesthe effort and flow variable at the bonded ports to be the same:

vB(t) = vR(t) and iB(t) = iR(t).

Also, notice that at the bond, the voltage source determines the effort variable. With the effortvariable determined, the resistor determines the flow according to iR(t) =

1RvB(t).

+_ R+

-

iR

vR+

-vB

iB

+_ R+

-

iB=iR

vR

+

-vB

Figure 2: Basic circuit illustration of a one port.

– Consider the circuit shown in Fig. 3. In the left portion of the figure, we see two disconnectedsubsystems. The current source has one set of effort and flow variables, vS and iS and therefore hasone port. The resistor also has one set of effort and flow variables, vR and iR and therefore has oneport. While disconnected, the effort and flow variables of these ports are independent. The rightportion of the figure shows the connected system. Once connected, the connection (or bond) forcesthe effort and flow variable at the bonded ports to be the same:

vS(t) = vR(t) and iS(t) = iR(t).

Also, notice that at the bond, the current source determines the flow variable. With the flow variabledetermined, the resistor determines the effort according to vR(t) = RiB(t).

R+

-

iR

vR

+

-vS

iS

R+

-

iS=i

R

vR

+

-vS

Figure 3: Basic circuit illustration of a one port.

– Fig. 4 shows system that involves effort and flow variables in the electrical, rotational, and fluidsenergy domains. Prior to their interconnection, the effort and flow variables in each energy domainare unconstrained. The voltage source has one port. The motor has two ports with power variablesrelated according to the following two equations:

ωm = kmvm and im = kmτm.

The pump has two ports with power variables related by the following two equations:

Qv = kpωp and τp = kpPp.


6.2 Power Connections (Karnopp Section 2.1) February 10, 2014

VoltageSource

vs

+

-

is

PumpQp

Pp

vm

imPT

QT

Tank

Motor+

-tm,wm

tp,wp

Figure 4: Disconnected electrically powered fluid transport system.

The connected system is shown in Fig. 5. After the system is connected, their is only one effortand flow variable at each interconnection point (i.e., bond). It is important to note that any bondconnects ports that are in the same energy domain. For example, the pump bonds are not compatiblewith the voltage source bonds.The purpose of the voltage source is to supply electrical power as represented by the effort and flowvariables vs(t) and im(t). The DC motor transforms the electrical power to mechanical power throughthe effort and flow variables τp(t) and ωm(t). The pump transforms the mechanical power to powerin the fluid domain through the effort and flow variables PT (t) and Qp(t).After the interconnection, the voltage source outputs the voltage vs, which is an effort variable. Sincethe voltage source has determined the effort vm = vs, using the DC motor equations above, therotational rate ωm = ωp has been determined by the motor. Since the DC motor has determinedωp = ωm, by the pump equations above, the flow rate Qv is determined. The accumulation of the flowQv in the tank creates a back-pressure on the pump that we have labeled PT . Using the equationsabove, we see that this determines τp because Pp = PT . We also see that τp determines im becauseτm = τp.Note that all effort and flow variables at all bonds have been determined (conceptually). Bond graphsprovide an efficient mechanism for proceeding through the above reasoning process in an orderlyfashion. State variable models can be determined directly form the bond graph.

VoltageSource

vs

+

-Pump Qp

im

PT

Tank

Motortp,wm

Figure 5: Connected electrically powered fluid transport system..



6.3 Word Bond Graphs (Karnopp Section 2.2)

Used to introduce concepts. Also useful for analyzing power flow.

� Represent each subsystem by a title word

� Represent each port interconnection by a single line or bond.

� The line is labeled with the names of the effort (top or left) and flow (bottom or right) variables.

� A half arrow is added to each line to indicate the direction of power flow at any time instant when the effortand flow variables are both positive.

� A full arrow is used to indicate environmental effects that occur with essentially zero power flow: gear selection,thermostat setting, etc. This is called a signal flow or active bond. It represents information flow at essentiallyzero power. These are one-way influences.

� On every bond, a causal stroke is used to indicate the port that determines the flow (i.e., has flow as its output).The port at the opposite end of the bond determines the effort.

� The half arrow power flow sign convention and the causal stroke notation are completely independent.

Subsequent lectures will introduce methods to enable development of detailed quantitative models.6

Examples.

� For the circuit shown in Fig. 2, the word bond graph is shown in Fig. 6.

– The bond is the horizontal line from the voltage source to the resistor. For this bond, the effortvariable is labeled as vB and the flow variable is labeled as iR.

– The half arrow indicates that for the voltages and currents as labeled on the schematic positive poweris flowing from the source to the resistor.

– The causal stroke shows that the resistor is determining the flow (current) and the source is deter-mining the effort (voltage).

Voltagesource Resistor

iR

vB

Figure 6: Word bond graph corresponding to the system in Fig. 2.

� For the circuit shown in Fig. 3, the word bond graph is shown in Fig. 7.

– The bond is the horizontal line from the voltage source to the resistor. For this bond, the effortvariable is labeled as vR and the flow variable is labeled as iR.

– The half arrow indicates that for the voltages and currents as labeled on the schematic positive poweris flowing from the source to the resistor.

– The causal stroke shows that the source is determining the flow (current) and the resistor is deter-mining the effort (voltage).

Current

sourceResistor

iR

vR

Figure 7: Word bond graph corresponding to the system in Fig. 3.

� Fig. 8 shows the word bond graph corresponding to the system in Fig. 5. Compare the bond graph tothe system and confirm the application of the causal strokes.

6Assign HW on power flow, energy storage, half arrows, subsystem inputs and outputs, causal strokes.



vsQpim

PTtpVoltSupply

Motor Pump Tankwm

Figure 8: Bond graph for the system in Fig. 5.

6.4 Exercises

Exercise 6.1 – Consider the system drawn in Figure 9. The system has six subsystems: a battery to supply afixed voltage e, a motor equations are i(t) = 1

K τ(i) and ωm(t) = 1K e(t), a shaft to transmit the motor torque to a

rotational inertia (by twisting: τ = kθm), a rotational inertia J , a cable to transmit a force T = a(rθ−x) to a mass,and a mass M that will be lifted against the force of gravity. The schematic labels the effort and flow variables at

M

J Motore

+

_

i

wLqL tm

wmqm

T

T

x,v

tm

g

Gravity Mass Cable Inertia Shaft Motor Battery

T is the cable tension applied to the massand to the rotational inertial

tm is the torque applied to the shaft

Transl. Transl. Transl. Rotat. Rotat. Electrical

Figure 9: Schematic and word bond graph for a Crane. See Exercise 6.1

each port. Note the following:

1. The effort variable T at the two ends of the cable has equal magnitude, but opposite directions.

2. The flow variables at the two ends of the cable are different: v and rωL. The rate at which the cable stretchesis sT = rωL − v.

3. The effort variable τm at the two ends of the shaft has equal magnitude, but opposite directions.

4. The flow variables at the two ends of the shaft are different: ωm and ωL. The rate at which the shaft twists isθs = ωL − ωm.

� Consider the energy storage in each of the five subsystems. Describe each as chemical, potential, or kinetic. Ifpossible, write an equation for the amount of energy that is stored in terms of the effort and flow variables.

� Construct a word bond graph (BG).

– Connect the six subsystems with appropriate bonds.

– Apply the correct causal stroke to each bond.

– Label each bond with the effort and flow variables (in the correct locations) to correspond with your causalstrokes.



Exercise 6.2 – Consider the system drawn in Figure 10. The system has five subsystems: two masses, a shockabsorber modeled as F1 = s(x1−x2), a spring modeled as F2 = s(x2−x3), and a velocity source that determines vs.

M1

Shock

Absorb

er

M2

Spring

g

Velocity source

F1

F1

F2

F2

x1,v1

x2,v2

xs,vs

Figure 10: Schematic for a shock absorber system intended to isolate the motion of M1 from the motion of ve. SeeExercise 6.2

The schematic labels the effort and flow variables at each port. Note the following:

1. The effort variable F1 at the two ends of the shock absorber has equal magnitude, but opposite directions.

2. The effort variable F2 at the two ends of the spring has equal magnitude, but opposite directions.

3. The flow variables at the two ends of the shock absorber are different: v1 and v2. The rate at which it stretches(i.e., stores potential energy) is v1 − v2.

4. The flow variables at the two ends of the spring are different: v2 and v1. The rate at which the spring stretchesis v2 − v3.

� Consider the energy storage in each of the five subsystems. Describe each as potential or kinetic. If possible,write an equation for the amount of energy that is stored in terms of the effort and flow variables.

� Construct a word bond graph (BG).

– Connect the five subsystems with appropriate bonds.

– Apply the correct causal stroke to each bond.

– Label each bond with the effort and flow variables (in the correct locations) to correspond with your causalstrokes.



Exercise 6.3 – This problem asks energy and power related question based on facts shown on the 2013 HooverDam brochure.

1. The height of the water behind the Dam in 2013 was near 337m. The average water height is typically 358m.The minimum water height is 305 m. Compute the potential energy per cubic meter of water at each of theseheights.

2. If the area of Lake Mead, behind the Dam, is 640 km2, how much total energy is lost when the lake heightchanges from its typical average height to its 2013 height.

3. Compute the pressure created by h meters of water. This is the force of the column of water per square meter.

4. Assuming 100% efficiency, derive the formula for the power output P of the flow Qv through the Dam. Thisshould be expressed as a formula containing P , Qv, h, and two physical constants.

5. Evaluate the power formula in Part 4 for the typical water height when the volume flow rate is 9.2× 109m3

Y r .

6. Discuss the sequence of different energy domains in this problem and the physical elements that transformpower from one domain to the next.

Note and carefully show unit and unit conversions throughout this problem.


7 BOND GRAPHS: LECTURE 8 February 10, 2014

7 Bond Graphs: Lecture 8

Bond Graphes use bonds between one ports, two ports, and multiports to represent power flow within a system.From bond graphes we are able to check causality and derive state space models.

This lecture reviews bonds and introduces one port elements for bond graph models (See Sect. 3.1 in Karnopp[2] or Ch. 6 in Ljung [3]). This lecture will work through one energy domain at a time while filling in the entries inTable 2.

Energy Dissipate Potential KineticDomain Energy Energy Energy

Generic R : e = Rf C : q = Ce I : p = If Se Sf

Electrical Resistor Capacitor Inductor Voltage Source Current SourceTranslational Friction Linear Spring Mass Force Source Velocity SourceRotational Friction Torsional Spring Inertia Torque Source Angular rate sourceFluid Resistance Tank Pipe Pressure source Flow rate sourceThermal Heat transfer Heat capacity *** Temperature source Heat flow source

Table 2: Physical analogies between components (one port elements) in different energy domains.

Bonds:

� Represent the transfer of energy where power is the product of flow times effort.

� The bond is labeled with effort (on the top or left) and flow (on the bottom or right) variables.

� The half arrow indicates the direction of positive power flow and assigns a sign convention.

� The half arrow is usually on the side with the effort.

� The causal stroke indicates the element that determines the flow.

� The half arrow and causal stroke are completely independent.

� Full arrows indicate signal flow (i.e., essentially zero power)

Basic One-ports:

� Components with a single power port (i.e., a single set of effort and flow variables).

� Can represent simple components (e.g., resistor, capacitor, mass).

� Can represent complex subsystem (e.g., wall outlet or constant pressure pump).

� The five types of one-ports are described below. In addition, they are illustrated with a single bond inFig. 11.

R-element:

1. Element in which the effort and flow are related by a static (algebraic) function.

R IC

Se Sf

e

f

ecfI

f

e

Figure 11: Simple bond graphs for the five types of one ports. Passive one ports (R, I, C) are on the top row.The half arrow is always into the element because the elements always absorb net power. The causal stroke of theelements that are capable of storing energy are drawn to indicate integral causality. The active elements (Se, Sf ) aredrawn in the second row. These may either supply or absorb net power, so no half arrow is indicated. The causalstroke indicates their natural causality.


February 10, 2014

2. When this expression has the linear form e(t) = Rf(t), then the power is

p(t) = Rf2(t) =1

Re2(t).

Since R > 0, a resistance element always dissipates power. Most often this power is wasted asheat.

3. In Fig. 11, the R element bond graph is drawn with a half arrow (absorbing power), but no causalstroke (it has no natural causality).

C-element:

1. Element in which the effort and displacement are related by a static function

2. When this expression has the linear form q(t) = Ce(t) or f(t) = C ddte(t), then the energy stored

in a C element is

EC(t) =

∫efdt

=

∫eCde

=1

2Ce2(t) =

1

2Cq2(t).

Therefore C elements can store energy. Since the stored energy is proportional to q(t), which isthe integral of flow, a capacitor is a flow storage device.

3. The stored flow is q(t), which is proportional to effort e(t). Because the C element energy isproportional to both q(t) and e(t), neither q(t) nor e(t) can change in a discontinuous fashion.Therefore, the natural or integral causality of a C element is to determine the effort e(t).

4. In Fig. 11, the C element bond graph is drawn with half arrow and causal stroke (w/ integralcausality). The half arrow is always into the C element as the C element can never supply morepower than has previously been stored into it.

I−element:

1. Element in which the momentum and flow are related by a static function.

2. When this expression has the linear form p(t) = If(t) or e(t) = I ddtf(t), then the energy stored

in a I element is

EI(t) =

∫fedt

=

∫fIdf

=1

2If2(t) =

1

2Ip2(t).

Therefore I elements can store energy. Since the energy is proportional to p(t), which is theintegral of effort, an inductor is a effort storage device.

3. The stored effort is p(t), which is proportional to flow f(t). Because the I element energy isproportional to both p(t) and f(t), neither p(t) nor f(t) can change in a discontinuous fashion.Therefore, the natural or integral causality of an I element is to determine the flow f(t).

4. In Fig. 11, the I element bond graph is drawn with half arrow and causal stroke (w/ integralcausality). The half arrow is always into the I element as the I element can never supply morepower than has previously been stored into it.

Se:

1. A source of effort.

2. Ideally supplies the indicated, possibly time varying effort, no matter what flow is required.

3. In Fig. 11, the bond graph is drawn for a source of effort Se with a causal stroke showing itsnatural causality (determining effort). The half arrow is not shown because the source may eithersupply or absorb power.

Sf :

1. A source of flow.


7 BOND GRAPHS: LECTURE 8 February 10, 2014

2. Ideally supplies the indicated, possibly time varying flow, no matter what effort is required.

3. In Fig. 11, the bond graph is drawn for a source of flow Sf with a causal stroke showing itsnatural causality (determining flow). The half arrow is not shown because the source may eithersupply or absorb power.


February 10, 2014

8 Multiports: Lecture 9

Basic Multi-port Junctions: There are two types of multi-port junctions: These are used to connect one-portsand two-ports in arbitrary configurations.

� The junctions represent the only two ways that components can be connected.

� In electrical systems, the corresponds to series and parallel connections.

Flow, Common effort, or 0-junction:

0 0 0

2

1 3

4

2

1 3

4

Figure 12: Example zero junction.

� Can have any number of ports.

� The effort variable is the same at all ports: e1 = e2 = e3 = e4.

� Power conserving:

e1f1 − e2f2 + e3f3 + e4f4 = 0

e1f1 − e1f2 + e1f3 + e1f4 = 0 → f1 − f2 + f3 + f4 = 0

f1 + f3 + f4 = f2.

� ‘zero’: sum the flow (using sign convention supplied by half arrows) all efforts are the same.

Electrical- Represents Kirchoff’s current law at a node where four branches join.

Mechanical- Represents geometric compatibility for a situation involving a single force and for velocitiesthat algebraically sum to zero.

Fluid- Represents conservation of volume flow rate at the junction of four pipes.

Example figures-

2

1 30

e1 e2 e3

i1 i3

i2

+

-

+

-

+

-

Q1 Q3

Q2

x3

v1 v2

v3 = x3

.

F F

Figure 13: Example zero junction.


8 MULTIPORTS: LECTURE 9 February 10, 2014

Effort, common flow, 1-junction:

1 1 1

2

1 3

4

2

1 3

4

Figure 14: Example one junction.

� Can have any number of ports.

� The flow variable is the same at all ports: f1 = f2 = f3 = f4.

� Power conserving:

e1f1 − e2f2 − e3f3 + e4f4 = 0

e1f1 − e2f1 − e3f1 + e4f1 = 0 → e1 − e2 − e3 + e4 = 0

e1 + e4 = e3 + e2.

� ‘one,’ sum the efforts (using sign convention supplied by half arrows) all flow are the same.

Electrical- Represents Kirchoff’s voltage law around a loop with a single loop current and four voltage drops.

Mechanical- Represents the sum of four forces at a point with a well-defined velocity. If the point has mass, thisis Newton’s law.

Fluid- Represents the sum of pressures around a flow circuit with four pressure drops.

2

1 31

F1

F2

F3M

v = x.

e1 e3i

+

-

+ -

+

- e2

Figure 15: Example one junctions.


February 10, 2014

Example figures- Bond Graph Simplifications:

� A junction with only two inputs, with half-arrows pointing the same direction, can be replaced witha similarly directed half-arrow.

1 ==>e1

f

e2

f

e1=e2

f

0 ==>e

f1

e

f2

e

f1 = f2

� Two adjacent junctions of the same type can be combined.

1 1 ==> 1

2 3

1 4

5

2 3

1 4

5

0 0 ==> 0

3

1 4

2 5

2 3

1 4

5


8 MULTIPORTS: LECTURE 9 February 10, 2014

8.1 Exercises

Exercise 8.1 – For each of the following 0-junctions, write the equations relating the efforts and flows.

2

1 30

2

1 30

2

1 3

4 5

0

a) c)b)

Exercise 8.2 – For each of the following 1-junctions, write the equations relating the efforts and flows.

2

1 31

2

1 31

2

1 3

4 5

1

a) c)b)

Exercise 8.3 – If possible, simplify the bond graphs in Fig. 16.

0

R:a

R:b

R:d C:c

I:g

(a) Reduce.

00

R:a

R:b

R:d C:c

I:g 1 0

(b) Reduce.

00

R:a

R:b

C:c

I:g 1 0

I:h

(c) Reduce.

0

R:a

R:d C:c

I:g 00

Sf:u

R:b

R:n C:m

1 0

(d) Reduce.

1

R:a

R:d

I:g 10 Se:u

R:n

1 0

(e) Reduce.

1 IC

R

se:u

1

(f) Reduce.

Figure 16: Possibly unreduced bond graphs


February 10, 2014

9 Causality Propagation and State Space Models: Lecture 10 – 11

Causality Propagation:

Procedure: (walking the bond graph)

1. Choose a source and mark its automatic causality. Assign a source variable name.

2. Propagate this causality and variable as far into the graph as possible.

3. Repeat 1 and 2 for all remaining sources.

4. Choose an I or C element and mark its natural causality. Assign a (state) variable to the naturaloutput having integral causality.

5. Propagate this causality and state variable as far into the graph as possible.

6. Repeat 4 and 5 for all remaining I and C elements.

7. Choose an R element and mark an arbitrary causality. Assign a variable name.

8. Propagate this causality and variable as far into the graph as possible.

9. Repeat 7 and 8 for as many R elements as necessary to complete the causality graph.

Normally, the procedure would halt at step 6.If the algorithm fails due to a causality conflict, then the graph is ill-posed (has derivative causality). Thisis a topic of a future lecture.

Rules:

1-junction Rule – All bonds except one will have the causal stroke at the 1-junction.

0-junction – Exactly one of the bonds will have the causal stroke at the 0-junction.

State variable models from bond graphs: After completing the above procedure for walking the bond graph:

1. The number of energy storage devices (with integral causality) is the number of states (i.e., order).

2. A valid definition of the state vector is the vector of output power variables (i.e., efforts or flows) associatedwith each energy storage devices having integral causality

� Effort variables for C-elements

� Flow variables for I-elements

Define a vector containing these variables.

3. For each energy storage element with integral causality, write the differential equation for the correspondingstate variable.

4. Write the vector state equations.

9.1 Examples

Example 9.1 — This example develops the state space model for the bond graph on the top left of Fig. 17. Sincethis is the first example, we go through the method with considerable discussion, normally this will not be the case.For this example, the outputs are the flows through the R and C elements.

0 IC

R

0 IC

R

ec

ecec

0 IC

R

ec

ecec

0 IC

R

ec

fI-fR-fI

fRec/R

fR=ec /R

Figure 17: Propagation of state variables through a bond graph. See Example 9.1. Note that we defined the symbolfr = ec

R just for the convenience of fitting the equations onto the figure.


9 CAUSALITY PROPAGATION AND STATE SPACE MODELS: LECTURE 10 – 11 February 10, 2014

Because the bond graph has no sources, we begin at step 4, by selecting the C element. Its integral causalitydetermines an effort variable that we have named ec. Because the C element determined an effort, the causal strokeon the bond is placed at the port connected to the 0-junction. The result of step 4 is shown in the bond graph atthe upper right. In step 5, we propagate the result of step 4 as far as we can. Because all efforts on a 0-junctionare equal and the C element determines one of the efforts to be ec, the efforts on all the other bonds connected tothis 0-junction must also be ec. Because the 0-junction determined the effort on the bonds connected to the R andI elements, the causal stroke is placed at the end of the bonds connected to the R and I elements. By the definitionof a linear R element (e.g., Ohm’s law), the flow determined by the R element is fR = ec

R . The result of step 5 isshown in the bond graph at the lower left. In step 6 we check whether there are still energy storage devices withunspecified output power variables. Because the flow of the I element is unspecified, we return to step 4 and namethis variable fI . In step 5 we propagate the flow through the 0 element to the flow of the C element by summingthe flows according to the sign convention defined by the half arrows. The result is the bond graph on the bottomright with all efforts and flows defined and each bond having exactly one causal stroke.

Because the bond graph has two energy storage devices and both have integral causality, the number of statesis n = 2. A valid definition of the state vector is the output power variables of these energy storage devices:

x =[ec fI

]⊤. The state space model has two parts: (1) a vector algebraic equation relating the vector of outputs

to the state vector and input vector and (2) one first order differential equation for each state variable.

The outputs are y1 = ecR and y2 = ec

R + fI . The output vector is y =

[y1y2

]. The output vector is written as a

vector function of the state as

y =

[1R 01R 1

] [ecfI

]=

[1R 01R 1

]x.

The first-order differential equation for each state is read off the bond graph.

1. For the first state x1 = ec; therefore, x1 = ec. The effort ec is the power output variable of an C element. Bythe definition of a generic C element: e = 1

C f . For the C element in Fig. 17, the flow is fR + fc. Therefore,

ec =−1

C(fR + fc)

= − ecCR

− 1

Cfc

x1 = − 1

CRx1 −

1

Cx2

which is the first order ODE for x1.

2. For the second state x2 = fI ; therefore, x2 = fI . The flow fI is the power output variable of an I element. Bythe definition of a generic I element: f = 1

Le. For the I element in Fig. 17, the effort is ec. Therefore,

fI =1

Iec

x2 =1

Ix1

which is the first order ODE for x2.

Because this system is linear, it can be written in the standard matrix form as

x =

[ −1RC

−1C

1I 0

]x

y =

[1R 01R 1

]x.

△


9.1 Examples February 10, 2014

Example 9.2 — This example develops the state space model for the bond graph on the left of Fig. 18. The leftimage shows the result of first labeling the output u of the source of effort with its causality; then labeling the outputpower variable of the C element as eC along with its causality and propagating it as far as possible; finally, labelingthe output power variable of the I element as fI along with its causality. The image on the right shows the resultof propagating fI through the bond graph. Along the way, the symbol eR = fiR was defined, just so that the bondgraph did not get too crowded.

IC

R

ec

eR=f

IR

se:u

u

1

fI

eR

u - eC- e

R

fI

fI

fI

1 IC

R

se:u

fI

ec

u

Figure 18: Propagation of state variables through a bond graph. See Example 9.2. Note that we defined the symboleR = fiR just for the convenience of fitting the equations onto the bond graph.

For this example, the output y is the effort across the R element.The bond graph contains two energy storage devices: C and I. Both have integral causality. Therefore the state

dimension is two. A valid definition of the state vector is the output power variables of these energy storage devices:

x =[fI ec

]⊤(It is only coincidence that the names are the same as in Example 9.2.). From the definition of an

I element (f = 1I e) we have

fI =1

I(u− eC − eR)

= −1

IeC − R

IfI +

1

Iu

x1 = −R

Ix1 −

1

Ix2 +

1

Iu

which is the first order ODE for x1. From the definition of an C element (e = 1C f) we have

eC =1

CfI

=1

Cx1

which is the first order ODE for x2.The effort across the R element is eR = RfI , so y = Rx1.Because this system is linear, it can be written in the standard matrix form as

x =

[−R

I − 1I

1C 0

]x+

[1I0

]u

y =[

R 0]x.

△



Example 9.3 — This example develops the state space model for the bond graph on the left of Fig. 19. The leftimage shows initial bond graph. By combining the adjacent bond 1-junctions the bond graph reduces to that shownon the right, which is that same as the starting point of Example 9.3; therefore, the state space model is the same.

1 IC

R

se:u

fI

ec

u

1 IC

R

se:u

1

Figure 19: Reduction of a bond graph. See Example 9.3.

△

Example 9.4 — For the bond graph on the left of Fig. 20. The reader should be able to walk the causality andthe variables labeled on the left bond graph through to obtain the bond graph on the right.

For this example, the outputs are the efforts across the two I elements: y =[ea eb

]⊤..

fb

ec

u

0 C:cI:a

R:d

se:v

1

I:b

fa

fb

ec

u

0 C:cI:a

R:d

se:v

1

I:b

fa f

a

fa

ec

ec

ecu-ecfa-f

b-ec/d

ec/d


Given the bond graph on the right:

� There are three energy storage devices. All three have integral causality. Therefore, the number of state

variables is n = 3 and a valid definition of the state vector is x =[fa fb ec

]⊤.

� For this definition of the state vector, the output vector is computed as

y =

[u− x3

x3

].

� From the definitions of the I and C elements, the state equations are

fa =1

a(u− ec)

fb =1

bec

ec =1

c

(fa − fb −

1

dec

)� The state space model, since it is linear, can be written in matrix form as

x =

0 0 − 1a

0 0 1b

1c − 1

c − 1cd

x+

1a00

u

y =

[0 0 −10 0 1

]x+

[10

]v.

△



Example 9.5 — For the bond graph on the left of Fig. 21, the output is are the flow across the C : c element:y = fc.

The reader should be able to state a reason why each of the red or green circle items can be removed.

fb

uu-f

bece

a

0

C:c

sf:u 1 I:b0

C:a

0 11

1

0

C:c

sf:u 1 I:b0

C:a

0 11

1

C:c

sf:u I:b

C:a

0 1ea

ea

fb

fb

ea-e

c


The resulting simplified bond graph is shown with inputs and state variables, as well as causal strokes, propagatedthrough out. The reader should be able to show the following:

� There are three energy storage devices. All three have integral causality. Therefore, the number of state

variables is n = 3 and a valid definition of the state vector is x =[ea fb ec

]⊤.

� For this definition of the state vector, the output vector is computed as

y =[x2

].

� From the definitions of the I and C elements, the state equations are

ea =1

a(u− fb)

fb =1

b(ea − ec)

ec =1

cfb

� The state space model, since it is linear, can be written in matrix form as

x =

0 − 1a 0

1b 0 −1

b0 −1

c 0

x+

1a00

u

y =[

0 1 0]x.

△



9.2 Exercises

Exercise 9.1 – Propagate the listed variables and the power output variable of the C element through the bondgraph to show that the following two bond-graphs are equivalent in the sense that the efforts and flows on eachelement and at the output on the right hand side are identical.

C:c

0

11

0

R:d

e

f

1C:c R:d

0e

f

Exercise 9.2 – Propagate the listed variables and the power output variable of the I element through the bondgraph to show that the following two bond-graphs are equivalent in the sense that the efforts and flows on eachelement and at the output on the right hand side are identical.

I:b

0

1

1

0

R:d

f

e

1

I:b R:d0

f

e

Exercise 9.3 – For each of the following,

1. Walk the causality and state variables through the bond graph.

2. Define the state vector.

3. Define the state variable model.

4. If possible, write the state space model in matrix form.

For each bond graph, the output y of the system is the effort across the I-element.

I:L

Se 0 R:R2

R:R1

I:L

Se 1 C:C



Exercise 9.4 – For each of the bong graphs in Fig. 22:





For each bond graph, the output y of the system is the effort across the R : R1 element.

I:L

Sf 0 C:C

R:R1

Sf 1 C:C

R:R1

Figure 22: Bond graph for Exercise 9.4.

Exercise 9.5 – After simplifying the bond graphs in Figs. 16 (see page 35):

1. Walk the causality, input variables, and state variables through the bond graph.

2. Write the state space model. Consider the output in each case to be the vector of flows through the energystorage devices.

Exercise 9.6 – For each of the bond graphs in Fig. 23:





For each bond graph, the output of the system y is the vector containing the effort across each I element.

fb

ec

u

0 C:cI:a

R:d

se:v

1

I:b

fa

fb

ec

u

0 C:cC:a

R:d

se:v

1

I:b

ea

Figure 23: Bond graph for Exercise 9.6



Exercise 9.7 – For each of the bond graphs in Fig. 24. Consider the output y of the system to be the vector offlows through the R-elements.

1. Simplify the BG if possible.

2. How many energy storage devices?

3. How many inputs? What are the input signal variables?

4. How many outputs? What are the output signal variables?

5. Walk the inputs and state variables through the BG while applying the causal strokes.

6. Derive the state space model.

00

R:a

Sf:u

C:c

L:g 1 0

L:h

0

R:a

C:cL:g

Sf:u

01

R:a

Sf:u

R:d C:c

I:g 1 0

I:h


Exercise 9.8 – For the bond graph in Fig. 25, with the vector of currents through each C element and the sourceas output vector y.

1 0

sf:v

I:b

v e2

R:d

C:c

f1

C:a

e1



2. How many independent7 outputs? What are the output signal variables?


4. What is the state order?


7The variables are independent if and only if there is no way to form a linear sum (with nonzero coefficients) that adds to zero. Forexample, the currents through resistors connected in series are not independent.



Exercise 9.9 – For each of the bond graphs in Fig. 26. Consider the output y of the system to be the vector offlows through the I-elements.

1. Simplify the BG if possible.

2. How many energy storage devices?


4. How many outputs? What are the output signal variables?



01

R:d R:dC:1/k

1 0

I:M

0

I:M

Sf:u1 Sf:u2

01

R:d R:dC:1/k

1 0

I:M

0

I:M

Se:u1 Se:u2




Exercise 9.10 – For the bond graph in Fig. 27, with input u(t) and output y(t) = 1kFk(t).

0 1

I:M

v

Fksf:u

R:d

C:1/k

C:1/h

Fc



2. How many independent outputs? What are the output signal variables?




Exercise 9.11 – For the bond graph in Fig. 28, with the currents through each I element and the source asoutputs.

1 0

se:vs

I:d

vs vmia

R:b 1 R:k

C:mI:L

io

C:C

vc



2. How many independent outputs? What are the output signal variables?





February 10, 2014

10 Electrical Systems & Bond Graphs: Lectures 12

� The current through and voltage across an ideal linear resistor are related by Ohm’s law:

eR(t) = RiR(t)

where R is the resistance measured in Ohms. A resistor has a static relationship between an effort variableand a flow variable; therefore, according to Table 2 this is an R type element. No real resistor is linear. Someresistors, like a diode, are designed to be nonlinear.

� The current through and voltage across an ideal linear capacitor are related by

iC(t) = Cd

dteC(t)

where C is the capacitance measured in Farads. Integration of this equation yields∫iC(t)dt = = CeC(t)

QC(t) = CeC(t).

A capacitor has a static relationship between a displacement variable and an effort variable; therefore, accordingto Table 2 this is a C type element.

� The current through and voltage across an ideal linear inductor are related by

eL(t) = Ld

dtiL(t)

where L is the inductance measured in Henries. Integration of this equation yields∫eL(t)dt = = LiL(t)

λL(t) = LiL(t).

An inductor has a static relationship between a momentum variable and a flow variable; therefore, accordingto Table 2 this is an I type element.

� The effort source is a voltage source.

� The flow source is a current source.


11 BOND GRAPH MODELING (CIRCUITS, FLUIDS, THERMAL): LECTS. 12 – 14 February 10, 2014

11 Bond Graph modeling (Circuits, Fluids, Thermal): Lects. 12 – 14

This section first presents a modeling procedure. Then it presents several examples of the application of the method.

11.1 Modeling Procedure

The procedure described below is useful for electrical, fluid, and thermal system modeling.

Procedure:

1. Draw a schematic diagram and assign a power convention.

(a) Label positive voltage drops.

(b) Label positive current directions.

(c) For R, I, and C elements ensure that the current enters the terminal that is assumed to be positive.This is the positive power is absorbed sign convention.

2. Label each node voltage and insert a (labeled) 0-junction to represent each node.

(a) Repeat ground node if convenient.

(b) Use nodes even between series elements.

3. Establish the positive voltage drops between nodes using 1-junctions.

(a) Attach elements between nodes using 1-junctions.

(b) Use half-arrow sign conventions to correctly define voltage drops.

(c) For an output voltage connect a Sf with f = 0.

4. Remove all bonds having zero power.

(a) The ground voltage is zero. Propagate this through to the extent possible.

(b) Remove all elements with zero power (i.e., zero effort).

5. Simplify the bond graph (using identities on next page).

6. Walk causality, inputs, and state variables through the BG

7. Read off the state space model.


11.2 Electrical Circuit Examples February 10, 2014

11.2 Electrical Circuit Examples

This subsection presents several examples of the bond graph modeling of circuits. The purpose is to find the statespace model.

Example 11.1 — The top portion of Fig. 29 shows a circuit schematic fully labeled to define the power convention.Each element has defined voltage and current. For R, I, and C elements the voltage and currents are defined suchthat the elements are absorbing power. The schematic also has all nodes labeled. The ground node is labeled g.

For this circuit, the input is u = iS . The design parameters are R, C, and L. For the purpose of the example,we choose the output to be y = vL.

The second portion of Fig 29 shows the result of Step 2. Each node is represented by a 0-junction. The node labelis above the 0-junction to facilitate the matching process. To facilitate the next steps, the 0-junctions representingsome nodes are replicated. When multiple 0-junctions are used to represent the same node, they are connected by abond, to ensure that the duplicated 0-junctions have the same effort and flow. Because the g node is the ground, itseffort variable is defined to be 0 volts.

In the third step, the circuit elements are inserted between the nodes. From the circuit schematic, Kirchoff’svoltage law gives the following equations:

vg + vs = va, vg + vR = va, vg + vL = vb and vb + vC = va.

These equations are used to determine the directions of the half-arrows surrounding each of the 1-junctions.

� Starting with the current source, because the current iS is entering the node assumed to be negative, thecurrent source is supplying power to the circuit; therefore, its bond has the half arrow entering the 1-junction.Then, using the equation vg + vs = va and the properties of 1-junctions (the sum of efforts on entering halfarrow is equal to the sum of the efforts on exiting half arrows), the va half arrow is leaving the 1 and the vghalf arrow is entering it.

� R elements can only absorb power. In the previous step, we were careful to define the variables such that thisphysical condition is satisfied. Because the R element is absorbing power, the half arrow connecting it to the1-junction is entering the R element. Then, using the equation vg + vR = va and the properties of 1-junctions(the sum of efforts on entering half arrow is equal to the sum of the efforts on exiting half arrows), the vg halfarrow is leaving the 1-junction and the va half arrow is entering it.

� Similarly I elements can only absorb power. Therefore, the half arrow connecting it to the 1-junction is enteringthe I element. Then, using the equation vg + vL = vb and the properties of 1-junctions, the vg half arrow isleaving the 1-junction and the vb half arrow is entering it.

� C elements can only absorb power. Therefore, the half arrow connecting it to the 1-junction is entering the Celement. Then, using the equation vb + vC = va and the properties of 1-junctions, the vb half arrow is leavingthe 1-junction and the va half arrow is entering it.

This portion of the figure also shows the bonds that have zero power, which will be removed from the bond graph.The second to the bottom portion of the figure shows the resulting bond graph after removing the nodes and

bonds that have zero power. The red circles on the bond graph indicate that the two 0-junctions representing nodea can be merged and that three 1 junctions and one 0 junction that can each be replaced by a bond.

The result of the bond graph simplification is shown in the bottom portion of the figure. The causal strokes,inputs, and state variables have been propagated through the network.

From this bond graph:

� There are two independent energy storage devices, so the state dimension is n = 2.

� A valid definition of the state vector is defined by the power output variables of the energy storage devices

which is x =[vc iL

]⊤� The state space model, since it is linear, can be written in matrix form as

x =

[0 1

C

− 1L −R

L

]x+

[0RL

]u

y =[−1 −R

]x+

[R

]u.



R

C

L

+ -

is+_vs

+_

+_

vC

vLvR

iC

iL

iR

a a b

ggg

a a b

ggg

0 0 0

0 0 0

a a b

ggg

0 0 0

0 0 0

1 R 1 1 I:L

1

sf:is C0

0 00 0

a a b

0 0 0

1 R 1 1 I:L

1

sf:is C

0 1sf:is

C

I:L

R

is iLiL

iL vCis-iLvR

vR=R(is-iL)vR vR vR-vC

Ste

p 1

Ste

p 2

Step 3

Steps 4-5

0 0

y+_

Figure 29: Figure used in Example 11.1 to illustrate the steps in the bond graph modeling of a circuit.


11.2 Electrical Circuit Examples February 10, 2014

△

Example 11.2 — The top portion of Fig. 30 shows a circuit schematic fully labeled to define the power convention.Each element has defined voltage and current. For R and I elements the voltage and currents are defined such thatthe elements are absorbing power. The schematic also has all nodes labeled. The ground node is labeled g.

R L

is+_vs

+_

+_vLvR

iL=y

iR

a a a

ggg

a a a

ggg

0 0 0

0 0 0

1 R 1 1 I:Lse:vs0

0 00 0

Ste

p 1

Ste

ps 2

-3

0se:vs I:L

R

vsiLiR-iL

iR=vs/R

Ste

ps 4

-6 vs

vs iR


For this circuit, the input is u = vS . The design parameters are R and L. For the purpose of the example, wechoose the output to be y = iL.

The second portion of Fig. 30 shows the result of Steps 2 and 3. Each node is represented by a 0-junction.Because the g node is the ground, its effort variable is defined to be 0 volts. In the third step, the circuit elementsare inserted between the nodes. From the circuit schematic, Kirchoff’s voltage law gives the following equations:

vg + vs = va, vg + vR = va, and vg + vL = va.

These equations are used to determine the directions of the half-arrows surrounding each of the 1-junctions. Thisportion of the figure also contains a red lasso indicating the junctions and bonds that have zero power, which willbe removed from the bond graph.


From the final bond graph:

� There is one independent energy storage device, so the state dimension is n = 1.

� A valid definition of the state vector is defined by the power output variable of the energy storage device:x =

[iL

]� The state space model, since it is linear, can be written in matrix form as

x =[0]x+

[1L

]u

y =[1]x.

△



Example 11.3 —The top portion of Fig. 31 shows a circuit schematic fully labeled to define the power convention. Each element

has defined voltage and current. For R, C and I elements the voltage and currents are defined such that the elementsare absorbing power. The schematic also has all nodes labeled. The ground node is labeled g.

is+_vs

LvLiL

R+_vR

iR

a

gg

a b

gg

0

0 0

1se:vs0

00

Ste

p 1

Ste

ps 2

-3S

teps 4

-6

C

+ -vCiC

+ - b

I:L

R1

01

1

C:C

0se:vs

I:L

R01

1

C:C

vs

vs

vs

vC

vs-vC

vs-vC

vs-vC

iR=(vs-vC)/R

iR

vC iL

iL iL

iL-iRiL-iRiL-iR

y+_


For this circuit, the input is u = vS . The design parameters are R, C and L. For the purpose of the example, wechoose the output to be y = vR.

The second portion of Fig. 31 shows the result of Steps 2 and 3. Each node is represented by a 0-junction.Because the g node is the ground, its effort variable is defined to be 0 volts. In the third step, the circuit elementsare inserted between the nodes. From the circuit schematic, Kirchoff’s voltage law gives the following equations:

vg + vs = va, vg + vR = vb, vb + vC = va, and vb + vL = va.

These equations are used to determine the directions of the half-arrows surrounding each of the 1-junctions. Thisportion of the figure also contains a red lasso indicating the junctions and bonds that have zero power, which willbe removed from the bond graph.


From the final bond graph:

� There are two independent energy storage devices, so the state dimension is n = 2.

� A valid definition of the state vector is defined by the power output variable of the energy storage devices:

x =

[iLvC

]� The state space model, since it is linear, can be written in matrix form as

x =

[0 1

L1C − 1

RC

]x+

[01

RC

]u

y =[

0 − 1]x+

[1]u.

△



11.3 Exercises

Exercise 11.1 – For each of the following circuits:

1. Find the simplified bond graph model.

2. Propagate the causality, inputs, and state variables through the model.


4. Define the state space model. If the system is linear, write the model in matrix form.

c+_

e

d

b

au(t)

y(t)

(a) See instructions for Exercise11.1.

u(t) R L c

dy(t)

+_

(b) See instructions for Exercise 11.1.

R

C

+ -+_vs +

_ +_

vC

AvRvR

iC

i3

iR

a a b

ggg

is+_ +_

vL+ -

(c) See instructions for Ex-ercise 11.1. The outputy(t) = vc(t).

+_u +_

ve

b

y+_a +_ Ave

c

d

(d) See instructions for Exercise 11.1.

R C

u y+_

RL

+ -

(e) See instructions for Exercise11.1. To get y on the bond graph atthe indicated open circuit location,you can insert a source of flow withvalue zero. Or, strategically placeyour ground node between the Rand C elements.

Figure 32: Circuits for state space modeling.

Exercise 11.2 – For each of the bond graphs in Exercise 9.6, find the circuit that would produce it.



Exercise 11.3 –

1. Show that the circuit in Figure 33 yields the bond graph in Figure 25 on p. 43.

f1+_

a

e1

u b

+ -

d ce2

+_

Figure 33: Circuit for state space modeling Exercise 11.3.

2. Define the output to be the flow through the two capacitors. Use the bond graph to show that the state spacemodel with state x = [e1, f1, e2]

⊤ is

x =

0 0 00 0 1

b0 − 1

c − 1cd

x+

1a01c

u

y =

[0 0 00 −1 1

d

]x+

[11

]u

Exercise 11.4 –

1. Show that the circuit in Figure 34 yields the bond graph in Figure 28.

ia

c

vc

b

+ -

d

mvm

+_

+_u

k

a

io


2. Use the bond graph to show that the state space model with state x = [vc, vm, ia, io]⊤ is

x =

0 0 1

c 00 0 0 1

m

− 1a 0 − b

a 00 − 1

d 0 −kd

x+

001a1d

u

y =[0 0 0 1

]x



Exercise 11.5 – For the circuit shown in Figure 35:

R

C

+ -+_vs

+_

+_

vC

AvRvR

iC

i3

iR

a a b

ggg

y+_

is+_ +_

vL+ -

RL

io

1 0se:vs

I:L

vs

vCis

vC=(1-A)vR

vL AvR

R:R2

1 0 R:RL

C:C se:AvR

is is

AvRAvRio

io=AvRRL

AvR+vCAvR+vC

AvR+vC

vL=vs- vC1

(1-A)

iR

iR=AvR+vC

R2

is-iR is-iRis-iR is-iR-io


1. Show that it can be modeled using the bond graph indicated.

2. Find the state space model when the state vector is defined as x =

[iSvC

]3. Show thathe state space model, since it is linear, can be written in matrix form as

x =

[0 −1

L(1−A)1C

−1RC(1−A)

]x+

[1L0

]u

y =[

0 A1−A

]x+

[0

]u.

Exercise 11.6 – Fig. 36 shows a half-wave rectifier circuit. This circuit converts an alternating current voltagesource u(t) into an approximately constant voltage source.

L L

Ru vo

+

_

v3+

_

+_ C2

v1

i1+

_+

_

C1

v4

ic

i2

v2

i3

vi

+ _


For this half-wave rectifier circuit:

1. Find the equivalent bond graph. Note that a diode is a nonlinear resistor because the voltage (effort) andcurrent (flow) are algebraically related.




5. Simulate the state space model for a sinusoidal input: u(t) = 10 sin(2π60t). Consider and explain the temporalvariation of i5, vi, v3, and v0.

Each diode should be modeled as a nonlinear resistor with current and voltage related by the (algebraic) functionid(t) = g(vd(t)), where g represents a known nonlinear function, such as id = IS

(e40vd − 1

)with IS = 10−20 Amps.

In this expression, β is a large positive real number.Example 18.4 on page 93 analyzes a full wave rectifier circuit.


12 EXAM 1 February 10, 2014

12 Exam 1

For 2013, Exam 1 covers material to here.The topics of the exam can be read directly from the content of the lectures. For example, some topics are: ODE

to TF, TF to ODE, poles and zeros, frequency response, State space modeling, linear state space models, matrix andvector multiplication, simulation, energy domains, power connections, bonds and their properties, 1-ports and theirproperties, multi-ports and their properties, causality, bond graph simplification, causality and variable propagationthrough bond graphs, state space models from bond graphs, electrical circuit modeling via bond graphs.


February 10, 2014

13 Bond Graph modeling – Two Ports: Lecture 15

Basic (Ideal) Two-ports:

� Power conserving: e1(t)f1(t) = e2(t)f2(t). Power absorbed equals power delivered.

� Non-ideal models constructed by combination with R, L, and C elements.

There are two types of two ports:

Transformer:

� Constitutive law: e1 = me2 and mf1 = f2 with transfrmer modulus m.

� Physical examples of transformers are shown in Figure 37.

TFe1f1

e2f2

F2

a

bF1

v1

v2

e1

i1

e2

i2

+

-

+

-

F

v

QP

A

t1

w1 t2

w2

QP

t1

w1

Figure 37: Examples of two-port transformers

Lever: Massless, rigid, frictionless.

ω =v2b

=v1a

→ b

av1 = v2.

aF1 = bF2 → F1 =b

aF2

m =a

b

Gear Train: Massless, rigid, frictionless.

vc =ω1

r1=

ω2

r2→ r1

r2ω1 = ω2.

Fc =τ1r1

=τ2r2

→ F1 =r1r2

F2

m =r1r2

Electrical Transformer: Lossless.

→ Ni1 = i2.

→ e1 = Ne2

m = N

Hydraulic Ram: Massless, frictionless.

→ Av = Q.

→ F = AP

m = A


13 BOND GRAPH MODELING – TWO PORTS: LECTURE 15 February 10, 2014

Gyrator:

� Constitutive law: e1 = rf2 and rf1 = e2 with gyrator modulus r.

– Similar to resistor equation, relating effort to flow.

– Except effort and flow are at two ports.

– Power conserving.

� Physical examples of transformers are shown in Figure 38.

GYe1

f1

e2

f2

e1

i1

e2

i2

+

-

+

-

F2

F1

v1v2

e1

i1

+

-

t2

w2

Figure 38: Examples of two-port gyrators

Mechanical Gyrator: Rigid, frictionless.

→ F1 = rv2.

→ F2 = rv1

DC Motor: No friction, no resistance, no inductance, no inertia.

→ τ2 = Ti1.

→ e1 = Tω2

r = T



13.1 Examples

Example 13.1 — Fig. 39 shows the initial and completed bond graph for a system containing a transformerwith inputs, state variables, and causality propagated throughout. Note that power in equals power out for the

VT=vs-R1f/N-v1

VL=vT/N-R2f-v2

1se:vs TF

I:L

1 C:C2

C:C1

R:R1 R:R2

:

N

vs

v1

v2

f

R1f/N f/N

f/N f f

R2f f

VT VT/N

f/N

f/N VL

1se:vsTF

I:L

1 C:C2

C:C1

R:R1 R:R2

:

N

vs

v1

v2

f

Figure 39: Example transformer bond graph model.

transformer.For this system:

� The input is u(t) = vs(t), the output is y(t) = fN , the design parameters are N , R1, R2, L, C1, and C2.

� There are three independent energy storage devices, so the state dimension is n = 3.

� A valid definition of the state vector is defined by the power output variables of the energy storage devices:

x =[v1 v2 f

]⊤.

� Assuming linear elements, the state space model is

x =

0 0 1C1N

0 0 1C2

− 1NL − 1

L − 1L

(R2 +

R1

N2

)x+

001

NL

u

y =[

0 1 0]x.

△



Example 13.2 — Fig. 40 shows the bond graph of a system containing a gyrator with inputs, state variables, andcausality propagated throughout. For this system:

1se:vsGY

I:L

1 C:C2

C:C1

R:R1 R:R2

:

N

vs

v1

v2

f

fN

f f

vR vR/R2

vR=vS-v1-Nf

vL=NvR/R2-R2f-v2

vR/R2

vR/R2

vR/R2

NvR/R2

R2f f

vL

Figure 40: Example gyrator bond graph model. Consider the output to be the flow through R1.

� The input is u(t) = vs(t), the output is y(t) = 1R2

(vS − v1 −Nf), the design parameters are N , R1, R2, L,C1, and C2.

� There are three independent energy storage devices, so the state dimension is n = 3.


x =[v1 v2 iL

]⊤.

� Assuming linear elements, the state space model is

x =

− 1C1R2

0 NC1R2

0 0 1C2

− NR2L

− 1L − 1

L

(R2 +

N2

R2

)x+

1C1R2

0N

R2L

u

y =[

− 1R2

0 − NR2

]x+

[1R2

]u.

△

The reader should compare the models that resulted from Examples 13.1 and 13.2. The only change in the bondgraphs is the replacement of the transformer with a gyrator.



13.2 Exercises


1. Clearly state the inputs, outputs, and design parameters.

2. Propagate the causality, inputs, and state variables through the bond graph.

3. Write the state space model.

1se:vsGY

I:L

1 C:C2

R:R1 R:R2

:

N

The outputs are the currentsthrough the R elements.


1se:vs0 GY C:C2

C:C2

R:R1

:

N

The output is the currentthrough the R element.

(b) See instructions for Exer-cise 13.1.

1se:vsGY C:C2

R:R1

:

N


(c) See instructions for Exer-cise 13.1.

1se:vsGY

I:L

1 C:C2

R:R1 R:R2

:

N


(d) See instructions for Exercise13.1.

Figure 41: Gyrator bond graphs for state space modeling.







1se:vsTF

I:L

1 C:C2

R:R1 R:R2

:

N



1se:vs0 TF I:a

C:C2

R:R1

:

N

The outputs is the currentthrough the R element.

(b) See instructions for Exer-cise 13.2.

1se:vsTF C:C2

R:R1

:

N

The outputs is the currentthrough the R element.

(c) See instructions for Exer-cise 13.2.

1se:vsTF

I:L

1 C:C2

R:R1 R:R2

:

N


(d) See instructions for Exercise13.2.

Figure 42: Transformer bond graphs for state space modeling.





1se:vsGY

C:C

R:R

:

G


TF I:a:

N


1se:vsGY

I:C

R:R

:

G


TF I:a:

N

(b) See instructions for Exercise13.3.

Figure 43: Gyrator-Transformer bond graphs for state space modeling.



Exercise 13.4 – For the bond graph in Fig. 44:

1. Clearly state the inputs, energy storage devices, and design parameters.


3. Write the state space model with the output being the flow through the I-element with value M .

1 se:tTFI:J2 :

R:d2

R2 R1

0 TF:0 1

I:J1I:M R:d1

Figure 44: Transformer bond graphs for Exercise 13.4.


14 BOND GRAPH MODELING OF NETWORKS: LECTURE 16 February 10, 2014

14 Bond Graph modeling of Networks: Lecture 16

14.1 Examples

Example 14.1 — Fig. 45 shows three steps in the bond graph derivation of a state space model of a circuitcontaining a transformer. The top portion of the figure shows the circuit schematic with nodes, voltages, and currentslabeled. For this example, all elements are modeled as linear elements. Note that the left side of the transformer as itis labeled is absorbing power and the right side is supplying. The middle portion shows the unsimplified bond graph.The bottom portion shows the simplified bond graph after completing the causality and state variable propagation.From the simplified bond graph,

R1

C1

L

is+_vs

vC

vLv1

iCiL

a b c

ggg

y+_

+_

+ - v2+ -

d h

g

+ -

e1 e2+_

+_

+_

1

se:vsC:C11

0a

0g

0b

0b

R:R1

1

0g

0h

0c

0g

0g

1 TF 1

1

I:L

0d

R:R2

1

C:C2 1

0g

1se:vs0 TF

I:L

1

R2

C2

C:C2

C:C1

R:R1 R:R2

vs vC vC NvC

vS-vC iS

iS=(vS-vC)/R1

iSiS iLiL

R2iL iL

NiLvC iS-NiL

y

vL iL

:

N

N

:

N

vL=NvC-y-R2iL

Figure 45: Examples a circuit containing a Transformer

� The input is u(t) = vs(t), the output is y(t), the design parameters are N , R1, R2, L, C1, and C2.

� There are three independent energy storage devices, so the state dimension is n = 3. It is important to notethat the transformer is modeled as ideal, having no resistance or inductance. The inductance and resistanceare accounted for by the series insertion of L and R2.


x =[vC y iL

]⊤.

� The state space model, since it is linear, can be written in matrix form as

x =

− 1R1C1

0 − NC1

0 0 1C2

NL − 1

L −R2

L

x+

1R1C1

00

u

y =[

0 1 0]x.

△

Exercise 14.1 – Fig. 46 shows the circuit schematic and reduced bond graph for a simple AC to DC power supplycircuit. In this circuit, the triangular element with voltage vd and current id is a diode, which is a nonlinear resistiveelement with voltage and current related by id = g(vd). The function g is positive and monotonically increasing;therefore, its inverse exists: vd = g−1(id) for ic > 0.

1. Derive the reduced bond graph for the circuit schematic.

2. Use the bond graph to derive a state space model for the system.



is+_vs

vd

y+_

+ -

+_

1se:vs0TF

I:Lf

1

C:Cf

I:LT R:RL

:

N

N

LT LF

CF

id

RL

R:diode

1

a

g

b f

e1 e2+_

+_

vC+_

Figure 46: Examples a circuit containing a Transformer

Explanation of the circuit:

� In the United states, the voltage supply between nodes a and g is typically 120 volts at 60 Hz.

� The portion of the circuit between nodes a and b is an electrical transformer. The purpose of the transformeris to reduce the input voltage to the specified level of the load.

� The portion of the circuit between nodes b and f is a half-wave rectifier and filter. The purpose of the diodeis to prevent id from being negative when e2(t) < vc(t) + vd(t). When e2(t) > vc(t) + vd(t), id(t) charges thecapacitor and supplies the load. When e2(t) < vc(t) + vd(t), the capacitor discharges to supply current to theload.

Overall, the intent of the circuit is to convert an alternating and noisy input voltage to a constant DC value for aresistive load.


15 FLUIDS: LECTURES 17-18 February 10, 2014

15 Fluids: Lectures 17-18

A fluid is a substance that cannot support shear stress. Liquids and gases are each fluids.

� Liquids are (essentially) non-compressible.

� Gases are compressible8.

Hydraulic systems operate using a liquid operating fluid. Pneumatic systems operate using a gaseous operating fluid.Why use hydraulic/pneumatic systems?

Hydraulic/Pneumatic:

1. high power density (1000-8000 lbsin2 )

2. high bandwidth with high power

3. automatic cooling

4. very nonlinear dynamics

Electromechanical:

1. low power density (<200 lbsin2 ) due to magnetic saturation

2. low bandwidth with high power due to large amount of ferrous metal required for high power

3. requires external cooling

4. fairly linear dynamics

8Compressible means that the density is a function of pressure.


15.1 Fluid Systems February 10, 2014

15.1 Fluid Systems

The following two tables were presented in previous lectures.

Domain Effort Momentum Flow Displacement Power, N-m/secGeneric e p f q P , Joules/sec.Electrical e, Volt λ, Volt-s i, Amp Q, Coul. e(t)i(t)Translational F , Newton p, Newton-sec. v, m/s X, m F (t)v(t)Rotational τ , Newton-m pτ , Newton-m-s ω, rad/sec θ, rad τ(t)ω(t)Fluid P , Newton/m2 pP , Newton-s/m


Thermal T , ◦Kelvin ****** QH , Joules/sec EH , Joules T (t)QH(t)


Domain


Electrical Resistor Capacitor Inductor Voltage Source Current SourceTranslational Friction Linear Spring Mass Force Source Velocity SourceRotational Friction Torsional Spring Inertia Torque Source Angular rate sourceFluid Resistance Tank Pipe Pressure source Flow rate sourceThermal Heat transfer Heat capacity *** Temperature source Heat flow source

Table 4: Physical analogies between components in different energy domains.

R-Elements are used to model (memoryless) flow restrictions that dissipate energy. Because they are memoryless,they can be modeled by algebraic relations between the effort Pd and flow QVd

:

Pd(t) = g(QVd(t)).

Examples are orifices or valves. Typically, the function g will be nonlinear.

For linear analysis, the simplified model is often assumed:

Pd(t) = RQVd(t).

As illustrated in Fig. 47, Pd(t) = P1(t) − P2(t) is the pressure difference along a pipe of length L withcross-sectional area A. The shape of the cross-section also affects the resistance, for example:

Circular cross-section: Rc =128µLπ d4

Square cross-section: Rs =32µLw4

Rectangular cross-section: Rr = 8µL(w h2

1+ hw

)

where µ is the absolute viscosity, h is the height, w is the width, and dh is the hydraulic diameter.

A P1

P2

L

QV

Figure 47: Pipe Resistance.

Figure 48 shows a section of pipe that changes cross-sectional area. Assuming that the volume flow rate isconstant throughout the length of pipe, then the velocity in each section of pipe is related as

Qv = A1v1 = A2v2 = A3v3.



A2A

1

Q

v1

v2

A3v

3

Figure 48: Fluid orifice.

C-Elements are used to model elements for which the pressure (the effort variable)) is an algebraic function ofthe volume in the element. The volume of fluid in the element (a displacement) is the integral of the netinflow. One example is an open tank filled with a liquid of density ρ. The volume of liquid in the tank isthe integral of the net volume flow rate: V (t) =

∫Qv(t)dt. For a tank with constant cross-sectional area A,

the height of fluid in the tanks is h = V/A. The weight of fluid in the tank is W (t) = ρ g V (t). The forceat the bottom of the tank is the sum of the weight of the fluid plus the force due to atmospheric pressure:F (t) = APa + ρ g V (t) = APa + ρ g Ah. The pressure at the bottom of the tank is P = Pa + ρgh which canbe expressed as

Pg(t) =gρ

AV (t) or V (t) =

A

gρPg(t),

where Pg(t) = P (t)−Pa is the gauge pressure. The liquid tank has a static relationship between a displacementvariable V (t) and an effort variable P (t); therefore, according to Table 2 this is a C type element, whereCT = A

gρ . Note that this expression assumes an open tank of uniform cross-sectional area. Another example isderived on page 68.

h(t)P(t)Qv(t)

C: A rg

Pa

Figure 49: Illustration of a liquid filled open tank.

Fluid tanks store potential energy

Other types of fluids or other types of tanks can result in nonlinear capacitance equations.

I-Elements: Consider the flow of a fluid of density ρ through a tube of length L and uniform cross-sectional areaA, is illustrated in Fig. 50. A pressure difference PL(t) = P1(t)−P2(t) between the ends of the tube results in

A P1

P2

L

QV(t)v(t) I:rL

A

Figure 50: Illustration of a liquid filled open tank.

a force PA on the mass ρAL of fluid between the ends of the tube. Newton’s law states that

PL(t)A = ρALd

dtv(t)

where v(t) is the velocity of the fluid and Qv(t) = Av(t). Therefore,

P (t)A = ρLd

dtQv(t)

P (t) =ρL

A

d

dtQv(t)

P (t) = Lfd

dtQv(t)

where Lf = ρLA is the fluid inertance with units kg/m4. Integration of this equation yields

pP (t) = LfQv(t).


15.2 Additional Examples February 10, 2014

This is a static relationship between a momentum variable and a flow variable; therefore, according to Table 2an inertance is a I type element.

Fluid inertors store kinetic energy in the form of a fluid mass moving, for example through a pipe.

The effort source is a pressure source.

The flow source is a volume flow rate source.

In fluid systems it is important to understand the difference between absolute pressure and gauge pressure. Gaugepressure measures the difference between the pressure at a specific point inside a vessel and the atmospheric pressureoutside the vessel at a nearby point. This is analogous to a voltmeter measuring the effort in a circuit between twopoints.

15.2 Additional Examples

This section contains a few more examples related to fluid elements.

15.2.1 Bulk Modulus

The bulk modulus characterizes the compressibility, change in pressure as a function of density, of a fluid. Bulk

modulus is defined as β = ρ0∂P∂ρ

∣∣∣P0, T0

. Typical values are 442× 109 Pa for diamond, 50× 109 Pa for glass, 2.2× 109

Pa for water, and 1.0× 105 Pa for air.Note that

dρ

dt=

dρ

dP

dP

dt=

ρ

βP . (29)

Example 15.1 — This example considers the capacitance of a closed container holding a gas with bulk modulusβ. The closed tank in Fig. 51 has constant volume V and net inward volume flow rate Qv(t). By the conservation

P(t)

Qv(t)

C: V b

Figure 51: Capacitance of a closed tank.

of mass,

ρ(t)Qv(t) =d

dt(ρ(t)V )

= ρ(t)V + ρ(t)V

The equation

Qv(t) =ρ(t)

ρV + V (30)

is referred to as the Continuity Equation.Because the container has a constant volume, V = 0. Using eqn. (29), this simplifies as

Qv(t) =ρ(t)

ρ(t)V

Qv(t) =V

βP .

Therefore, the capacitance of a closed tank containing a gas with uniform bulk modulus β is C = Vβ . △



15.2.2 Absolute and Gauge Pressure

Absolute pressure is the pressure measured relative to absolute vacuum. At the surface of the earth, this pressure isatmospheric pressure.

Most gauges, such as the simple gauge illustrated in Fig. 52, measure the pressure P of the device connected toit relative to atmospheric pressure. This is referred to as gauge pressure. The left end of the gauge connects to alever mechanism moves the gauge dial. The pressure difference across the diaphragm is balance by the spring force.Due to an openings in the left side of the canister, the pressure difference is P = P −Pa. Therefore, x = 1

k (P − Pa)

P

Pa

xQv(t)

Figure 52: Pressure gauge.

Example 15.2 — This example considers the diaphragm mechanism in Fig. 52. The left half of the mechanism ismechanical. The spring stores potential energy. The right half of the mechanism is a fluid capacitor storing the flowQv(t) in the chamber with volume V (t) = Ax(t). Assuming that the piston is massless, the force balance equationis

P (t)A = kx(t)

=k

AV (t)

Therefore,

V (t) =A2

kP (t).

Starting from the continuity equation of eqn. (30) on page 68 and using the bulk modulus relation of eqn. (29):

Qv(t) =ρ(t)

ρV + V

=A

βx(t)P (t) +

A2

kP (t)

=

(A

βx(t) +

A2

k

)P (t).

The capacitance of this device, neglecting the mass, is nonlinear with expression C(x) =(

Aβ x+ A2

k

). △

15.2.3 Massless fluid ram

Consider the fluid ram illustrated in Fig. 53. For this example, the ram is considered to be massless and frictionless.These non-idealities will be discussed later. Due to the rigid shaft, the velocities v1 and v2 are equal: v = v1 = v2.

P2

Pa

Q2

Pa A

2

A1

v1

v2

Q1 P1

Figure 53: Fluid ram.

Because the volume flow rates are related to the shaft velocity as Q1(t) = A1v(t) and Q2(t) = A2v(t), we have

v(t) =Q1

A1=

Q2

A2or

A2

A1Q1 = Q2.


15.2 Additional Examples February 10, 2014

Similarly, according to Newton’s law:mrv = P1A1 − P2A2

for a massless ram (mr = 0), the forces on its two ends will be equal and opposite; therefore,

P1A1 = P2A2 or P1 =A2

A1P2.

Because

P1 = mP2

mQ1 = Q2

with m = A2

A1the fluid ram is a form of transformer as defined in Section 13.



15.3 Fluids Bond Graph Modeling Procedure

The method to develop a bond graph for a fluid system is exactly the same as that stated on p. 47 with pressureand volume flow rate replacing voltage and current. Often, to aid the process, it is convenient to sketch an analogouselectrical circuit.

Procedure:

1. Draw a fluid schematic diagram and assign a power convention.

(a) Label positive pressure drops.

(b) Label positive volume flow rate directions.


(d) Clearly label atmospheric pressure Pa

2. Draw an analogous electrical circuit schematic diagram and assign a power convention.

(a) Label positive pressure drops.

(b) Label positive volume flow rate directions.


(d) Atmospheric pressure Pa plays the role of the ground voltage.

There should be a direct correspondence between each element, flow, and effort labeled in the fluid andelectrical schematics. At each node, the pressures and volume flows in the fluid schematic should satisfyexactly the same equations as they do in the electrical schematic.

After this point, the methods are identical.

15.4 Examples

The following examples clarify the procedure.

Example 15.3 — Fig. 54 shows a fluid system schematic, the analogous electrical circuit, and the reduced bondgraph with inputs, state, and causality propagated throughout.

In the equivalent electrical circuit,

� The three absolute pressures Pp, Pt, and Pa are explicitly labeled.

� The pipe from the pump to the tank is modeled as a R : b element in series with and L : a element. These twoelements are in series, not in parallel, as there is one flow Qi in the pipe. The pressure drop Pp − Pt along thepipe is the pressure necessary to overcome the resistance of the pipe and the inertance of the fluid in the pipe.The actual values of a and b would be determined by the fluid and the dimensions of the pipe. In some caseseither value may be approximated as being zero.

� The flows Qi and Q0 are explicitly labeled.

� The net flow into the tank or into Ct is Qt = Qi −Q0.

An implicit assumption of the model is that the tank does not overflow.From the bond graph:

� There are two energy storage devices. Both have integral causality. Therefore, there are two independentenergy storage devices and the state dimension is n = 2.


x =[Pt Qi

]⊤.

� The input is u(t) = Pp(t) and the model parameters are the pipe parameters a and b, the tank capacitance Ct,and the valve setting Rv. The model parameters are themselves functions of the physical system parameters:pipe length, diameter, and shape; fluid density ρ; and tank cross sectional are. The output is y(t) = h(t).



ConstantPressure

Pump

Infinite LiquidResevoir

Pp

h

Valve

QoPtQi

Pa

+_Pp

a b

CtRv

Pa

QoQiQt

Pt+ -v

1 v2

+ -

I:a

Se:Pp

QiP

p1 0

Qi Qi

C:CtR:b

bQi QiP

t

Pt

Pt R:RvP

t/R

v

Qi-P

t/R

v

Pp-P

t-bQi

Figure 54: Fluids example. The output of interest is h(t) the height of fluid in the tank.

� The state space model can be written as

x =

[− 1

RvCt

1Ct

− 1a − b

a

]x+

[01a

]u

y =[ 1

ρg 0]x

where g is the acceleration due to gravity.

△

Example 15.4 — Fig. 55 shows a fluid system schematic, an equivalent electrical circuit schematic, and thereduced bond graph with inputs, state, and causality propagated throughout.

0

R:R12

C:C1

C:C2

SF:Qi

P1

P2

R:R2

P2

Q2

Q1

=P1

/R1

Q2

=P2

/R2

P1

Q1

Qi

Qi-Q

1Q

1-Q

2

P1

P2

Qi

Q1

Q2

Constant

Flow Source

valve

valveC1

R1

Pa

Q1

Qi

Qc1

P1

C2R2

Pa

Q2

Q1

Qc2

P2

P1

SF:Q1

Q1

P2 0

Figure 55: Fluids example. The output of interest is Q2(t).

The model makes the following assumptions:

1. The inertance of each pipe is small and has not been modeled.

2. The tanks do not overflow.

3. All fluid resistance models are linearized about the steady-state flow.

From the bond graph:

� There are two energy storage devices. Each has integral causality. The state dimension is n = 2.


x =[P1 P2

]⊤.



� The input is u(t) = Qi(t) and the model parameters are C1, C2, and the valve settings R1 and R2. These modelparameters are themselves functions of the physical system parameters: pipe length, diameter, and shape; fluiddensity ρ; and tank cross sectional are. The output is y(t) = Q2(t) =

1R2

P2(t).


x =

[− 1

R1C10

1R1C2

− 1R2C2

]x+

[1C1

0

]u

y =[

0 1R2

]x.

△

Example 15.5 — Fig. 56 shows a fluid system schematic and the reduced bond graph with inputs, state, andcausality propagated throughout.

P1

P2

Q1

Qp

Q2

ConstantPressureSource

valve

Reservoir

1 0 1 0

R:RpR:R12

C:C1 C:C2

SE:Pp

P1 P2

Pp R:Rv

I:Mw

P1P1P2P2Q2

Q2=P2/Rv

Qp

QpQp

QpR12 QpPp-P1 Q1 Q1=(Pp-P1)/Rp

Q1Q1

Q1-QP P1-P2QP-Q2

Figure 56: Fluids example. The output of interest is Q2(t).

The model makes the following assumptions:

1. For the pipe from the pump into tank 1, only the resistance to fluid flow has been modeled.

2. For the pipe connecting the two tanks, both resistance and inertance have been modeled.

3. The tanks do not overflow.

4. All fluid resistance models are linearizations about the steady-state flow.

From the bond graph:

� There are three energy storage devices. Each has integral causality. The state dimension is n = 3.


x =[P1 Qp P2

]⊤.

� The input is u(t) = Pp(t) and the model parameters are RP , C1, Mw, R12, C2, and the valve setting Rv.These model parameters are themselves functions of the physical system parameters: pipe length, diameter,and shape; fluid density ρ; and tank cross sectional area. The output is y(t) = h(t).


x =

− 1RpC1

− 1C1

01

Mw0 − 1

Mw

0 1C2

1RvC1

x+

1RpC1

00

u

y =[

0 0 1Rv

]x.

△



15.5 Exercises

Exercise 15.1 – For the fluid system in Fig. 57:

1. Find and label the analogous circuit.

2. Find the state space model for the fluid system. Consider the output to be the height of the fluid in each ofthe tanks.

P1

P2Q

1Qp

Q2

ConstantPressureSource

valve

ConstantFlow

Source

Figure 57: Fluid system for Exercise 15.1.

Exercise 15.2 – For the fluid system in Fig. 56, find and label the equivalent circuit.

Exercise 15.3 – For the respirator system in Fig. 58:

1. Find and label the equivalent circuit. Account for accumulation of flow in the tube plus each of the two lungs.

2. Find the state space model for the fluid system. Consider the output to be the pressure of the fluid in thelungs.

PT

PL

Q1

Flow

Source


Exercise 15.4 – Draw and label the schematic for a fluid system that is analogous to the circuit in Fig. 33 onpage 53.

Exercise 15.5 – Draw and label the schematic for a fluid system that is analogous to the circuit in Fig. 34 onpage 53.

Exercise 15.6 – For the fluid system of Fig. 54, there is a circular pipe (inner diameter of 0.02m and length 50 m)between the pump and tank is modeled as storing kinetic energy in I : a and dissipating energy through R : b. Thetank is storing energy in C : CT .

The fluid has viscosity µ = 1.0× 10−2N secm2 , bulk modulus β = 1.4× 109 N

m2 , and γ = ρg = 8.7× 10−3 Nm3 .

Use the parameters provided above to show that a = 1.3×108N secm5 , b = 2.8×108N sec2

m5 , and CT = 5.6×10−7m5

N .Check that the units work out for the efforts and flows summed on the various bonds in the bond graph.



Exercise 15.7 – For the fluid system in Fig. 59, consider the Pump to be source of pressure:


2. Find the state space model for the fluid system. Consider the output to be the height of the fluid in each ofthe tanks.

Pump


Pp

h

PtQi

Pa


Exercise 15.8 – For the fluid system in Fig. 60, makes the following assumptions:

P1

P2

Q1

Qp

Q2

FlowSource

valve

Reservoir

Figure 60: Fluids example. The outputs are the heights of the fluids in the two tanks.

� For the pipe from the pump into tank 1, only the resistance to fluid flow has to be modeled.

� For the pipe connecting the two tanks, both resistance has to be modeled.

� The tanks do not overflow.

� All fluid resistance models are linear.

� The cross-sectional areas of the two tanks are A1 and A2.


2. Find the state space model for the fluid system.


February 10, 2014

16 Thermal Systems: Lecture 19-20

Modeling of thermal systems using bond graph methods produces pseudo-bond graphs, not bond graphs. Seediscussion of pseudo-bond graphs in Karnopp.

Pseudo-bond graph:

Effort - Temperature (dimensionless), T . Quantifies the average energy of microscopic molecular motion.

� There are various temperature scales: Fahrenheit TF , Celsius TC , and Kelvin Tk.

� Because TK = TC + 273.15, temperature differences are the same in the Kelvin and Celsius scales:

∆TK = TK1 − TK2

= (TC1 + 273.15)− (TC2 + 273.15)

= TC1 − TC2

= ∆TC

Flow - Heat flow rate (Joules/s=Watts), QH

� Product of temperature and heat flow rate is not power. Heat flow rate already has units of power.

� Because the thermal effort (temperature) and heat flow do not multiply to give power, they are not powervariables and the bond graph method applied to thermal systems, while useful, does not yield a bond graph.

� Bond graph methods are applicable to pseudo-bond graphs using methods comparable to circuit analysis.

� Bond graphs and pseudo-bond graphs cannot be connected.

Methods of Heat Transfer:

1. Radiation - Heat transfer without the presence of a surrounding medium.

2. Conduction - The process of heat transfer through a solid.

3. Convection - The process of heat transfer between the surface of a solid and a fluid that is in contact withthe surface of the solid.

(a) Conduction of heat from solid to fluid in immediate contact with surface

(b) Replacement of heated fluid with fresh fluid due to fluid flow.

Heat Storage: Is achieved by changing the temperature of an object. See definition of temperature.

General Thermal Modeling:

� Nonlinear

� No I-elements. Heat flow can stop instantaneously.

� Partial differential equations: (Laplace Equation)

∂2T

∂x2+

∂2T

∂y2+

∂2T

∂z2=

1

α

∂T

∂t

T - temperature (x, y, z) Cartesian coordinatesα = k

ρCp- Thermal diffusivity

k - thermal conductivity ρ - densitycp - specific heat

� Linear lumped-parameter models are first order approximations.

� Accurate modeling requires knowledge of

– thermodynamics

– heat flow

– fluid mechanics


16 THERMAL SYSTEMS: LECTURE 19-20 February 10, 2014

16.1 Steady State (static) Thermal Systems

Special Case - Steady state 1-d conductive heat flow:

Assumptions:

1. One dimensional flow (ideal insulation on remaining sides)

2. Steady state (no time dependent change in temperature, no energy stored)

3. Thermal conductivity k is independent of position and temperature

T0

T1

x

z

y

} DT

Dx

T0

Qk

Rk= L

kA

T1

R:Q

k

DT LkA

Qk

Figure 61: Illustration of 1-d conductive heat flow.

Fourier Equation:

Qk = −kAdT (x)

dx

� A is the area through which heat transfer occurs (i.e., normal to x)

� The negative sign in the Fourier equation indicates that heat flows toward the lower temperature.

� For a single layer model, as shown in Fig. 61, assuming that insulation prevents heat flow in the xand y directions,

Qk =kA

L(T0 − T1)

or

T0 = T1 +L

kAQk.

� The solid could be subdivided into n layers each with resistance Rk(n) =L

nkA . If the temperature onthe front of the block is Tn, then the increasing temperatures moving backwards through the blockare related by

Ti−1 = Ti +Rk(n)Qk for i = n, . . . , 1.

Thermal Conductive resistance: Rk = LkA .

Caveats: Temperature change is actually continuous along the medium. This equation

1. Allows computation of Q when T1, T2, and all parameters are known.

2. Allows computation of T2 when T1, Q, and all other parameters are known.

Table 5: Thermal constants for selected solids: ρ - density, cp - specific heat capacity, k - thermal conductivity.

ρ, kgm3 cp,

Jkg ◦K k, W

m ◦K

Copper 8933 383 399Aluminum 2702 896 153Silicon 2330 703 0.19

Plexiglass 1180 1500 0.19Glass 2800 800 0.81


16.1 Steady State (static) Thermal Systems February 10, 2014

Special Case - Convection: The process of heat transfer between the surface of a solid and a fluid that is incontact with the surface of the solid.

1. Conduction of heat from solid to fluid in immediate contact with surface

2. Replacement of heated fluid with fresh fluid due to fluid flow.

Solid ThermallyConductive Material

Fluid flowProfile

BoundaryLayer

Ts

TF

R:

TS

Qh

Rh= 1

hcA

TF

Qh

1hcAQ

h

DT

Figure 62: Illustration of 1-d convective heat flow.

Newton’s Law of Cooling:Qh = hcA(TS − TF )

Resistance due to convection:

Rh =1

hcA

Coefficient of convection hc: depends on the fluid density and nature of flow.

� Free convection - Flow is induced due to temperature induced density gradients.

� Forced convection - Flow is forced by a pump or fan.

Table 6: Thermal coefficients of convection for selected fluids: h - thermal conductivity.Air Air Water WaterFree Forced Free Forced

h, Wm2 ◦C 6-30 30-600 60-300 300-6000



Example 16.1 — How much heat is lost per square meter of plexiglass window?

Assumptions:

� Ti = 25◦C, To = 0◦C,

� k = 0.195 Wm ◦K , h = 20.0 W

m ◦K , thickness = 0.006 m, A = 1.0m2.

� The four edges of the window perpendicular to the heat flow are perfectly insulated.

Variables:

� Inputs: Inside temperature Ti, outside temperature To

� Outputs: Temperature of window inner surface TSi , Temperature of window outer surface TSo

QTi To

TSiTSo

InsideAir

OutsideAir

Plexiglass

Window

+- +-Ti To

TSoTSi

RhiRho

Rk

Q

1

Se:To

Rhi

Rk

Rho

Se:Ti

Figure 63: Illustration of 1-d heat flow example for a plexiglass window.

Schematic, Equivalent Circuit, Pseudo-BG

Parameters:

Rk =L

kA=

0.006m(0.195 W

m ◦K

)(1.0m2)

= 0.0308◦K

W

Rh =1

hA=

1(20 W

m2 ◦K

)(1.0m2)

= 0.0500◦K

W

Solution:

1. To only find QH :

QH =Ti − To

Rhi +Rk +Rho

2. To find QH , Tsi , and Tso :

QHRhi = Ti − Tsi

QHRk = Tsi − Tso

QHRho = Tso − To

Rhi 1 0Rk −1 1Rho 0 −1

QH

Tsi

Tso

=

2500

QH = 191W, Tsi = 15.4◦C, Tso = 9.5◦C.

Cost: The energy usage per day is E =∫ 24hrs

0QHdt = 4.58kWHrs. If energy costs $0.07 per kWHr, then

the cost is $0.32 per day or $117 per year.

△


16.1 Steady State (static) Thermal Systems February 10, 2014

Example 16.2 — Safe operation of an electrical component requires Tc < 75 ◦C. The component is ratedfor 0.25 W without a heat sink when the ambient temperature is Tc = 25 ◦C. How large of an area of a copperheat sink is required to operate at 1.0 W?

Heat sink

component

dAQ=0.25 W

Rk

Rht RhsRhb

Qht QhsQhb

Qc

Ths

14

Tc

Ta

Figure 64: Illustration of static heat sink example.

The equivalent circuit is shown in Fig. 64. The component is producing Qc = 1W of heat flow. The heatflow must first be conducted from the component into the solid Rk, then convect through either the top Rht,bottom Rhb, or sides Rhs into the ambient air.

Let A = L2 denote the area of one surface of the heat sink, which is assumed to be square with edges of lengthL. There are two surfaces with this area. Each will have convective resistance Rht = Rhb =

1hcA

◦CW .

In addition, there are the four edges. Because the plate will be very thin (d ≪ L), the area of each sideAs = Ld will be very small in comparison to A. Therefore, the convective heat resistance through each sideRhs =

1hcAs

◦CW will be very large in comparison to the either Rht or Rhb.

The total resistance to heat flow is Rk plus the parallel combination of Rht, Rhb, and14Rhs. Assuming that

the convective heat resistance is dominant (i.e., Rh >> Rk), we can neglect Rk. Also, the parallel combinationof the six conductive heat paths is approximately

Rh =1

2hcA

◦C

W.

Assume that hc = 25 Wm2 ◦C .

Tc = Ta +QRh

75 > 25 + (1W )1

2hcA

◦C

W

50A >1

2hc=

1

50.0= 0.02

A > (0.02)2m2

A square copper plate that is L = 2cm per edge would be sufficient.

For this area, the resistance due to convection is Rh = 50◦CW and Rk =. Assuming that the heat sink is d = 0.001

m thick, then with the component located at the center of the heat sink, the resistance due to conduction is

approximately Rk = L/2400Ld ≈ 1.25

◦CW , which verifies the convection dominance assumption (Rh ≫ Rk). △



Example 16.3 — A solid-state electronic component inside a case must dissipate heat Q = (Vi−Vo)I = 5W(i.e., I = 3A and (Vi − Vo) = 1.67V ). Assume the device is in steady state.

For safe operation, the junction temperature Tj must remain less than 125 ◦C. The ambient temperature is

25 ◦C. The junction to case resistance is 3◦CW .

Heat sink

device

Silicon

Case

Thermal compound

Ta

Ta

Tj

gas

T1 T2

+-Tj Ta

Rjc RhsRtc

Q

T1 T2

Q +-

1

Se:TaSf:Q

Rjc

Rtc

Rhs

Figure 65: Illustration of static heat sink example

For the heat to leave the active portion of the device, it must conduct from the junction to the case, from thecase through a thermally conductive adhesive to the heat sink, and convect from the heat sink into the ambientair. That is the path of least resistance. Heat can also convect directly from the case to the air, but due tothe small surface area, that path is less significant. The thermally conductive adhesive is critical. If it was notincluded, then the small gap of trapped air would insert a convective resistance between the case and the heatsink.

Use the following notation:Rjc Heat resistance from junction to case

(parallel paths through gas and silicon)Rtc Heat resistance of thermal compoundRhs Heat resistance of heat sink

(including conduction and convection)

From the bond graph in Fig. 65, the junction temperature can be computed as

Tj = Ta +Q (Rjc +Rtc +Rhs) (31)

The constraint Tj < 125 ◦C can be achieved by

1. decreasing the ambient temperature Ta (rarely possible)

2. decreasing the heat flow (rarely possible)

3. decreasing Rjc (not possible by the end user)

4. decreasing Rtc

� Use a high quality thermally conductive adhesive to ensure a low conductivity path for heat flow andto avoid trapped air (an insulator) in the gap.

5. decrease Rhs (usually possible).

Based on the above data, eqn. (31), and the constraint that Tj < 125 ◦C:

125 > Ta +Q (Rjc +Rtc +Rhs)

100 ◦C

5 W− 3

◦C

W> (Rtc +Rhs)

17◦C

W> Rtc +Rhs.

Many heat sinks satisfying this constraint can be found from electronic suppliers. Many electronic componentapplication notes include discussion of specification of a heat sink that are similar to this example. △


16.2 Dynamic Thermal Systems: Lecture 20 February 10, 2014

16.2 Dynamic Thermal Systems: Lecture 20

Heat Capacity: The ability to hold or store heat by a change in temperature

cpmdT

dt= QHi −QHo +QHg +

dW

dt(32)

where cp is the specific heat(

Jkg ◦K

)and m is the mass. Note that eqn. (32) is in the standard form for a

C-element (Ce = f) where the value of the C element is

Cp = cpρV. (33)

Major Assumption: Heat flow is small enough that temperature inside the mass can be considered uniform (i.e.,no temperature gradient inside mass). This assumption will be approximately true is the process is convectiondominated:

Rh >> Rk.

This can always be achieved by subdividing the mass into n layers in the direction of heat flow.

Biot Number: The Biot number is defined as

Bi(n) =Rk(n)

Rh.

If Bi(n) < 0.1, then the process is convection dominated and each layer can be modeled as constant temperature.Therefore, we would like to choose n such that the system is convection dominated. We find n as follows:

0.1 ≥ Bi(n) (34)

0.1 ≥(

LnkA

)(1hA

)0.1 ≥

(Lh

nk

)n ≥

(Lh

0.1k

). (35)

Choosing n to satisfy eqn. (35) yields a model where the convection dominated assumption is valid. The systemis then divided into n layers in the direction of heat flow with Rk(n) and Cp(n) computed appropriately:

Cp = cpρAL

n, Rk(n) =

L

nkA, Rh =

1

hA.



Example 16.4 — Continuing from Example 16.2 where A = (0.02m)2 and d = 0.001.

Parameters:ρ = 8933 kg

m3 density of coppercp = 383 J

kg ◦K heat capacity of copper

k = 400 Wm ◦K copper coefficient of conductivity

h = 20 Wm2 ◦K coefficient of convection

Biot Number Computations:

Rk =L

2nkLd=

1

n(800)(0.001)=

1.25

nRh =

1

hA=

1

2(20)(0.02)2= 62.5

◦K

W

Bi(n) =Rk

Rh=

1.25

62.5n=

0.02

n< 0.1 for n = 1.

Rk can be ignored and heat sink can be modeled accurately as having a uniform temperature.

Cp = ρcpV =

(8933

kg

m3

)(383

J

kg ◦K

)(0.022 m2 0.001 m

)= 1.37

J◦K

Model: Energy Balance: Energy stored = Heat flow in - Heat flow out.

ρcpVdTc

dt= Qi −Qo

CpdTc

dt= v(t)i(t)− 1

Rh(Tc − Ta)

1.37dTc

dt= v(t)i(t)− 1

62.5(Tc − Ta) Watts

dTc

dt=

−1

86Tc +

1

86Ta + 0.73v(t)i(t).

In steady state,Tc = Ta + 63v(t)i(t).

The time constant is τ = 86s.

△


16.2 Dynamic Thermal Systems: Lecture 20 February 10, 2014

Example 16.5 — Model the temperature distribution through a plate glass window [6].

Parameters:A = 0.88m2 Area of glasst = 7.5mm Thickness of glass

ρ = 2800 kgm3 Density of glass

cp = 800 Jkg ◦K Specific heat capacity of glass

k = 0.75 Wm ◦K Glass coefficient of conductivity

h = 30 Wm2 ◦K Coefficient of convection

Biot Number Computations:

Rk =L

nkA=

0.0075

n(0.75)(0.88)=

0.0114

n

◦K

WRh =

1

hA=

1

(30)(0.88)= 0.0379

◦K

W

Bi(n) =Rk

Rh=

0.0114

n0.0379=

0.3

n≤ 0.1 for n ≥ 3.

Cp(n) =ρcpV

n=

(2800

kg

m3

)(800

J

kg ◦K

) (0.88 m2 0.0075 m

)n

=14784

n

∣∣∣∣n=3

= 4928J◦K

Model: The glass slab is modeled as three layers. This involves five temperatures within the glass slab: TSi

is the temperature on the inside surface of the glass, TSo is the temperature on the outside surface of theglass, and Ti is the temperature at the center of the i-th glass slab for i = 1, . . . , 3. These temperaturesare illustrated in Fig. 66.

The equivalent circuit shows the inside and outside temperatures as constant temperature sources. Becausethe distance from Ti to Ti+1 is L

3 it has resistance Rk. On the two edges, because the distance from the

temperature at the edge to the center of the first slab is L6 it has resistance Rk

2 plus the resistance due toconvection.

Ti To

TSiTSo

Inside

AirOutside

Air

T1 T3T2

C C C

+- +-Ti To

TSo

TSi

RhiRho

RkRk Rk/2Rk/2 T1

T3T2

C C C

Rhi+Rk/2

Se:Ti 1 0

C

1

Rk

1 0

C

1 0

C

Rk

Se:ToT3

T3 T3 T0

T2

T2 T2

T1

T1T1

Rho+Rk/2

L/3

Figure 66: Illustration of dynamic window example

The reader should be able to derive the state space model as

x =

3Rk+2Rhi

Rk(2Rhi+Rk)C

1RkC

01

RkC− 2

RkC1

RkC

0 1RkC

− 3Rk+2Rho

Rk(2Rho+Rk)C

x+

2

C(Rhi+2Rk)

0

0 00 2

C(Rho+2Rk)

u

y =[0 1 0

]x

where x = [T1, T2, T3]⊤, the output is T2, and u = [Ti, To]

⊤. △



16.2.1 Summary

� The one dimensional heat transfer QH between two bodies having temperature difference T between them is:

T (t) = RQH(t)

where R is the resistance to heat flow (conductive or convective) measured in◦K

Watt .

� The rate of change of the temperature of a body as a function of net heat flow into the body is described by

QH(t) = Cd

dtT (t)

where C is the thermal capacity. Integration of this equation yields∫QH(t)dt = = CT (t)

EH(t) = CT (t).

which is a static relationship between a displacement variable and an effort variable; therefore, according toTable 2 this is a C type element.

� There is no I type element for thermal systems.

� The effort source is a temperature source.

� The flow source is a heat flow source.

� To simplify analysis (by avoiding partial differential equations), we divide a conductor into small enough piecessuch that each piece can be approximated to have a constant temperature. The Biot number provides a methodto select the number of layers required to make this assumption reasonable.

16.3 Exercises

Exercise 16.1 – Repeat Example 16.1 using glass instead of plexiglass. Compare the cost of the heat loss. Howdoes this cost change as a function of the temperature difference Ti − To (e.g., constant, linear, quadratic )?

Exercise 16.2 – For the two pane window of Fig. 67, assume the system is in steady state. Be careful with units.

Insulation

Insulation

Ple

xig

lass

Win

dow

Ple

xig

lass

Win

dow

Air Gap

Ti To

Q

Ti=25 C, Ti=-5 C,k = 0.195 W/(m K),h = 20.0 W/(m2 K),A = 2 m2t = 0.006 m (per pane)

Figure 67: Two pane window for Exercise 16.2

The stated convection coefficient is for the ambient free flowing air outside of the window. Select a value for theconvection coefficient in the airgap and discuss why you choose it.

� Draw the equivalent circuit.

� Give numeric values with units for all model parameters.

� Draw the bond graph.

� Compute the heat flow Q. Estimate the cost of this heat flow on an annual basis.

� Sketch an equivalent fluid system.



Exercise 16.3 – Similar to example 16.5, model the one dimensional temperature distribution through a plate withthe following parameters:

Parameters:

A = 1.00m2 Areat = 7.5mm Thickness

ρ = 2500 kgm3 Density

cp = 1000 Jkg ◦K Specific heat capacity

k = 0.50 Wm ◦K Coefficient of conductivity

h = 20 Wm2 ◦K Coefficient of convection

Be careful with units. The solution should have n = 4.Similar to Fig. 66,


� Draw the thermal schematic.



� Find the state space model.


Exercise 16.4 – Consider the system shown in Fig. 68. The cube of material is insulated on five sides, with asource of heat in the center of the left edge.

Insulation

Insulation

L=0.09 m

ToQ(t)

Insula

tion

To= 0 C,k = 10.0 W/(m K),h = 20.0 W/(m2 K),A = 2 m2 (right surface)cp = 383 J/(kg K)r = 5000 kg/(m3)


� Draw the thermal schematic.


� Use the Biot number to compute the number of required layers.



� Find the state space model.




Exercise 16.5 – A can of soda with initial temperature 35 C is emersed into an ice water bath, as illustrated inFig. 69. The ice water should be modeled as a constant temperature source.

Use the parameter values listed below and list any additional assumptions:

1. The heat can follow two paths out from the soda. Compare the total resistance of each path. Which path willcarry the majority of the heat? Neglect the other path in the subsequent analysis.

2. Compute the heat capacitance of the soda and the can. Which is dominant? Why?

3. Draw an equivalent circuit that includes a single capacitor for the heat capacity of the soda, a single temperaturesource for the ice water bath, and two resistors representing convective heat resistance through the soda andthrough the ice water.

4. Develop a single state dynamic model for the temperature of the soda.

5. Identify the time constant in minutes. How long would it take for the temperature to reach steady-state? Doesthis match your real-life experience?

Tw=0 CTs

Ta=20 C

BE CAREFUL WITH UNITS.


Parameters: r = 1.3in (can radius), t = 0.004in (can thickness), l = 4in (can height), hw = 60 W(m2 ◦K) coef.

convection for water, ha = 6 W(m2 ◦) coef. convection for air, hs = 60 W

(m2 ◦) coef. convection for soda, kal = 236 W(m ◦)

coef. conductivity for aluminum, cpal= 896 J

(kg ◦) specific heat of aluminum, cps = 4180 J(kg ◦) specific heat of soda,

ρal = 2702 kg(m3) density of aluminum.


February 10, 2014

17 Exam 2

Exam 2 covers to here in 2013.


18 CAUSALITY PROBLEMS: LECTURE 21 February 10, 2014

18 Causality Problems: Lecture 21

The examples and exercises up to this point have been designed such that the state variable and causality propagationprocedure on page 36 ended prior to Step 7 and only involved integral causality. In general use, two slightly morecomplicated situations may arise.

Derivative Causality – While propagating causality through a bond graph, an energy storage device may beforced, by prior causality assignments, to have derivative causality. Such energy storage devices are dependenton at least one other input of energy storage device in the bond graph.

Algebraic Loops – After assigning causality to all inputs (Steps 1-3) and to all energy storage devices (Steps 4-6),there may still be R elements without assigned causality. When this occurs, there is at least one algebraic loopin the system. Steps 7-9 describe how to continue with the causality propagation.

Each of these cases is described in greater detail, with examples, in the following subsections.Each of these issues is created by design and modeling decisions, possible made without realization. These issues

can be removed by alteration of the design or modeling decisions.At least in the case of derivative causality, the designer should pause and reconsider whether the design and

model are reasonable.

18.1 Derivative Causality

Fig. 70 shows I and C elements in derivative causality. For the I element, the graph to which the I elementis connected determines the flow fw and the I element determines the effort Ifw, which exhibits the derivativecausality. For the C element, the graph to which the C element is connected determines the effort ew and the Celement determines the flow Cew, which exhibits the derivative causality. In either case, the causality propagationcontinues normally.

CIfw

Ifw

.ew

Cew

.

Figure 70: Energy storage elements with derivative causality.

A valid definition of the state vector is still given by the output power variables of the energy storage devicesthat have integral causality. The energy storage elements with derivative causality still store energy, but the amountof energy is determined by those elements that have integral causality.

The process is easily clarified with a few examples.

Example 18.1 — Fig. 71 shows a circuit and the corresponding reduced bond graph. The black, green, andblue variables show the propagation of the causality and state variable up to the point where the C : b element isrecognized as having derivative causality. The red variables show the propagation of the results of the derivativecausality through the bond graph. Given that the C : b input is effort e1, the C : b element determines the flow tobe be1, then the 0-junction determines the flow at C : a. From the bond graph:

� There are three energy storage devices, but one has derivative causality. Therefore, there are two independentenergy storage devices and the state dimension is n = 2.


x =[e1 f

]⊤.

� The input is u(t) and the design parameters are L, a and b. The output is y(t) = x1(t).

� The differential equations for each state are found by the standard method:

f(t) =1

Lu(t)− 1

Lx1(t)

ae1(t) = f(t)− be1(t)

x1 =1

(a+ b)x2(t)


18.1 Derivative Causality February 10, 2014

1se:u

I:L

u

fff C:b0

C:a

e1

e1e1

u-e1

.be1.

f-be1

a

L

f+_u e1

vL

i1

y+_

+_

+ -

+_

i2

b

Figure 71: Bond graph with derivative causality.

where (a+ b) is the equivalent capacitance of the two parallel capacitors.


x =

[0 1

(a+b)

− 1L 0

]x+

[01L

]u.

The derivative causality should motivate the designer to carefully consider why the circuit dcontains two capacitorsin parallel. △

Example 18.2 — The top portion of Fig. 72 shows a bond graph where the I :L element is defined with integralcausality. As the power output variable f and causality is extended through the bond graph, the C element is foundto have derivative causality. The bottom portion of the figure uses derivative causality for the C element to definethe flow as g

k f and continues with the variable propagation throughout the graph. From the bond graph:

1se:u GY

I:L

:

gI:J0

C:1/k

u

f

ff

gf

gf

gf

1se:u GY

I:L

:

gI:J0

C:1/k

u

f

ff

gf

gf

gf

gf/k. vv+gf/k

.g(v+gf/k).

v

u-g(v+gf/k).

Figure 72: Bond graph with derivative causality.

� There are three energy storage devices, but one has derivative causality. Therefore, there are two independentenergy storage devices and the state dimension is n = 2.


x =[v f

]⊤.

� The input is u(t) and the design parameters are L, k, g and J . The output is has not been specified.

� The differential equations for each state are found by the standard method:

v(t) =g

Jf(t)

Lf(t) = u− gv(t)− g2

kf

f =1(

L+ g2

k

) (u− gv(t))

f =1

Le(u− gv(t))



where Le = L+ g2

k is the equivalent I value.


x =

[0 g

J−kg

Lk+g2 0

]x+

[0k

Lk+g2

]u.

△

18.2 Algebraic Loops

When the causality propagation does not end at Step 6, the bond graph is said to have incomplete causality. Each timethat the modeler assumes an arbitrary causality for an R element at Step 7 implies the existence of one algebraicloop. Propagation of the arbitrary causality through the bond graph will ultimately provide the formula for thealgebraic constraint.

Example 18.3 — The top portion of Fig. 73 shows a circuit schematic. The bond graph at the completion ofStep 6 of the causality propagation procedure is shown in the middle section of the figure. Note that all energystorage and input devices have their causality and power output variables assigned and that the causality has beenpropagated as far as possible, yet at least some R elements do not yet have their effort, flow, and causality assigned.

Risvs+_

+_

vC

v2i2

a a b

gggSchem

atic

Ste

ps 1

-6S

teps 7

-9

+ _ i1C

Rv1

+_

0 1sf:is is

R

R

C

vs

0 1sf:is is

R

R

C

vci2

is-i2 is-i2is-i2

R(is-i2)vc+R(is-i2)vc+R(is-i2)

Figure 73: Circuit with incomplete causality. The output is y(t) = v1(t).

At Step 7, the modeler arbitrarily selects an R element and assigns its causality. The bottom portion of thefigure shows the result of selection the left most R element to define the flow i2 and propagating that choice as far aspossible through the bond graph. The effort v2 has purposefully been left blank. Note that there are two definitionsfor v2. From the properties of the 0-junction we have that

v2 = vc +R(is − i2)

and from the properties of the R element we have that

v2 = i2R.

These two definitions of the same variable supply the algebraic constraint:

i2R = vc +R(is − i2)

which yields 2Ri2 = vc +Ris or i2 = 12 is +

12Rvc.

From the bond graph,

� There is one independent energy storage devices, so the state dimension is n = 1.


18.2 Algebraic Loops February 10, 2014

� A valid definition of the state vector is defined by the power output variable of the energy storage device:

x =[vC

]⊤.

� The input is u(t) = is(t) and the design parameters are R and C. The output is

y(t) = v1(t) = R(is(t)− i2(t)) = R(is(t)−1

2is(t)−

1

2Rvc(t))

=1

2Ru(t)− 1

2x(t).

� Assuming linear storage devices, the state space model can be written as

x =1

C(is − i2)

=1

C

(is −

(1

2is +

1

2Rvc)

))x = − 1

2RCx+

1

2Cu.

△



Example 18.4 — The full wave rectifier circuit of Fig. 74 is one of the most commonly used circuits. It convertsan alternating voltage into an approximately constant voltage. The amount of voltage variation is determined bythe circuit parameters.

L L

Ru

vo

+

_v3

+

_+_C2

+ _

v1

i1+

_

v6

i6

+ _

v7

i7+

_

+

_

+ _

+ _

C1

v5

i5v4

i4

i2

v2

i3

x1

i8

ag

f

c

d

b

b

iu

Figure 74: Figure used in Example 18.4.

This example is a work in progress.

1

0

se:u

R:d4

R:RC:C1

00

0 1

11

1 I:L

C:C2

I:L

R:d5

R:d6u

x1

x2x1

x1

x1x2 x2

x3

x3 x3

x1-x3

x4

x4

x4Rx4x2-x4

x3-Rx4

v6

Def. Aux. Var. vc=(v6+x1) i7=g(-vc) va=u+vc i5=g(-va) i4=g(va-x1)

a

c

fb

va

va i5va

va-x1 i4

i4

i4i4-i5

i5+i7-x2

i5+i7-i4

1

vcvc

-vai5

R:d7 1

vc-vci7

i7

i4-i5

i4-i5 i5+i7-i4

*****

Constraints at ***** i6=g(v6) i6=i5+i7-i4

Figure 75: Figure used in Example 18.4.

△



18.3 Exercises

Exercise 18.1 – For the bond graph in Fig. 76, with (effort) input u(t) and output y(t) = 1kFk(t).

0 1

I:M

v

Fkse:u

R:d

C:1/k

C:1/h

Fc


1. How many energy storage devices are there?


3. Identify any instances of derivative causality.

4. Give an example of a type (e.g., sinusoid, step, ...) of finite magnitude input u(t) that would require infinitepower to apply. Explain why.








1se:u TF

I:L

:

gI:J

(a) See instructions for Exercise 18.2.

1se:u TF

I:L

:

gI:J0

C:c

C:b

(b) See instructions for Exercise 18.2.

1I:a C:b

R:d

GY:g

(c) See instructions for Exercise 18.2.

Figure 77: Bond graphs exhibiting derivative causality.

Exercise 18.3 – For each of the following, using the variables and indicated causality:


2. Propagate the causality, inputs (where specified), and state variables through the bond graph.


0R:a R:b

I:d

(a) See instructions for Ex-ercise 18.3. The output isthe effort across the I ele-ment.

1R:a R:b

C:c

(b) See instructions for Ex-ercise 18.3. The output isthe flow through the C ele-ment.

Figure 78: Bond graphs exhibiting algebraic loops.



Exercise 18.4 – For the circuit in Fig. 79:

1. Find the simplified bond graph model.

2. Propagate the causality, inputs, and state variables through the model.


4. Define the state space model. If the system is linear, write the model in matrix form.

Clearly indicate the location at which you arbitrarily assign an effort or flow. You should only need to do this once.State the constraint that results.

R C

u

y+_

RL

+ -+_ R

Figure 79: See instructions for Exercise 18.4.

Exercise 18.5 – For the fluid system in Fig. 80, consider the Pump to be source of (volume) flow:

1. Find and label the analogous circuit. Account for both frictional loss and kinetic energy stored in the pipe.

2. Find the state space model for the fluid system. Consider the output to be the height of the fluid in each ofthe tank.

Pump


Pp

h

PtQi

Pa


Clearly indicate any elements that have derivative causality.


19 LINEAR DEPENDENCE, DETERMINANTS, MATRIX INVERSE: LECTURE 22 February 10, 2014

19 Linear Dependence, Determinants, Matrix Inverse: Lecture 22

19.1 Linear Dependence

Given vectors ui ∈ ℜn for i = 1, . . . ,m, the vectors are linearly dependent if there exists αi ∈ ℜ, not all zero, suchthat

∑mi=1 αiui = 0.

� If this is only true when all αi = 0, then the set of vectors is linearly independent.

� Example: Consider u1, u2, u3 ∈ ℜ3 where

u1 =

111

, u2 =

011

, u3 =

001

.

Then,

α1u1 + α2u2 + α3u3 = 03

α1

111

+ α2

011

+ α3

001

=

000

.

The first row implies that α1 = 0. The second row implies that α2 = 0. The third row implies that α3 = 0.Therefore, these three vectors are linearly independent.

� Example: Consider v1, v2, v3 ∈ ℜ3 where

v1 =

111

, v2 =

011

, v3 =

100

.

Then,

α1v1 + α2v2 + α3v3 = 03

α1

111

+ α2

011

+ α3

100

=

000

.

This equation is satisfied for α1 = 1, α2 = −1, α3 = −1, Therefore, these three vectors are linearly dependent.

� We would like to have a convenient tool to check whether a set of vectors is linearly independent.

– A set of m > n vectors in ℜn is always linearly dependent.

– A set of n vectors ui ∈ ℜn, i = 1, . . . , n can be arranged as a square matrix U = [u1, u2, . . . , un] ∈ ℜn×n.If this set of vectors is linearly independent, then we say that the matrix U is nonsingular.


19.2 Determinants, Minors, and Cofactors February 10, 2014

19.2 Determinants, Minors, and Cofactors

� If A ∈ ℜn×n, then the determinant of A is denoted by |A| or det(A). The determinant is a scalar real number.

� If |A| = 0, then A is nonsingular. The vectors forming the rows (and columns) of A are linearly independent.

� If |A| = 0, then A is singular. The vectors forming the rows (and columns) of A are linearly dependent.

� We need an organized method for computing |A|, since it involves a summation with n! terms.

Minors – Mij is the minor associated with aij . Mij is the determinant of the matrix formed by dropping thei-th row and j-th column from A. The elements Mij define a matrix M .

Minor example: For A =

1 5 40 7 −13 2 7

,M11 =

∣∣∣∣ 7 −12 7

∣∣∣∣ = 7 ∗ 7− 2 ∗ (−1) = 51.

M21 =

∣∣∣∣ 5 42 7

∣∣∣∣ = 5 ∗ 7− 2 ∗ 4 = 27.

Cofactors – The cofactor associated with aij is

cij = (−1)i+jMij .

Cofactor example: For A =

1 5 40 7 −13 2 7

,c11 = (−1)1+151 = 51.

c21 = (−1)2+127 = −27.

Determinant Formulae – Two formulae are possible:

Row expansion: |A| =∑n

j=1 akjckj .

Column expansion: |A| =∑n

j=1 ajkcjk.

Select k to make your life easy.

Example 19.1 — Is the matrix A singular?

A =

1 0 01 1 01 1 1

Using row one.

|A| = 1

∣∣∣∣ 1 01 1

∣∣∣∣+ 0

∣∣∣∣ 1 01 1

∣∣∣∣+ 0

∣∣∣∣ 1 11 1

∣∣∣∣ = 1

Not singular. Column (and rows) are linearly independent. △

Example 19.2 — Is the matrix B singular?

B =

1 0 11 1 01 1 0

Using column three.

|B| = 1

∣∣∣∣ 1 11 1

∣∣∣∣+ 0

∣∣∣∣ 1 01 1

∣∣∣∣+ 0

∣∣∣∣ 1 01 1

∣∣∣∣ = 0

Singular. Columns (and rows) are linearly dependent. △



19.2.1 Properties of Determinants

For A,B ∈ ℜn×n

1. |AB| = |A| |B|

2. |A| = |A⊤|

3. If any row or column of A is entirely zero, then |A| = 0.

4. If any two rows (or columns) of A are linearly dependent, then |A| = 0.

5. Interchanging two rows (or two columns) of A does not change the determinant.

6. Multiplication of a row of A by α ∈ ℜ, yields α|A|.

7. A scaled version of one row can be added to another row without changing the determinant.

19.3 Matrix Inversion

Scalars: For scalars, a, b, x ∈ ℜ, the solution to the equation ax = b is found by division x = ba when a = 0. This

approach does not extend to matrices, as the idea of matrix division is not defined.

We can consider multiplication by the scalar a ∈ ℜ to be an operation. The inverse operation is multiplicationby the scalar a−1 = 1

a ∈ ℜ which is well-defined when a = 0. We know that the real number 1 is the unityoperator for scalar multiplication. We know that multiplication by 1

a is the inverse multiplicative operation tomultiplication by a, because a

(1a

)= 1. This idea extends to matrices.

Matrices: For A,B ∈ ℜn×n, with |A| = 0, we will call B the inverse of A if

BA = AB = I

where I is the identity matrix in ℜn×n.

� Note that the matrices A and B must be square.

� The matrix A must be nonsingular and by the properties of determinants,

|AB| = |A| |B||I| = |A| |B|1 = |A| |B|

|B| =1

|A|.

Since |A| is well-defined (not 0 or ∞), the |B| is also well-defined. The matrix B is nonsingular.

� Typically, we use the notation B = A−1, which is read “B is the inverse of A.”

Computation of A−1: When A is a square nonsingular matrix,

A−1 =C⊤

|A|

where C is the cofactor matrix for A and C⊤ is called the adjoint of A.


19.3 Matrix Inversion February 10, 2014

Example 19.3 —

Let A =

1 0 60 4 −33 5 −1

. Find A−1.

First we compute |A|. If the determinant is zero, then there is no need to try computing A−1, as it would notexist. We use row expansion with row 1:

|A| = 1c11 + 0c12 + 6c13

= (−1)2∣∣∣∣ 4 −35 −1

∣∣∣∣+ 0 + 6(−1)4∣∣∣∣ 0 43 5

∣∣∣∣= (−4 + 15) + 6(−12) = −61

Since |A| = 0, the inverse will exist and we calculate it.

M11 =

∣∣∣∣ 4 −35 −1

∣∣∣∣ = 11 M12 =

∣∣∣∣ 0 −33 −1

∣∣∣∣ = 9 M13 =

∣∣∣∣ 0 43 5

∣∣∣∣ = −12

M21 =

∣∣∣∣ 0 65 −1

∣∣∣∣ = −30 M22 =

∣∣∣∣ 1 63 −1

∣∣∣∣ = −19 M23 =

∣∣∣∣ 1 03 5

∣∣∣∣ = 5

M31 =

∣∣∣∣ 0 64 −3

∣∣∣∣ = −24 M32 =

∣∣∣∣ 1 60 −3

∣∣∣∣ = −3 M33 =

∣∣∣∣ 1 00 4

∣∣∣∣ = 4

C11 = (−1)1+1(11) C12 = (−1)1+2(9) C13 = (−1)1+3(−12)C21 = (−1)2+1(−30) C22 = (−1)2+2(−19) C23 = (−1)2+3(5)C31 = (−1)3+1(−24) C32 = (−1)3+2(−3) C33 = (−1)3+3(4)

C =

11 −9 −1230 −19 −5

−24 3 4

; therefore, A−1 =C⊤

|A|=

11 30 −24−9 −19 3−12 −5 4

−61

Check:

AA−1 =

1 0 60 4 −33 5 −1

11 30 −24−9 −19 3−12 −5 4

1

−61

=

11− 72 30− 30 −24 + 24−36 + 36 −76 + 15 12− 12

33− 45 + 12 90− 95 + 5 −72 + 15− 4

1

−61

=

1 0 00 1 00 0 1

△



Example 19.4 —

Symbolic Example (microphone feedback):

Let (sI −A) =

s −1 0a s+ b −c

−d 0 s+ f

. Find (sI −A)−1.

Compute |A|- Using a row 1 expansion, with the cofactors as computed below:

|sI −A| = sC11 +−1C12 + 0C13

= s(s+ b)(s+ f)− cd+ a(s+ f)

= s3 + (b+ f)s2 + (bf + a)s+ (af − cd)

Compute minor-

M11 =

∣∣∣∣ (s+ b) −c0 (s+ f)

∣∣∣∣ = (s+ b)(s+ f) M12 =

∣∣∣∣ a −c−d (s+ f)

∣∣∣∣ = a(s+ f)− cd

M13 =

∣∣∣∣ a (s+ b)−d 0

∣∣∣∣ = d(s+ b)

M21 =

∣∣∣∣ −1 00 (s+ f)

∣∣∣∣ = −(s+ f) M22 =

∣∣∣∣ s 0−d (s+ f)

∣∣∣∣ = s(s+ f)

M23 =

∣∣∣∣ s −1−d 0

∣∣∣∣ = −d

M31 =

∣∣∣∣ −1 0(s+ b) −c

∣∣∣∣ = c M32 =

∣∣∣∣ s 0a −c

∣∣∣∣ = −cs

M33 =

∣∣∣∣ s −1a (s+ b)

∣∣∣∣ = s2 + bs+ a

Compute cofactor-

C =

(s+ b)(s+ f) −a(s+ f) + cd d(s+ b)(s+ f) s(s+ f) d

c cs s2 + bs+ a

Form inverse-

(sI −A)−1 =

(s+ b)(s+ f) (s+ f) c−a(s+ f) + cd s(s+ f) cs

d(s+ b) d s2 + bs+ a

s3 + (b+ f)s2 + (bf + a)s+ (af − cd)

Check: (sI −A)(sI −A)−1 = I △


19.3 Matrix Inversion February 10, 2014

Two dimensional Examples: The special case of 2×2 matrices is useful to consider as it frequently occurs in ex-amples.

For A ∈ ℜ2×2:

A =

[a11 a12a21 a22

]M =

[a22 a21a12 a12

]C =

[a22 −a21

−a12 a12

]

A−1 =

[a22 −a12

−a21 a12

]a11a22 − a12a21

This result will be used frequently without proof.

Example 19.5 — If (sI − B) =

[(s+ 2) −3−2 (s+ 3)

], then (sI − B)−1 =

(s+ 3) 32 (s+ 2)

s2+5s . Note

that all elements have the same denominator prior to pole-zero cancellations. Note that the denominatorhas the same order as the original matrix. △

Example 19.6 — If (sI − C) =

[s 0−1 (s+ 4)

], then

(sI − C)−1 =

[(s+ 4) 0

1 s

]s2 + 4s

=

[ 1s 01

s(s+4)1

s+4

].

Note that all elements have the same denominator prior to pole-zero cancelations. After pole-zero can-celations, this fact may not be as apparent. Typically it is better to leave the answer in the first form.△

Check both of the above to ensure the inverse is correct.



19.4 Exercises

Exercise 19.1 – Each of the following specifies a matrix A. In each case:

� Find A−1 using the cofactor method.

� Multiply A−1A to check the result.

Show your work.

1) A =

[−3 0−2 −2

]2) A =

[1 1

−1 −2

]

3) A =

−3 0 0−2 −2 00 0 1

4) A =

1 0 01 0 01 1 2

Exercise 19.2 – For the state space system in Example 15.4 on p. 72:

1. Show that

(sI −A) =

(s+ 1

R1C1

)0

− 1R1C2

(s+ 1

R2C2

) 2. Compute (sI −A)

−1.

Exercise 19.3 – For the state space system

x =

0 1 00 0 10 0 −5

x+

001

u

y =[1 0 0

]x.

1. Find (sI −A).

2. Compute (sI −A)−1

.


February 10, 2014

20 Eigenvalues and Eigenvectors: Lecture 23

� Let A ∈ ℜn×n. The set of (complex) scalars λi ∈ C and (nonzero) vectors vi ∈ Cn satisfying

Avi = λivi or (λiI −A)vi = 0n

are the eigenvalues and eigenvectors of A.

� We are only interested in nontrivial solutions. The vector vi = 0n ∈ ℜn is always a solution of the aboveequations. It is referred to as the trivial solution and is not of interest.

� The variable λ is a dummy variable. The solutions to |λI −A| = 0 and |sI −A| = 0 are exactly the same.

� Nontrivial solutions vi exist only if (λI − A) is a singular matrix. Therefore, the eigenvalues of A are thevalues of λ such that A is singular. Using the properties of the determinant, this is the values of λ such that|λI − A| = 0. The equation |λI − A| = 0 is an an n-th order polynomial in λ that is called the characteristicequation of A. Because |λI − A| = 0 is an n-th order polynomial in λ, there are n (possibly non-unique)eigenvalues.

Example 20.1 — Let B =

[0 01 −4

]:

(sI −B) =

[s 0−1 (s+ 4)

]and |sI −B| = s2 + 4s = s(s+ 4).

The eigenvalues of B are s = 0 and s = −4.The eigenvector corresponding to λ1 = 0, is a nonzero vector v1 such that

(λiI −B)v1 = 02[0 0

−1 4

]v1 =

[00

]which is satisfied by v1 =

[4, 1

]⊤or any scalar multiple of this vector.

The eigenvector corresponding to λ2 = −4, is a nonzero vector v2 such that

(λiI −B)v2 = 02[−4 0−1 0

]v2 =

[00


[0, 1


This inverse of (sI −B) is

(sI −B)−1 =

[(s+ 4) 1

0 s

]s(s+ 4)

.

Note that (sI −B)−1 is a matrix of transfer functions. Before any simplification (pole-zero cancelations) all of thesetransfer functions have the same denominator, which is the characteristic equation of B. Therefore, the poles ofthese transfer functions are the same as the eigenvalues of B. △

When a matrix has complex eigenvalues, the eigenvalues and eigenvectors will occur in complex conjugate pairs.

Example 20.2 — Find the eigenvalues and eigenvectors of A =

[0 1

−25 −6

]:

(sI −A) =

[s −125 (s+ 6)

]and |sI −A| = s2 + 6s+ 25 = (s+ 3 + 4j)(s+ 3− 4j).

The eigenvalues of A are s = −3± 4j.


20 EIGENVALUES AND EIGENVECTORS: LECTURE 23 February 10, 2014

The eigenvector corresponding to λ1 = −3 + 4j, is a nonzero vector v1 such that

(λiI −A)v1 = 02[−3 + 4j −1

25 3 + 4j

]v1 =

[00


[−(3 + 4j), 25


The eigenvector corresponding to λ2 = −3− 4j is v2 = v∗1 =[−3 + 4j, 25

]⊤. △

The following example computes eigenvectors for a matrix of dimension higher than n = 2. Students typicallyfind this a bit harder.

Example 20.3 — Let A =

0 0 10 0 10 2 −1

. Performing the Laplace expansion along the first column, it is straight-

forward to show that

|sI −A| = s

∣∣∣∣ s −1−2 s+ 1

∣∣∣∣ = s(s+ 2)(s− 1).

Therefore the eigenvalues of A are 0, 1, and -2.

For s = 0: We need to find a nontrivial vector v1 such that

(sI −A)|s=0 v1 = 03 (36) 0 0 −10 0 −10 −2 −1

v1 = 03. (37)

Because the first column is zero, choosing v1 =[1 0 0

]⊤satisfies the equation.

For s = 1: We need to find a nontrivial vector v2 such that

(sI −A)|s=1 v2 = 03 (38) 1 0 −10 1 −10 −2 2

v2 = 03. (39)

Choosing v2 =[1 1 1


For s = −2: We need to find a nontrivial vector v3 such that

(sI −A)|s=−2 v3 = 03 (40) −2 0 −10 −2 −10 −2 −1

v3 = 03. (41)

Choosing v3 =[1 1 −2


△


20.1 Matrix Diagonalization: Extra material February 10, 2014

20.1 Matrix Diagonalization: Extra material

If a matrix A ∈ ℜn×n has n distinct eigenvalues λi, then it also has n distinct and linearly independent eigenvectorsvi. Form the diagonal matrix D = diag(λi) and the matrix M = [v1, . . . , vn]. The matrix A can be decomposed as

A = MDM−1.

A more general version of this result applies to matrices with repeated eigenvalues. The interested reader shouldinvestigate the Jordan canonical form.

Example 20.4 — Consider the matrix A =

[5 −2

−2 2

]. The determinant of (sI −A) is

|sI −A| = s2 − 7s+ 6 = (s− 6)(s− 1);

therefore, the eigenvalues of A are s = 1 and s = 6.The eigenvector corresponding to λ1 = 1, is a nonzero vector v1 such that

(λiI −A)v1 = 02[4 −2

−2 1

]v1 =

[00


[1, 2


The eigenvector corresponding to λ2 = 6, is a nonzero vector v2 such that

(λiI −A)v2 = 02[−1 0−1 −4

]v2 =

[00


[2, −1


Therefore,

M =

[1 22 −1

]and M−1 =

1

5

[1 22 −1

].

Note that

M−1AM =1

5

[1 22 −1

] [5 −2

−2 2

] [1 22 −1

]=

[1 00 6

],

which demonstrates the Jordan decomposition (or eigenvector-eigenvalue decomposition). △

This decomposition shows that the eigenvalues and eigenvectors of a matrix contain all the information neededto describe the matrix. The eigen-decomposition of A is also useful for certain types of theoretical analysis.

Example 20.5 — Continuing from Example 20.3, we define D =

0 0 00 1 00 0 −2

and M =

1 1 10 1 10 1 −2

. Direct

multiplication confirms that A = M DM−1. △

20.2 Similar Matrices

Two square matrices A and B are similar if there exists a nonsingular matrix S such that B = S−1AS.It is straightforward to show that any two similar matrices have the same eigenvalues. Let A and B be similar

matrices and let λ be an eigenvalue of B. Starting from the definition of eigenvalues we have that

|λI −B| = 0 (42)

|λI − S−1AS| = 0, by the definition of similarity (43)

|λS−1S − S−1AS| = 0, by the definition of the matrix inverse (44)

|S−1 (λI −A)S| = 0, (45)

|S−1| | (λI −A) | |S| = 0, by the definition of the determinant (46)

| (λI −A) | = 0, because S is nonsingular; (47)

Therefore, λ is an eigenvalue of A.Note for example, in Section 20.1, A and D are similar matrices.


20 EIGENVALUES AND EIGENVECTORS: LECTURE 23 February 10, 2014

20.3 Exercises

Exercise 20.1 – For a square matrix A that has non-repeated eigenvalues, use the results of Section 20.1, to showthe following:

1. A2 = MD2M−1

2. Am = MDmM−1 for any integer m.

Exercise 20.2 – For a square matrix A, the matrix exponential is defined as

eAt = I +At+1

2!(At)2 +

1

3!(At)3 + · · · .

Assuming that A that has non-repeated eigenvalues, use the results of Section 20.1 and Exercise 20.1, to show thefollowing:

1. eAt = M(I +Dt+ 1

2! (Dt)2 + 13! (Dt)3 + · · ·

)M−1

2. diag(eλt) =(I +Dt+ 1

2! (Dt)2 + 13! (Dt)3 + · · ·

).

Note that for scalar λ ∈ C : eλt = I + λt+ 12! (λt)

2 + 13! (λt)

3 + · · · which is the standard definition of the exponentialfunction.

Exercise 20.3 – For A =

−1 0 00 −5 −60 1 0

:1. Find the eigenvalues and eigenvectors of A.

2. Form the eigenvalues and eigenvectors of A into the matrices M and D and show that A = MDM−1.

3. Find eAt.

4. Find the solution x(t) = eAtx(0) of x(t) = Ax(t) with initial condition x(0) =[1 −1 0

]⊤.

Do the computations of this problem by hand. Only use Matlab to check and debug your answer.

Exercise 20.4 – For A =

0 0 0 00 −1 0 00 0 −6 −80 0 1 0

:1. Find the eigenvalues and eigenvectors of A.

2. Form the eigenvalues and eigenvectors of A into the matrices M and D and show that A = MDM−1.

3. Find eAt.

4. Find the solution x(t) = eAtx(0) of x(t) = Ax(t) with initial condition x(0) =[1 −1 −1 1

]⊤.

Do the computations of this problem by hand. Only use Matlab to check and debug your answer.


February 10, 2014

21 State Space to Transfer Function: Lecture 23

A linear state space model has the general form

x = Ax+Bu

y = Cx+Du.

In most physical systems D = 0.The goal of this lecture is to discuss computation of the transfer function corresponding to the above state space

system, when the initial conditions are assumed to be zero. To derive the transfer function, we take the Laplacetransform of each of the above equations, solve for and eliminate X(s), and then solve for Y (s) as a function of U(s):

L{Ix} = L{Ax+Bu} L {y} = L{Cx+Du}sIX(s) = AX(s) +BU(s) Y (s) = CX(s) +DU(s)sIX(s)−AX(s) = BU(s) Y (s) = CX(s) +DU(s)(sI −A)X(s) = BU(s) Y (s) = CX(s) +DU(s)

X(s) = (sI −A)−1

BU(s) Y (s) = CX(s) +DU(s)

Y (s) = C (sI −A)−1

BU(s) +DU(s)

Y (s)

U(s)= C (sI −A)

−1B +D

Note that the C in this equation is from the state space model, it is not the cofactor matrix of A.Useful facts:

1. (sI −A)−1

is a matrix of transfer functions. The denominator of each transfer function is the polynomiald(s) = |sI −A| .

� The poles are the solutions to d(s) = |sI −A| = 0.

� The eigenvalues of A are the solutions of |sI −A| = 0.

2. The eigenvalues of A and the poles of the transfer function from U(s) to Y (s) are identical, before pole-zerocancellations.

3. The Matlab function ‘ss2tf’ performs this operation.


21 STATE SPACE TO TRANSFER FUNCTION: LECTURE 23 February 10, 2014

Example 21.1 — Consider the state space system

x =

[0 1

−6 −5

]x+

[01

]u

y =[1 1

]x.

(SI −A) =

[s −16 s+ 5

]→ |sI −A| = s2 + 5s+ 6 = (s+ 3)(s+ 2)

eigenvalues: − 2,−3

Y (s)

U(s)= C (sI −A)

−1B +D =

[1 1

] [ s −16 s+ 5

]−1 [01

]

=[1 1

][

s+ 5 1−6 s

](s+ 3)(s+ 2)

[01

]=

Y (s)

U(s)=

(s+ 1)

(s+ 3)(s+ 2)

Zeros: s = −1 and Poles: s = −2, − 3 △


v =

[−2 00 −3

]v +

[−11

]u

y =[1 2

]v.

(SI −A) =

[s+ 2 00 s+ 3

]→ |sI −A| = s2 + 5s+ 6 = (s+ 3)(s+ 2)

eigenvalues: − 2,−3

Y (s)

U(s)= C (sI −A)

−1B +D =

[1 2

] [ s+ 2 00 s+ 3

]−1 [ −11

]

=[1 2

][

s+ 3 00 s+ 2

](s+ 3)(s+ 2)

[−11

]=

Y (s)

U(s)=

(s+ 1)

(s+ 3)(s+ 2)

Zeros: s = −1 and Poles: s = −2, − 3 which are identical to the previous example. △

A single transfer function has an infinite number of equivalent state space transformations. If x is a valid statevector and P is a nonsingular matrix, then v = Px is also a valid state vector. Note that in the previous two

examples, v = Px where P =

[−3 −12 1

]. While the state space representations of Examples 21.2 and 21.2 are

different, they have the same transfer functon.


21.1 Equivalent State Space Representations February 10, 2014


v =

0 1 00 0 10 −1 −2

v +

001

u

y =[0 1 0

]v.

(SI −A) =

s −1 00 s −10 1 s+ 2

→ |sI −A| = s3 + 2s2 + s = s(s+ 1 + 1j)(s+ 1− 1j)

eigenvalues: 0,−1,−1

Y (s)

U(s)= C (sI −A)

−1B +D =

[0 1 0

] s −1 00 s −10 1 s+ 2

−1 001

=[0 1 0

] (s2 + 2s+ 1) (s+ 2) 1

0 s(s+ 2) s0 −s s2

s(s+ 1)(s+ 1)

001

=

Y (s)

U(s)=

s

s(s+ 1)(s+ 1)=

s

s(s2 + 2s+ 1)

Zeros: s = 0 and Poles: s = 0,−1,−1. △

Be extremely careful with pole-zero cancelations. Never cancel an unstable pole.

Example 21.4 — Find the transfer function for the following system

x =

0 1 0−a −b cd 0 −f

x+

001

u

y =[1 0 0

]x.

Using the results from Example 19.4 on page 101, the transfer function is

Y (s)

U(s)=

c

s3 + (b+ f)s2 + (bf + a)s+ (af − cd).

△

21.1 Equivalent State Space Representations9

9Not finished.


21 STATE SPACE TO TRANSFER FUNCTION: LECTURE 23 February 10, 2014

21.2 Exercises

Exercise 21.1 – For each of the following state space systems, find the input to output transfer function:

a) x =

[−3 0−2 −2

]x+

[10

]u with y =

[0 1

]x

b) x =

[−1 00 1

]x+

[11

]u with y =

[1 1

]x

c) v =

0 1 00 0 10 −1 −2

v +

001

u with y =[0 0 1

]v

d) v =

0 1 00 0 10 −1 −2

v +

001

u with y =[0 1 0

]v

e) v =

0 1 00 0 10 −1 −2

v +

001

u with y =[1 0 0

]v

Exercise 21.2 – For the state space system in Example 15.4 on p. 72, show that the transfer function from Qi toQ2 is

Q2(s)

Qi(s)=

(1R2

)(1

R1C2

)(1C1

)(s+ 1

R1C1

)(s+ 1

R2C2

) .See also Exercise 19.2 on page 103.

Exercise 21.3 – Find the transfer function for each of the following:

1. Exercise 15.8. Give expressions for the undamped natural frequency, the DC gain, and the decay rate.

2. The state space system in Example 25.1 on page 122.

3. The state space system in Example 25.2 on page 123.


February 10, 2014

Domain Effort Momentum Flow Displacement Power, N-m/secGeneric e p f q P , Joules/sec.Electrical e, Volt λ, Volt-s i, Amp Q, Coul. e(t)i(t)Fluid P , Newton/m2 pP , Newton-s/m


Thermal T , ◦Kelvin ****** QH , Joules/sec EH , Joules T (t)QH(t)Translational F , Newton p, Newton-sec. v, m/s X, m F (t)v(t)Rotational τ , Newton-m pτ , Newton-m-s ω, rad/sec θ, rad τ(t)ω(t)


22 Mechanical Systems: Lecture 24

This section discusses both rotational and translational motion of mechanical systems. Both types of systems usethe same bond graph modeling methodology, which is distinct from the method used for electrical, fluid, and thermalsystems.

Tables 7 and 8, which were presented in previous lectures, summarize the variable classifications and the one portelements.

Domain


Electrical Resistor Capacitor Inductor Voltage Source Current SourceFluid Resistance Tank Pipe Pressure source Flow rate sourceThermal Heat transfer Heat capacity *** Temperature source Heat flow sourceTranslational Friction Linear Spring Mass Force Source Velocity SourceRotational Friction Torsional Spring Inertia Torque Source Angular rate source

Table 8: Physical analogies between components in different energy domains.

22.1 Velocities

In the analysis of mechanical systems it will be necessary to clearly distinguish between absolute and relative velocities.

Absolute velocity is the velocity at which a point moves with respect to an inertial reference frame.

Relative velocity is the velocity at which one point moves with respect to another point.

An inertial frame is one that for the purpose of the analysis can be considered to be non-accelerating and non-rotating.For an automobile, the analyst might consider the roadway as the frame-of-reference. Although the roadway isattached to the earth, which is rotating and accelerating, for the purpose of the analysis, earth’s acceleration androtation might be considered negligible. The velocity of the center of a mass of the car with respect to the roadway’sinertial frame defines an absolute velocity. The velocity of any point on the body of the car, for example a wheel,with respect to another point on the car defines a relative velocity.

kv2

v1

hv3v1

Figure 81: Relative Ve-locities

Relative velocities are very useful in the analysis of springs and dampers. In Fig. 81,v1, v2, and v3 represent absolute velocities.Each absolute velocity definition includes aname and an assumed positive direction. Relative velocities are the difference between twoabsolute velocities. The rates of compression and expansion of the spring with constant kare, respectively:

vc = v1 − v2 and ve = −v1 + v2.

The rates of compression and expansion of the spring with constant h are, respectively:

vc = v1 + v3 and ve = −v1 − v3.

In applications, the definitions of the (assumed) positive absolute velocities will be imposedby the application, then the relative velocities are computed accordingly.

In each application, regardless of the direction of the positive direction of the absolutevelocity, the analyst must define whether they are defining relative velocities to be positivewhen compressing ‘+C’ or stretching ‘+T ’.


22 MECHANICAL SYSTEMS: LECTURE 24 February 10, 2014

d

v1 v2

d

v1 v2

+T

+C vC=-v1-v2

vT=v1+v2 +T vT=-v1+v2

+C vC=v1-v2

Figure 82: Relative Velocities

22.2 Mechanical: Translation Systems

This section briefly introduces the translational 1-ports.

R-elements: The force Fb applied to compress a damper is related to the rate of compression vb of the damper byan algebraic equation: Fb(t) = g(vb(t)). For a linear (viscous damper), this simplifies to

Fb(t) = bvb(t)

where b is the coefficient of viscous damping in Newtons-s/m. Because dampers have a static relationshipbetween an effort variable and a flow variable, according to Table 8, dampers are R-elements. Dampers areused to model friction between objects in relative motion. Most friction is inherently nonlinear. Fig. 83 showsa few different methods of depicting viscous friction in mechanical schematic diagrams.

C-elements: The force Fk applied to the two ends of a spring and the relative compression of the ends of the springXk are related by

Fk(t) = kXk(t) or Xk(t) =1

kFk(t)

where k is the spring constant in Newtons/m. A spring has a static relationship between a displacement variableand an effort variable; therefore, according to Table 8 springs are C type elements. Note that the capacitanceof a spring is Ck = 1

k .

I-elements: The velocity vM of and net force F applied to a mass M are related by

F (t) = Md

dtvM (t)

where M is the inertial mass in kg. In this relation ship, the positive direction of the applied force and positivedirection of the velocity are assumed to be in the same direction; otherwise, a negative sign is required in theequation. Integration of this equation yields∫

F (t)dt = = MvM (t)

p(t) = MvM (t).

A mass has a static relationship between a momentum variable and a flow variable; therefore, according toTable 8 this is an I type element.

The effort source is a force source.

The flow source is a velocity source.


22.2 Mechanical: Translation Systems February 10, 2014

M

x

Md

x

Md

x

xxxxxx

Friction with

coefficient d

No friction

Friction with

coefficient d

Figure 83: Methods for depicting vis-cous friction in mechanical schematics.


22 MECHANICAL SYSTEMS: LECTURE 24 February 10, 2014

22.3 Mechanical: Rotation Systems

� The torque τb applied to the two ends of a linear angular rate damping element is related to the angular velocitydifference ωb (i.e., flow) between the ends of the damper by:

τb(t) = bωb(t)

where b is the coefficient of viscous damping in Newton-m-s/rad. A damper has a static relationship betweenan effort variable and a flow variable; therefore, according to Table 8 this is an R type element.

� The torque τk applied to the two ends of a torsional spring and the relative angular displacement of the endsof the spring θk are related by

τk(t) = kθk(t) or θk(t) =1

kτk(t)

where k is the spring constant in Newtons-m/rad and Ck = 1k . A torsional spring has a static relationship

between a displacement variable and an effort variable; therefore, according to Table 8 this is a C type element.

� The angular velocity ωJ of and net torque τ applied to an inertia J are related by

τ(t) = Jd

dtωJ (t)

where J is the rotational inertia in kg −m2. Integration of this equation yields∫τ(t)dt = = JωJ (t)

pτ (t) = JωJ(t).

A mass has a static relationship between a momentum variable and a flow variable; therefore, according toTable 8 this is an I type element.

� The effort source is a torque source.

� The flow source is a angular velocity source.

22.4 Right hand rule

Explain:

v = ω ×R (48)

τ = r × F (49)

Need figures.10

10Unfinished


February 10, 2014

23 Bond Graph modeling–Mechanical Systems: Lecture 25

For both translational and rotational systems, the following procedure facilitates the derivation of a state spacemodel.

23.1 Mechanical Modeling Procedure

1. Assign a power convention to schematic.

(a) Define positive absolute velocities.

(b) Define sign convention for R & C-elements (Tension or Compression)

(c) Define zero absolute velocity.

2. Use 1-junctions to represent each distinct point with a well-defined absolute velocity.

(a) Attach I-elements to 1-junctions that have inertia

(b) Attach sources as necessary to 1-junctions.

(c) Use a 1-junction to establish the zero absolute velocity.

(d) Add half-arrows.

3. Add 0-junctions between 1-junctions as necessary to define relative velocities.

(a) Do not add half-arrows yet.

4. Add R and C-elements.

(a) Power is entering.

(b) Use 1-junctions for R and C-elements acting between the same points (summed efforts are in series).

5. Add half-arrows to 0-junctions from Step 3.

(a) Relative velocity must be defined consistent with the power convention assumed in Step 1b (i.e., Tensionor Compression).

(b) Be careful here. This step is the key.

6. Eliminate zero power bonds.

7. Simplify.

8. Apply causality and walk variables through graph.

9. Write state space model.

24 Bond Graph modeling – Translation Examples: Lecture 26-27

24.1 Sign convention examples

Step 1b of the procedure described above often initially causes confusion. This section derives the state space modelfor a simple physical system by two different methods. The only difference is the assumed sign convention. Inthe final comparisons, the state space models are different; however the input-output behavior as described by thetransfer functions are identical.


24 BOND GRAPH MODELING – TRANSLATION EXAMPLES: LECTURE 26-27 February 10, 2014

Example 24.1 — In this version of the example, the“+T” or tension sign convention is assumed. This meansthat the force applied to the elements while they arestretched is considered to be positive. The rate of stretchof the spring and damper is vT = v1 − v0.

For this example, the output is the velocity of themass: y(t) = v1(t).

Mk

d F

v1v0=0 +T

Sf:0 1 1 I:mv0 v1

Sf:0 1 0 1 I:mv0 v0 v1 v1vT=

v1-v

0

Sf:0 1 0 1 I:m0 0 v1 v1

vT=

v1-v

0

1C:1/k R:dFkvT vT

dv1

I:m

v1

1C:1/k R:dFkv1 v1

Se:F

Se:F

Se:F

Se:F

F v1

F-(Fk+dv1)

Figure 84: Translation Example 24.1: In Tension.

Fig. 84 shows the steps in drawing and simplifyingthe bond graph for the translational system shown in theupper left.

The second to last portion of the simplification is notobvious. The spring and damper act between the sametwo locations (the wall and the mass). Each element couldbe modeled separately as in the bond graph on the leftside of Exercise 9.1 on the page 41. Exercise 9.1 showsthat that bond graph can be equivalently represented asshown here.

A valid definition of the state vector is

x =

[v1Fk

]

Application of the element defining equations at each en-ergy storage device yields the following ODE’s:

At I : mv1 = F − Fk − dv1

At C :1

kFk = v1

Solving for the state variable derivatives yields:

v1 =1

mF − 1

mFk − d

mv1

Fk = kv1

Theerefore, the state-space ODE is

x1 =1

mF − 1

mx2 −

d

mx1

x2 = kx1

which can be written in matrix form as

x =

[− d

m − 1m

k 0

]x+

[1m0

]F

y =[

1 0]x.

The resulting transfer function is

Y (s)

F (s)=

s

ms2 + ds+ k.

If a second output of interest is the displacement of thespring from its equilibrium position, p1(t) = 1

kFk, thenthe readers should be able to derive that The resultingtransfer function is

P1(s)

F (s)=

1

ms2 + ds+ k.

△


24.1 Sign convention examples February 10, 2014

Example 24.2 — In this version of the example, the“+C” or compression sign convention is assumed. Thismeans that the force applied to the elements while theyare compressed is considered to be positive. The rate ofcompression of the spring and damper is vC = v0 − v1.

M

k

d F

v1

v2=0+C

Sf:0 1 0 1 I:mv0 v0 v1 v1

1C:1/k R:dfCvC vC

vC=

v0

-v1

Sf:0 1 1 I:mv0 v1

Se:F

Sf:0 1 0 1 I:mv0 v0 v1 v1

vC=

v0

-v1

Se:F

Se:F

-dv11C:1/k R:d

fC

-v1

fC-dv1

-v1

-v1

0 1 I:mv1

fC-dv1

Se:F

F+fC-dv1v1

Figure 85: Translation Example 1: Compression.

Note that in the final bond graph, the remaining 0-junction cannot be removed and inserts a negative signin the equations.

The state vector is defined as

x =

[v1fC

]Application of the element defining equations at each en-ergy storage device yields the following ODE’s:

At I : mv1 = F + fC − dv1

At C :1

kfC = −v1.

Solving for the state variable derivatives yields:

v1 =1

mF − 1

mfC − d

mv1

fC = −kv1.

Therefore, the state-space ODE is

x1 =1

mF +

1

mx2 −

d

mx1

x2 = −kx1

which can be written in matrix form as

x =

[− d

m1m

−k 0

]x+

[1m0

]F

y =[

1 0]x

The resulting transfer function is

Y (s)

F (s)=

s

ms2 + ds+ k.

△

Different definitions of the state result in different,but equivalent, state models. The state space modelsare equivalent in the sense that the have the same input-output response (i.e., transfer function).



Example 24.3 — This example considers a massless junction with output y = p2.

Se:Mg

1

0

1

0

1

v0

v1

v2

Sf:0

I:M

v1

v1

- v0

v2

- v1

v0

v2

Se:Mg

1

0

1

0

1

v0

v1

v2

Sf:0

I:M

v1

v1

- v0

v2

- v1

v0

v2

C:1/h

1

R:d

C:1/k

+T v0=0

M

kd

v1

v2

Mg

h

Se:Mg

0

1v2

I:M

v1 v

2 - v

1

v2

1

R:d

C:1/k

C:1/h

F1

F2

F2

F2

F2-F

1

v2

MgMg-F2

v1 v

1

v1

=F

2 - F

1

d

Figure 86: Translation Example 2.

Fig. 86 shows the steps in following the standard method. A valid definition of the state is x = [F1, F2, v2]⊤.

The element ODE’s are:1k F1 = 1

d (F2 − F1)1h F2 = v2 − 1

d (F2 − F1)Mv2 = −F2 +Mg

The state space model in matrix form is

x =

−kd

kd 0

hd −h

d h0 − 1

M 0

x+

00g

y =

[1k

1h 0

]x.

△



24.2 Exercises

Exercise 24.1 – For the translational system shown inFig. 87, assume that the input u(t) is an applied force.

1. Draw the simplified bond graph for the system.

2. Propagate the causality, inputs, and state variablesthrough the bond graph.

3. Define the state space model.

m1

h

d u(t)

v1(t) v2(t)

m2xxxxxxx xxxxxxx

b2b1

Figure 87: Mechanical schematic diagram for transla-tional Exercise 24.1.

Note that the damper between the masses represents fric-tion in the mechanism interconnecting the two masses.This viscous friction should be proportional to the rela-tive velocity between the masses. The cross-hatching un-derneath each mass represents friction between the massand the fixed reference frame.





m1

u(t)

v1(t)

v2(t)

m2

xxxxxxx

xxxx xxx x x xxxx x x xxx

b2

b1


The cross-hatching between the two masses representsfriction between them and should be proportional to theirrelative velocity. The cross-hatching underneath the bot-tom mass represents friction between the mass and thefixed reference frame.





m1

h

d u(t)

v1(t) v2(t)

m2xxxxxxx xxxxxxx

b2

k

b






m1

h

d u(t)

v1(t) v2(t)

m2xxxxxxx xxxxxxx

b2

k bv3(t)




Exercise 24.5 – For the translational system shown inFig. 91, assume that the inputs u1(t) and u2(t) are forcesapplied at the indicated locations.

1. Draw the simplified bond graph for the system. SeeExercise 9.9.



Md hM b

u2(t)u1(t)v1(t) v2(t)

Figure 91: Mechanical schematic diagram for transla-tional Exercises 24.5 and 24.6.

Exercise 24.6 – For the translational system shown inFig. 91, assume that the inputs u1(t) and u2(t) are ve-locities applied at the indicated locations.




Exercise 24.7 – For the translational system shown inFig. 92, assume that the input u(t) is a velocity appliedat the point labeled A.




Exercise 24.8 – For the translational system shown inFig. 92, assume that the input u(t) is a force applied atpoint A.



Mk

du(t)

p(t)

h

A

Figure 92: Mechanical schematic diagram for transla-tional Exercises 24.7 and 24.8.


4. Find the transfer function from the applied forceu(t) to the displacement p(t) of the mass M .


February 10, 2014

25 Bond Graph modeling – Rotation Examples: Lecture 27-28

The modeling procedure is the same as for translational systems. See page 116.

Example 25.1 — The schematic in Fig. 93 shows a rotational system. The torque τ is applied to a rotationalinertia at one end of a shaft. The twist of the shaft transmits a torque to the opposite end of the shaft, which causesthe inertia J2 to rotate. There is viscous friction between the shaft and a housing. The output is the shaft twist:y(t) = θ2(t)− θ1(t).

w2

J1J2

d2d1

k

w1

t

w0

+T

1 0 11 2

10

10

00

I:J2

I:J1

Sf:0

C:1/k R:d2

R:d1

1 0 1 I:J2

I:J1

C:1/k

R:d2

R:d1

Se:t

Se:t

t

w2w1 w1

w1

w1

d1w1

w2

w2d2w2

ts w2-w1

tstst+ts-d1w1 -ts-d2w2

Figure 93: Top: Schematic diagram for rotational Example 25.1. Middle: BG at the end of Step 5b (see p. 116).Bottom: Complete BG.

The middle section of Fig. 93 shows the bond graph for the mechanical system at the completion of the modelingprocedure that is specific to mechanical systems. The bottom section shows the completed bond graph.


x =[ω1 τs ω2

]⊤.

Application of the element defining equations at each energy storage device yields the following ODE’s:

At I : J1ω1 = −d1ω1 + τs + τ

At C :1

kτs = ω2 − ω1

At C : J2ω2 = −τs − d2ω2

Therefore, the state-space model can be written in matrix form as

x =

− d1

J1

1J1

0

−k 0 k

0 − 1J2

− d2

J2

x+

1J1

00

τ

y =[

0 1k 0

]x.

△


25 BOND GRAPH MODELING – ROTATION EXAMPLES: LECTURE 27-28 February 10, 2014

Example 25.2 — The schematic in Fig. 94 shows the ECP torsional system which contains two rotational inertiasmounted on a shaft such that there is no translational motion, but each inertial can rotate around the vertical axis.The twist in the shaft between J1 and the frame is modeled as a spring. The twist of the shaft between J1 and J2is also modeled as a spring. A motor attached at J2 cn apply a torque τ . Finally, there is viscous friction betweeneach disk and the frame at the points of support.

The output is the angle of each disk: y(t) = [θ1(t), θ2(t)]⊤.

w2

J1

J2

b

d

k

w1

t

0

+T

h

11

10

0

0

I:J1

Sf:0

C:1/k R:d

12

0

I:J2

R:bSe:t

C:1/h

1

w2

w1

w2w2

w2-w1

th

1I:J1

C:1/k

R:d

1

0

I:J2

R:b

Se:t

C:1/h

w1

w2

bw2

t

th

th

t-th-bw2

tkth-tk-dw1

dw1 w1

w1

Figure 94: Top left: Schematic diagram for ECP torsional plant in Example 25.2. Top right: BG at the end of Step5b (see p. 116). Bottom: Complete BG.

The top right section of Fig. 94 shows the bond graph for the ECP system at the completion of the modelingprocedure that is specific to mechanical systems. The bottom section shows the completed bond graph.


x =[ω1 τk τh ω2

]⊤.

Application of the element defining equations at each energy storage device yields the following ODE’s:

At I : J1 J1ω1 = −dω1 − τk + τh

At C :1

k

1

kτs = ω1

At C :1

h

1

hω2 = ω1 − ω2

At I : J2 J2ω2 = −τh − bω2 + τ

Therefore, the state-space model can be written in matrix form as

x =

− d

J1− 1

J1

1J1

0

k 0 0 0h 0 0 −h0 0 − 1

J2− b

J2

x+

0001J2

τ

y =

[0 1

k 0 00 1

k1h 0

]x.

△



25.1 Exercises

Exercise 25.1 – Find the state space model for the sys-tem in Fig. 95.

J1 d

w1

t

0

h

Figure 95: Rotational schematic for Exercise 25.1.

1. As depicted, with a source of torque applied at theleft side.

2. With the source of effort replaced by a source ofangular rate (i.e., flow).

Exercise 25.2 – Find the state space model for the sys-tem in Fig. 96.

w2

J1b d

k

w1

t

0

h


1. As depicted, with a source of torque applied at theleft side.

2. With the source of effort replaced by a source ofangular rate (i.e., flow).

In this exercise, there is no inertia at the point markedwith angular rate ω2.

Exercise 25.3 – Fig. 97 shows a clutch plate systemwhere friction between J1 and J2 is used to transmitpower (and angular motion). The motion only involvesrotation, no translation. Find the state space model forthe system.

J2

d

w2

th

J1

w1

J3w3



26 MECHANICAL COUPLING – SLIP OR NO SLIP: LECTURE 29 February 10, 2014

26 Mechanical Coupling – Slip or No slip: Lecture 29

There are two ways to model the interconnections between mechanical systems. These are referred to as slip and noslip.

Slip: In this modeling scenario, two surfaces move relative to each other each with different velocities at the pointof contact. The different relative velocities result in a friction force that is applied equal on both surfaces inopposite directions.

No slip: In this modeling scenario, a constraint force ensures the constraint that the relative velocity at the pointof contact is always zero. This contact force is applied equally, in opposite directions, to the two surfaces.

The no slip modeling approach is applicable to chains and gears. It is also sometimes used in situations where thefriction is large enough to motivate the no-slip assumption. Consider the following two examples to motivate themodeling differences.

Example 26.1 — Consider the schematic in the top portion of Fig. 98, wherein a (nonstretchable) belt movesrelative to rotational inertia J with radius r. The angular rate of the disk is ω, so the velocity of a point on the edgeof the disk has speed vc = Rω. The figure also shows two bond graphs, one for a source of flow and one for a sourceof effort. Note that the transformer separates the translational energy domain on the left from the rotational energydomain on the right.

Jd

wr v

xx

xx

x

TF:r I:J

R:d

Sf:v

v0

wrw

Fd v-rw

Fd=d(v-rw)

Fd

Fd

rFd

Input: Source of flow

TF:r I:J

R:d

Se:F 0

wrw

F vc

vc=F/d

F F rF

Input: Source of effort

vc-rw

Figure 98: Schematic for Example 26.1.

First consider the case where the input is a source of velocity. The state space model is

Jω = rd(v − rω).

The corresponding transfer function isω(s)

V (s)=

rd

Js+ dr2.

Next consider the case where the input is a source of force. The state space model is

Jω = rF.


F (s)=

r

Js.

△


February 10, 2014

Example 26.2 — Consider the schematic in the left portion of Fig. 99, wherein a (nonstretchable) belt movesrelative to rotational inertia J with radius r. Due to the no-slip constrain, v = rω. The figure also shows two bondgraphs, one for a source of flow and one for a source of effort. Note that the transformer separates the translationalenergy domain on the left from the rotational energy domain on the right.

J

w

r v

xx

xx

x

TF:r I:JSf:v

vNo slip v/r(J/r)v

.

(J/r2)v

.

Input: Source of flow

TF:r I:JSe:F

w

F rF

Input: Source of effort

rw

Figure 99: Schematic for Example 26.1.

First consider the case where the input is a source of velocity. In the corresponding bond graph, the I-elementhas derivative causality. Because the number of energy storage devices with integral causality is zero, there are nostates. The corresponding transfer function is

ω(s)

V (s)=

1

r.

Next consider the case where the input is a source of force. The state space model is

Jω = rF.


F (s)=

r

Js.

△

Comparison of Examples 26.1 and 26.2 shows that the type of input and choice of assumptions affects the resultingmodel. The analyst is responsible for ensuring that the assumptions made are applicable to the problem that is ofinterest.

The no-slip assumption can simplify analysis by reducing the order of the system. It also may result in modelshaving derivative causality. The derivative causality can be removed by altering the model assumptions. For example,accounting for stretch of the belt or friction between the disk and belt.

Example 26.3 — Fig. 100(a) shows the schematic of a mechanical system with two masses and on rotationalinertia (pulley). In this example, we consider the case where the cable is a massless and non-stretchable chain. Thepulley has teeth that prevent the cable from slipping over the pulley. The simplified bond graph is shown in Fig.100(b).

J

No slipw

r

vcx

x

xx

x

x x x x

F

m2g

m2

m1

v2

v1

b

(a) Schematic.

I:m2

1 I:JSe:F-m2g

FTF:r

v1v1 v1/rJv1/r

.

Jv1/r2

.

I:m1

m1v1 v1

.

F1 = -F+m2g-m1v1-Jv1/r2-bv1

..

v1

bv1

R:b

F1 v1

(b) Reduced bond graph.

Figure 100: Schematic and bond graph for Example 26.3.

Due to the assumptions that the cable cannot stretch and the cable does not slip, two of the inertias havederivative causality. The state space model is

(m1 +m2 +

Jr2

)v1 = −bv1 − F +m2g.

△


26 MECHANICAL COUPLING – SLIP OR NO SLIP: LECTURE 29 February 10, 2014

26.1 Exercises

Exercise 26.1 – Reconsider the mechanical system in Example 26.3. Derive the state space model in matrix formunder the following alternative assumptions.

1. The cable cannot stretch, but the cable can slide over the pulley.

2. The cable can stretch, the point vc cannot slide relative to the pulley.

3. The cable can stretch and can slip over the pulley.

In both cases 2 and 3, you should consider the cable as having two distinct pieces. The portion between v1 and vcand the portion between vc and v2. Each can stretch separately.

Exercise 26.2 – The schematic in Fig. 101 shows two pulleys connected by a wrap-around belt. Develop the bond

k

k

R1

R2J1

J2

w1

t1

w2

t2

Figure 101: Pulley schematic for Example 26.2.

graph and state space model for each of the following scenarios.

1. Consider the belt to be massless, non-stretching, and non-slipping.

2. Consider the belt to be massless and non-slipping, but being stretchable. The coefficient k is the same at bothindicated locations, but the amount of stretch will be different.

3. Consider the belt to have mass, to be capable of stretching, and to be stretchable.


February 10, 2014

27 Simple Mixed Systems: Lecture 29

27.1 Wheels

This section considers a few models of wheels. Significantly more involved nonlinear models exist.Fig. 102 shows the schematic and bond graph for a hard wheel that is assumed to not slide at the point of contact

with the surface. This example would apply to applications such as a cog railroad. The state space model is(m+

J

r2

)vc = F.

JNo slip

w

r

vc F

0

1 I:JSe:F

FTF:r

I:m

vc

vc vc vc/r

.

vcJ/r

.

F-vcJ/r2

.

vcJ/r2

Figure 102: Schematic and bond graph for a hard wheel on a non-slip surface.

Fig. 103 shows the schematic and bond graph for a hard wheel for which the point of contact can slide relative tothe surface. This relative velocity is a more realistic situation than no slip for most wheels. The state space model is(

m+J

r2

)vc = F − bvc.

Jb

w

r

vcF

01 I:JS

e:F

FTF:r

I:m

vc

vc vc vc/r

.

F-vcJ/r2-bvc

R:b

bvc vc

.vcJ/rvcJ/r2

Figure 103: Schematic and bond graph for a hard wheel on a surface allowing slip.

Fig. 104 shows the schematic and bond graph for a rubber wheel for which, at the the point-of-contact with theroad, the tire can stretch and slide relative to the road surface. This stretching is modeled by a spring and damper,as some energy is lost in the stretching process. The state space model now has three states x = [vc, Fk, ω]⊤:

x =

− b+dm

1m

rdm

k 0 −kr

− rdJ

rJ

r2dJ

x+

1m00

F

Jb

w

r

vc F

0 1 I:JSe:F

Fk

TF:r

I:m

vc vc

vs=vc-rw

R:b

bvc vc

0

1 R:dC:1/k

wrw

vs vs

dvs

Fk-dvs

F Fk-dvs Fk-dvs r(Fk-dvs)

F+Fk-dvs-bvc vc

Figure 104: Schematic and bond graph for a rubber wheel on a surface allowing slip.


27 SIMPLE MIXED SYSTEMS: LECTURE 29 February 10, 2014

27.2 Electric Water Pump

Modeling of the electrical pump depicted by the schematic in Fig. 105 requires a model coupling the electrical,rotational, and fluid energy domains. Figure 106 shows the simplified bond graph labeled with the physical variables

im+_vs

+_

Lma

g

em+_

J

wm

tm

Fluid tankInfinite Fluid

Resevoir

Rm

Pu

mp

PT

valve

Q1

Qp

Pp

Figure 105: Liquid pump schematic.

from the schematic. Candidate state variables are labeled at the energy storage device that would determine theirvalue using integral causality. We should notice that the I : Jp and I : Lp elements are not independent, which willresult in one of them having derivative causality. The cause of this issue is that, as modeled, the rate of angularrotation determines the volume flow rate. The derivative causality could be remove by various means, including:modeling the flex of the vanes internal to the pump, modeling leakage past the vanes in the pump, or neglecting oneof the two inertias. In this case, we choose to work the model through with the derivative causality. In doing this,we should clearly note our assumptions.

Assumption 27.1 – The fluid leakage past the pump vanes is negligible.

Assumption 27.2 – The flex of the pump vanes is negligible.

The other schematic variables em, τm, Pp, and Q1 that are not candidate state variables, but are computable fromthe input and state variables, are labeled at the appropriate locations.

1se:vs0GY

I:LpC:Cf

I:Lm

R:Rv:

K

R:Rp

1

R:Rm I:Jp

1

R:Dp

TF:

N Qp PT

Q1wm

em

imvs tm

Pp

Figure 106: Liquid pump bond graph labeled with physical variables.


27.2 Electric Water Pump February 10, 2014

Choosing the state vector to be x = [im, Qp, Pt]⊤, Fig. 107 shows the bond graph after walking the causality and

state variables up to the element that determines their value.

1se:vs0GY

I:LpC:Cf

I:Lm

R:Rv:

K

R:Rp

1

R:Rm I:Jp

1

R:Dp

TF:

N Qp PT

Q1

imvs

Rmim im

imim

Kim PTPT

Qp

RpQp Qp

Q1=PT/Rv

Qpwm

wm=NQpPp =Ntp

wm

wm

wmem

em=Kwm

Pptp

Dpwm

Jpwm.

tp=Kim-Dpwm-Jpwm.

Figure 107: Liquid Pump bond graph with state variables propagated through.

The resulting state space model is

Lmx1 = vs −Rmx1 −KNx2 (50)

Leqx2 = NKx1 −(N2Dp +Rp

)x2 − x3 (51)

Cf x3 = x2 −1

Rvx3, (52)

where Leq =(Lp +N2Jp

). The variables labeled on the schematic can be computed as model outputs:

ωm = Nx2 (53)

em = KNx2 (54)

τm = Kx1 (55)

Q1 =1

Rvx3 (56)

Pp = Nτp (57)

and τp = K(1− JpN

Leq

)x1 +

(JpN(N2Dp+Rp)

Leq−DpN

)x2 +

(JpNLeq

)x3.


27 SIMPLE MIXED SYSTEMS: LECTURE 29 February 10, 2014

27.3 Exercises

Exercise 27.1 –

1. Find the transfer function corresponding to Figs. 102 and 103.

2. In Figs. 102, repeat the bond graph analysis in terms of ω. This will have I : J in integral causality and I : min derivative causality. Find the state space model and transfer function. Compare with Step 1.

3. In Figs. 103, repeat the bond graph analysis in terms of ω. This will have I : J in integral causality and I : min derivative causality. Find the state space model and transfer function. Compare with Step 1.

Exercise 27.2 – Derive a bond graph and state space model for the electric water pump system discussed in Section27.2 for the case where the motor inductance and pipe inertance are sufficiently small that they can be neglected.

Exercise 27.3 – Fig. 108 shows the schematic of a hydraulic dam DC electric power generation system. The netinput flow is regulated so that the height of the fluid in the reservoir is constant. Therefore, you can assume thatthe fluid reservoir is a constant pressure source.

1. Find the state space model. The generator should be modeled in a manner similar to the approach used forthe electric pump in Section 27.2.

2. How would the model change if the flow Qi was considered as the input and the reservoir was modeled as anenergy storage device?

Lg

Fluid reservoir

RL

PT

Q1Gen.

Qi

eg

ig

+ -

Figure 108: DC Voltage Generator.

Exercise 27.4 – Fig. 109 depicts a very simple mechanical schematic for a front wheel drive car. The chassis movesparallel to the ground with velocity v(t). The front and back wheel have distinct rotational rates. Each wheel hasradius R, rotational inertia as indicated, and can slip relative to the fixed ground surface. Consider the attachmentof the axles to the chassis to be rigid. The input is an applied torque at the shaft of the front wheel. The output isthe velocity v(t) of the chassis.

1. Write the algebraic relation for the slip (relative velocity between the road and the bottom of the wheel) ateach wheel.

2. Draw the simplified bond graph for the system. See Exercise 13.4 on p. 62.





M

Jb Jf

wb(t) wf(t)v(t)

t(t)

Figure 109: Mechanical schematic diagram for translational Exercise 27.4.


28 STATE SPACE TO CONVOLUTION: INTERESTING LECTURE February 10, 2014

28 State Space to Convolution: Interesting Lecture

The purpose of this section is to derive the convolution integral from the state-space model. Most texts just give theconvolution integral without explanation. Many students have found the following explanation to be enlightening.

Starting from the state-space equation for the time derivative of x,

x(t) = Ax(t) +Bu(t) (58)

x(t)−Ax(t) = Bu(t) (59)

e−Atx(t)− e−AtAx(t) = e−AtBu(t) (60)

d

dt

(e−Atx(t)

)= e−AtBu(t) (61)∫ t

0

d

dτ

(e−Aτx(τ)

)dτ =

∫ t

0

e−AτBu(τ)dτ (62)

(e−Aτx(τ)

)∣∣t0

=

∫ t

0


e−Atx(t)− Ix(0) =

∫ t

0


x(t) = eAtx(0) + eAt

∫ t

0


x(t) = eAtx(0) +

∫ t

0

eA(t−τ)Bu(τ)dτ (66)

Before going further, the reader should ensure that they know exactly what happened in the transition from one lineto the next. If you do not, discuss with your peers or ask in class.

Starting from the state-space equation for the output, assuming that D = 0,

y(t) = Cx(t) (67)

Substituting eqn. (66) to replace x(t), yields

y(t) = CeAtx(0) +

∫ t

0

CeA(t−τ)Bu(τ)dτ (68)

The first term CeAtx(0) is the output response due to the initial condition.

The second term∫ t

0CeA(t−τ)Bu(τ)dτ is the forced response due to the input u(t). Note that if we define

h(t) = CeAtB, then

y(t) = CeAtx(0) +

∫ t

0

h(t− τ)u(τ)dτ. (69)

When the initial conditions are zero (i.e., x(0) = 0),

y(t) =

∫ t

0

h(t− τ)u(τ)dτ (70)

which is the convolution integral.It is important to realize that a practicing engineer rarely, if ever, actually computes a convolution integral.

Instead, the practicing engineer uses important ideas related to the convolution integral:

� The system response is determined in part by the initial conditions and in part by the applied inputs.

� The nature of the response is determined by the poles of the transfer function, which are identical to theeigenvalues of the A matrix. The transfer function zeros also affect the response.

� The designer can choose the system parameters, redesign the system, or use feedback control to determine theeigenvalues of A.

� When the eigenvalues of A are in the strict left-half-plane (real part less than zero), the system is stable.


February 10, 2014

29 System Design and Analysis: Lecture 30

Once we have derived the model of a dynamic system, we are interested in analyzing the system so that we canselect the system parameters to achieve some specified level of performance. The objective of this and the next fewlectures is to present a few system design and analysis concepts that are indispensable to system design engineers.

As engineers, there are a few different types of specifications: stability, steady state, and transient.

29.1 Stability Specifications

Almost always, the system is desired to be stable. The physical concept of stability and its different clarificationsis exemplified by a ball rolling on a surface subject to gravity, as depicted in Fig. 110. The figure shows threeequilibrium points labelled by A, B, and C. Remember that an equilibrium point is a position in state space wherethe state derivative is zero. The equilibria indicated by B and C are each unstable, since a small disturbance from the

AC

B

Figure 110: Physical depiction of the concept of stability.

initial condition can cause the system to diverge from that equilibrium. The equilibrium indicated as A is (locally)stable because, assuming that there is friction, if the ball starts near or is slightly disturbed from this equilibrium,it converges back to the equilibrium.

For more complicated physical systems it may not be straightforward to determine whether the system is stableby direct inspection of the schematic. In fact, we may be interested in determining the range of parameter valuesfor which a system is stable. In such situations, we will determine the stability properties of the system by analysisof the model of the system. For linear systems, the main condition of the system to be stable is that the poles ofthe transfer function, or equivalently the eigenvalues of the A matrix, must be strictly in the left half-plane. In allproblems in this class, we will consider stability to be a required specification.

Example 29.1 The transfer functionY (s)

U(s)=

Gω2n

s2 + 2ζωns+ ω2n

. (71)

has two complex poles located at s = −ζωn±ωn

√ζ2 − 1. Therefore, the system is stable as long as ζ > 0 and ωn > 0.

Example 29.2 A third order system has poles at −p1, −p2, −p3. Therefore, the system characteristic equationmust be

(s+ p1)(s+ p2)(s+ p3) = 0

s3 + (p1 + p2 + p3)s2 + (p1p2 + p2p3 + p1p3)s+ p1p2p3 = 0.

For the system to be stable, the poles must be in the left half plane, which requires that p1, p2, and p3 each be positive.Therefore, if the system is stable, then all the coefficients of the characteristic equation will be positive. If anycoefficient is negative, then the system is not stable. However, the converse is not true; there are (many) examples ofunstable systems where all coefficients of the characteristic equation are positive. Consider s3+0.1s2+0.2s+0.2 = 0.


29 SYSTEM DESIGN AND ANALYSIS: LECTURE 30 February 10, 2014

29.2 Steady State & Transient Specifications

When we model systems, we can derive transfer functions from the input to the output. The typical procedure woulduse the following steps:

1. Draw and label the schematic.

2. Draw and simplify the bond graph.

3. Walk the causality and state variables through the bond graph, then write down the resulting state spacemodel.

4. Use the state space model to derive the transfer function.

From the design perspective, all of the above is performed symbolically, since numeric values for the system specificparameters are not yet known. The job of the designer is to analyze the model and select appropriate values for thesystem specific values to achieve a performance specification.

Often the models written in terms of their system specific parameters appear very different when, in fact, thestructures are very similar. Therefore, it is useful to think in terms of standard forms of models. The following twosubsections will introduce the standard forms for first and second order systems.

It is important to note that an n-th order system can always be factored into products of first and second ordertransfer functions. Therefore, knowledge of the behavior of first and second order systems is a first prerequisite foreffective system design.

29.2.1 First Order Systems

0 1 1.5 2 2.5 3 3.5 4

0.05

0.14

0.37

1

t/t

exp(-t/t)

Figure 111: Typical first order system ini-tial condition response with illustration oftime constant.

The standard form for a stable, first order system transfer functionfrom input u to output y is

Y (s)

U(s)= H(s) =

G a

s+ a

which has DC gain limt→0 H(s) = G and a pole at s = −a. The timeconstant of the system is τ = 1

|a| .

For the system to be stable, the pole must be in the left half plane,which requires a > 0.

The unforced (u(t) = 0) or initial condition response is

y(t) = Ge−tτ x(0) for t ≥ 0. (72)

For a stable system, the final value of y for the unforced case is zero.Figure 111 contains an example of a first order initial condition responsewith Gx(0) = 1. Note the importance of the time constant in this plot.There are two graphical means to determine the time constant fromthe graph of the initial condition response of a first order system.

1. In one time constant, y(t) will change 63% of the distance from its initial to its final value. Therefore, theanalyst can simply find the time at which y(τ) = 0.63y(0).

2. Corresponding to eqn. (72), y(t) = −Gτ e

− tτ x(0). The line tangent to the graph of y(t) at t = 0 is

z(t) = y(0)− y(0)

τt

which is also sketched in Fig. 111. Note that z(t) = 0 at t = τ . Therefore, the time constant can be foundgraphically by drawing a line tangent to the initial condition response and measuring the time between thepoint of tangency and the intercept with the time axis.

The step response (u(t) = 1(t)) is

y(t) = Ge−tτ x(0) +G

(1− e−

tτ

)for t ≥ 0.


29.2 Steady State & Transient Specifications February 10, 2014

29.2.2 Second Order Systems

Following are a few examples of second order systems. For each example, students should be able to fill in the stepsthat have been skipped in the modeling procedure defined at the beginning of this section.

Example 29.3 Fig. 112 shows a system with state space model

x =

[0 k

− 1m − d

m

]x+

[01m

]u

p =[

1k 0

]x

where x = [Fk, v]⊤. The corresponding transfer function is

P (s)

F (s)=

1k

km

s2 + dms+ k

m

. (73)

An engineer designing this system would need to select values for the system specific parameters k, m, and d to meet

C:k 1 Se:F

I:M

R:d

mp p.. .

p.

p.

p.

Fk

Fd

FM

k

d F

p

Figure 112: Schematic and bond graph for a simple translational system.

a given performance specification.


x =

[− 1

R1C− 1

C

− 1L −R2

L

]x+

[1

R1C

0

]u

y =[0 R2

]x

where x = [vc, iL]⊤. The corresponding transfer function is

Y (s)

U(s)=

(R2

R1+R2

)(R1+R2

R1LC

)s2 +

(1

R1C+ R2

L

)s+ R1+R2

R1LC

. (74)

An engineer designing this system would need to select values for the system specific parameters R1, R2, L and C

C

R1 L

R2+_

+_y

+_u Se:u 1 0 1i1 i1 iL

C R2

I:LR1

u vc vc

vc-iLR2 iL

iLR2 iLvc i1-iL

u-vc i1

i1=(u-vc)/R1

Figure 113: Schematic and bond graph for a simple circuit.

to meet a given performance specification.


x =

[−R

L − pL

− pJ 0

]x+

[1L0

]u

y =[0 1

]x



where x = [i, ω]⊤. The corresponding transfer function is

ω(s)

U(s)=

(1p

) (pL

)2s2 +

(RL

)s+

(pL

)2 . (75)

An engineer designing this system would need to select values for the system specific parameters p, R, and L tomeet a given performance specification.

Se:u 1 GY I:Ji i w

I:L

R1

u pw ip

u-Ri-pw i

R i i

R1 L

+_+_u

+_e

i

J

t,w t=ipe=pw

Figure 114: Schematic and bond graph for a mixed electrical and rotational system.

Examples 29.3–29.5 each derived the state space and transfer function models for a second order system. Thetransfer function models in eqns. (73–75) appear very different because they are each written in terms of their systemspecific parameters. However, the structures are in fact very similar. The objective of the remainder of this sectionis to define the standard form of a second order system model and motivate its utility.

The standard form for the transfer function of a second order system with input u and output y is

Y (s)

U(s)=

Gω2n

s2 + 2ζωns+ ω2n

. (76)

For ζ ∈ [0, 1) the system has two complex conjugate poles located at s = −σ ± jωd where

σ = ζωn (77)

ωd = ωn

√1− ζ2. (78)

The parameters of the standard model have the following physical meanings:G DC Gain,ωn undamped natural frequency in radians per second,ωd damped natural frequency in radians per second,ζ damping factor,σ decay rate.

The definitions of the standard model parameters in terms of system specific parameters can be read off by comparingeqn. (76) with the transfer function of the system of interest. Following are three examples of the definition of thestandard model parameters in terms of specific system parameters. The reason for doing this is that we can directlyrelate the standard model parameters to time domain performance specifications.

Example 29.6 For the system of Example 29.3 with transfer function

P (s)

F (s)=

1k

km

s2 + dms+ k

m

(79)

the system specific parameters are related to those of the standard second order transfer function as follows:

ωn =

√k

mζ =

d

2√mk

σ =d

2mG =

1

k

where computation of ωd is left as an exercise.




Y (s)

U(s)=

R2

R1+R2

R1+R2

R1LC

s2 +(

1R1C

+ R2

L

)s+ R1+R2

R1LC

(80)


σ =1

2

(1

R1C+

R2

L

)ωn =

√R1+R2

R1LC G =R2

R1 +R2

where computation of ζ and ωd is left as an exercise.


ω(s)

U(s)=

(1p

) (pL

)2s2 +

(RL

)s+

(pL

)2 (81)


ωn =p

Lζ =

R

2p

σ =1

2

(R

L

)G =

1

p

where computation ωd is left as an exercise.

0

0.1

0.9

1

Mp

Tr

Tp

Ts

1%

Figure 115: Typical second order system step response with definitions of parameters for time domain specs.

The step response of a stable, low pass, unity DC gain, second order system with complex poles is

y(t) = G

[1− e−σt

(cos(ωdt) +

σ

ωdsin(ωdt)

)]for t ≥ 0.

Students can derive this result for themselves by use of partial fractions. A typical response is shown in Figure 115along with some time domain specifications of system performance.

Tr Rise Time Time to rise from 10% to 90% of final value.Ts Settling Time Time to settle to and permanently stay within 1% of final value.Tp Peak Time Time at which the first peak occurs.Mp Overshoot Amount by which y(Tp) exceeds its final value for 0 ≤ ζ ≤ 1.

Each of these time domain performance specifications directly relates to one of the parameters of the standard secondorder low pass transfer function:



Tr ≈ 1.8ωn

Ts =4.6σ

Tp = πωd

Mp = exp

(−ζπ√1−ζ2

)for 0 ≤ ζ ≤ 1.

When performance is specified in the time domain by the shape of the step response, the designer can

1. translate the time domain specifications into constraints on ωn, ωd, σ, and Mp;

2. if possible, select pole locations that satisfy these constraints;

3. define the corresponding characteristic equations; and

4. if possible, select the system specific parameters to achieve this characteristic equation.

Note that two of the steps have been introduced with the phrase ”if possible.” It is possible either for a set of timedomain specifications to be impossible or for a desired transfer function to be unattainable by a given physical system.When the designer encounters either of these situations, the designer must either renegotiate the specification orredesign the physical system.

Following are a few examples that transform time domain specifications into desired transfer functions.

Example 29.9 If a system is specified to be stable, with unity DC gain, and the transient specification is Tp = π sand Tr = 0.9s, then

ωn ≈ 1.8

Tr= 2.0

rad

sand ωd =

π

Tp= 1.0

rad

s.

Therefore, using eqn. (78), we have

1 = 2√1− ζ2

0.25 = 1− ζ2

ζ =√0.75 = 0.866

and σ = ζωn = (0.866)(2) = 1.73. Therefore, the desired pole locations are s = −1.73±j1.0. This yields the followingdesired transfer function

4

s2 + 3.46s+ 4.

Based on a damping factor of ζ = 0.866 we expect the overshoot to be almost imperceptible at Mp = 0.004(0.4%).Students are encouraged to simulate the step response of the system to verify that this transfer function does achievethe specification.

Example 29.10 If a system is specified to be stable, with a DC gain of 5.0, and the transient specification is Tp ≈ 0.03s with a 5% overshoot, then

ωd =π

0.03= 105

rad

s.

Solving for ζ with Mp = 0.05 yields ζ = 0.69. We can now compute

ωn =ωd√1− ζ2

= 145

σ = ζωn = 100

Therefore, the desired pole locations are s = −100± j105. This yields the following desired transfer function

104690

s2 + 200s+ 20940.

Students are encouraged to simulate the step response of the system to verify that this transfer function does achievethe specification.

Often, the specifications are given in terms of inequalities instead of equalities. The solution method is similaras shown in the following examples.



Example 29.11 If a system is specified to be stable, with a DC gain of 0.5, and the transient specifications 1 <Ts < 5, Tp ≈ 1.0 s and overshoot less than 5%, then we will have one equality and two inequality constraints. First,

Tp ≈ 1.0

ωd =π

1.0= 3.14.

Second,

Mp < 0.05

ζ > 0.69.

Finally,

1 ≤ Ts ≤ 5

1 ≤ 4.6σ ≤ 5

σ ≤ 4.6 ≤ 5σ

−σ ≥ −4.6 ≥ −5σ

which states that the real part of the pole location Re(p) = −σ must satisfy Re(p) ∈ [−4.60,−0.92].

Re

Im

s=-4.6 s=-1.0

s= 3.14j

s=-3.14j

Figure 116: Constraints on the pole locations.

To keep track of the constraints, it is often convenient to sketch the regions in the complex plane and then selectthe poles from the specified region. In Fig. 116, the complex conjugate pair of poles must be on the lines s = ±jπ tosatisfy the peak time constraint, must be to the left of dashed dotted lines to satisfy the overshoot constraint, and thereal part of the pole must be between the two dotted lines to satisfy the settling time constraints. The poles s = −4±jπsatisfy all three constraints. Therefore, a transfer function satisfying all three constraints is

12.94

s2 + 8.00s+ 25.87.

Each of the above examples has specified a transfer function that achieves the time domain specification. Thestudent should go back to the transfer functions in Examples 29.3–29.5 and, when possible, select the system specificparameters to achieve the specified transfer function.

29.2.3 Higher Order Systems

11

11Not Finished



29.3 Exercises

Exercise 29.1 – For the state space system

x =

[0 1

−a −b

]x+

[01

]u

y =[1 0

]x

1. Find the transfer function from U(s) to Y (s).

2. Express each of the following as a function of the model parameters (a and b): DC gain, ωn, ωd, ζ, and σ.

Exercise 29.2 – For the electric water pump system discussed in Section 27.2 on p. 129, for the case where themotor inductance and pipe inertance are sufficiently small that they can be neglected, and the valve Rv is closed:

1. Derive the state space model with input vs and output PT .

2. Find the transfer function for the system.

3. Find expressions for ωn, σ, ωd, and ζ as a function of the physical parameters.

4. Find an expression for the steady-state output when the input voltage is held constant.


February 10, 2014

30 Extra Lecture - Diagram Models (w/ Simulink)

There are also various types of visual signal-flow models:

Schematic Diagram: An illustration of the physical construction of a system.

� Useful for definition of sign convention.

� Connections between subsystems transport power through effort and flow variables.

– Each connection forces equality of one effort and one flow

– One subsystem determines the effort. The other determines the flow.

� The power flow is bidirectional. Discuss power flow in figure.

M

k

d F

x

0 2 4 6 8 10−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Pow

er, W

atts

Time, t, s

F is 1 N for t∈ [0,5]

Pk

Pd

PM

PF

Figure 117: Schematic diagram of a mass, spring, damper system.

Block diagram: A logical illustration of the decomposition of a system into subsystems, each of which is representedby a block containing a mathematical expression relating the block outputs to the block inputs.

1Ms2+ds+k

F(s) X(s)

Figure 118: Block diagram for a mass, spring, damper system.

� Arrows indicate how the system inputs and block outputs affect inputs to other blocks.

� The arrows represent information flow, not physical flow.

� The information flow is unidirectional.

� Block diagram to state space is straightforward.

� Block (subsystem) models must be determined by other means.

� Useful for finding model of a system composed of several subsystems.

� The analyst must be careful to consider loading effects.

Fig. 119 shows two circuit schematics to illustrate the loading issue. The triangle in the circuit on theright represents a unity gain voltage amplifier with very high input impedance. Before either circuit isconnected, the transfer function from u1 to y1 is G1(s) = 1

R+sC and the transfer function from u2 to

y2 is G2(s) = 1sL . When connections are made at the terminals indicated by the black dots such that

y1(t) = u2(t), then the circuit on the left has the transfer function

GL(s) =1

RLCs2 + Ls+R= G1(s)G2(s)


30 EXTRA LECTURE - DIAGRAM MODELS (W/ SIMULINK) February 10, 2014

while the circuit on the right has the transfer function

GR(s) = G1(s)G2(s) =1

(R+ sC)sL.

The transfer function for the circuit on the left is not the product of G1 and G2 because the derivation ofG1 assumes that the current i1 is zero. This assumption is false once the two sub-circuits on the left areconnected. The high input impedance of the amplifier in the circuit on the right maintains the correctnessof this assumption even after the sub-circuits are connected.

+_u1 C y1 u2 L

R+

-

+

- iL=y2

+_u1 C y1 u2 L

R+

-

+

- iL=y2

vc+

-icic

iRiR

i1

Figure 119: Loading example.

Signal Flow: A special form of block diagram composed only of basic components: integrators, sources, summers,gains, delays, etc.

� This is a model representation, but model equations must still be found by other methods.

� Signal flow to state space is easy.

� The variables output by the integrator blocks define a valid state vector.

� Illustrate signal flow diagram for a state space model.

� Simulink (part of Matlab) is a signal flow or block diagram simulation language.

1s

1s+

-d

-k

F(t) x(t) x(t) x(t).. .

Figure 120: Signal flow diagram for a mass, spring, damper system.


February 10, 2014

Bond graph: A diagram showing the power flow between the subsystems that compose a system.

� Components: R, L, C, 1, Se. The parameter is defined after the colon.

� Port: Defined by each unique pair of effort and flow variables.

� Bonds: Connect the ports of different components

– represent the flow of power

– labeled with effort and flow defining power flow

� Half arrow: Defines sign convention for assumed positive power flow.

� Causal stroke: Defines which component connected to the bond is determining (outputting) the flowvariable.

C:k 1 Se:F

I:M

R:d

mx x.. .

x.

x.

x.

Fk

Fd

F

Figure 121: Bond Graph for a mass, spring, damper system.

Our main interest will be on the construction of state space models using bond graphs as a means to unify ideasfrom various energy domains. We will used schematics to define variables and sign conventions. We will use blockdiagrams in Simulink for simulation implementation.

examples


REFERENCES February 10, 2014

References

[1] Wolfgang Borutzky, “Bond Graph Methodology: Development and Analysis of Multidisciplinary Dynamic Sys-tem Models,” Springer, 2010. (Note: This text is available online as a downloadable pdf. However, the notationthat it uses is slightly different than the standard notation that I try to follow in class.)

[2] Dean C. Karnopp, Donald L. Margolis, Ronald C. Rosenberg, “System Dynamics: Modeling and Simulation ofMechatronic Systems,” John Wiley, 2006. (Note: This text establishes the standard notation that I try to followin class.)

[3] L. Ljung, T. Glad, “Modeling of Dynamic Systems,” Prentice-Hall, 1994. (Note: The notation that this bookuses is slightly different than the standard notation that I try to follow in class.)

[4] N. H. McClamroch, “State models of dynamic systems : a case study approach”, Springer-Verlag, c1980.

[5] Shearer, Kulakowski, and Gardner, “Dynamic Modeling and Control of Engineering Systems”

[6] R. L. Woods and K. L. Lawrence, “Modeling and Simulation of Dynamic Systems”

[7] Vu and Esfandiara, “Dynamic Systems: Modeling and Analysis”

[8] Close, Frederick, and Newell, “Modeling and Analysis of Dynamic Systems,”John Wiley and Sons


February 10, 2014

A Vector & Matrix Review

This is a review of basic ideas from linear algebra. If you find types or errors, suggestions for improvements, or wouldlike to see additional items reviewed, please let me know.

A.1 Definitions

A matrix A ∈ ℜn×m is a tabular arrangement of real numbers with n rows and m columns. An example for n = 2and m = 3 is

A =

[0.20 0.00 −3.14

−1.71 0.00 2.00

].

Sometimes it is necessary to name the elements of a matrix individually. In this case, we state that A = [aij ] fori = 1, . . . , n and j = 1, . . . ,m. For example in the above example, a2,3 = 2.00.

We will treat a vector as a special matrix that only has one column. For example,

v =

0.002.00

−7.14

is an example of a vector in ℜ3. When we want to refer to the component in the i-th row of the vector v, we use thenotation vi instead of vi1.

A.2 Matrix and Vector Operations

There are several operations that facilitate system modeling.The transpose operation, which is indicated by a subscript ⊤, simply exchanges the rows and columns of a

matrix. Therefore, for B = [bij ] ∈ ℜn×m, B⊤ = [bji] ∈ ℜm×n. Two examples may clarify this operation. First,

A⊤ =

0.20 −1.710.00 0.00

−3.14 2.00

is a matrix in ℜ3×2. Second, v⊤ =

[0.00 2.00 −7.14

]is a matrix in ℜ1×3 which is often referred to as a row

vector.The addition of matrices is straightforward and is defined componentwise. When we say that C = A+B for

A, B, C ∈ ℜn×m we mean that cij = aij + bij for i ∈ [1, n] and j ∈ [1,m]. In order to add two matrices, theirdimensions must be identical. Therefore, with the matrices given previously A and v cannot be added. Subtractionof matrices is defined similarly. If we define

B =

1.00 1.710.03 2.000.14 2.14

then A and B cannot be added or subtracted, but

A⊤ +B =

1.20 0.000.03 2.003.00 6.14

.

Matrix multiplication is particularly important. When A ∈ ℜn×m and B ∈ ℜm×p, then their productC = AB ∈ ℜn×p is defined as cij =

∑mk=1 aikbkj . Note that the number of columns in A and the number of rows

in B, referred to as the inner dimension of the product, must be identical; otherwise the two matrices cannot bemultiplied. Using the vectors and matrices defined above, you should be able to show the following:

AB =

[−9.2200 −12.65764.2900 5.3559

]A⊤B = not a valid operation

Av =

[22.4196

−14.2800

]B⊤v =

[−21.3600−25.5596

]Bv = not a valid operation


A VECTOR & MATRIX REVIEW February 10, 2014

both by hand and in Matlab. Also, you should get comfortable with writing systems of linear equations in matrixnotation. Using the above definitions, you should be able to show the following equivalences:

1. A summation can be written as a matrix product:

n∑i=1

aixi = [a1, . . . , an]

x1

...xn

2. A set of linear equations can be written in matrix format. For example,

4 = 23x− 4y − 8s

8 = 42s+ 15t+ 16x

15 = 23s− 23t

16 = 42y − 8s

is equivalent to 481516

=

−8 0 23 −442 15 16 023 −23 0 0−8 0 0 42

stxy

.

Equations of this form are easily solvable in programs like Matlab using matrix algebra.

3. A set of first order ODEs can be conveniently written in matrix form. For example,

v = 23v − 4y − 8z

y = 42z + 15y + 16v

z = 42y − 8z

is equivalent to vyz

=

23 −4 −816 15 420 42 −8

vyz

.

This has the standard form x = Ax where x = [v, y, z]⊤ and

A =

23 −4 −816 15 420 42 −8

.

Please ensure that you are comfortable with the above ideas. EE105 does use basic linear algebra.In Matlab there is a special matrix operation “.*” for element-by-element multiplication defined for matrices with

the same dimensions. This means that for general matrices A,B,C ∈ ℜn×m, the operation C = A . ∗B is computedas cij = aij ∗ bij .

Using the specific matrices

A =

[0.20 0.00 −3.14

−1.71 0.00 2.00

]and B =

1.00 1.710.03 2.000.14 2.14

,

the product C = A . ∗B does not compute, due to the different dimensions. The product

C = A⊤ . ∗B ==

0.20 −1.710.00 0.00

−3.14 2.00

. ∗

1.00 1.710.03 2.000.14 2.14

=

0.20 ∗ 1.00 −1.71 ∗ 1.710.00 ∗ 0.03 0.00 ∗ 2.00

−3.14 ∗ 0.14 2.00 ∗ 2.14

.


A.3 Exercises February 10, 2014

A.3 Exercises

Exercise A.1 – By hand (no electronics), compute the following:

a) A ∗B b) B ∗Ac) A. ∗B d) B. ∗Ae) A+B f) A⊤

where A =

[1 2

−1 4

]and B =

[2 4

−1 1

].


Index

Biot number, 82Bond graph to state space model, 36Bulk modulus, 68

causality, 11Causality propagation, 36Characteristic equation, 104Continuity equation, 68Convolution integral, 133

Determinant, 98

Eigenvalues, 104Exponential, matrix, 107Exponential, scalar, 107

Heat capacity, 82

Matrix, identity, 99Matrix, inverse, 99Matrix, nonsingular, 98Matrix, Similarity, 106Matrix, singular, 98

Not finished, 106, 115, 140Not finished., 110

Power supply, 63

Response, Forced, 133Response, Initial Condition, 133

State space model, from bond graph, 36

149

Documents

EE105LectureNotes(6)