sc4026 Exercise Session 7 (with solutions) - TU Delftaabate/2012sc4026/2012sc4026_ex7_soln.pdf · sc4026 Exercise Session 7 (with solutions) Alessandro Abate [email protected] Marco

sc4026Exercise Session 7 (with solutions)

Alessandro Abate

[email protected]

Marco Forgione

[email protected]

Delft Center for Systems and Control, TU Delft

October 25, 2012

– Ac.Yr. 2012/13, 1e Sem. Q1 – Exercise Session 7 – sc4026

A. Abate, M. Forgione

Solution to state space models

Find the output response to a step input, where the system

is given by:

X = AX + BU

Y = CX +DU,

where

X(0) =

[1

0

], B =

[1

0

], C =

[1 0

], D = 0,

and where A is [−1 3

0 −4

].

Compute then the steady-state error to a step input, and verify

the correctness of the result by calculating the transfer function

of the model.

Solution:

– Ac.Yr. 2012/13, 1e Sem. Q1 – Exercise Session 7 – sc4026 1


The output is

Y (t) = CX(t) +DU(t)

= CeAtX(0) + C

∫ t

0

eA(t−τ)

BU(τ)dτ +DU(t).

It is easy to see that eAt =

[e−t e−t − e−4t

0 e−4t

].

Step input means u(t) =

{1, t > 0

0, t = 0.With reference to

the values in the matrices B,C,X(0), we obtain y(t) =

e−t + 1 − e−t = 1. Notice that the non-zero initial condition

“cancels out” the effect of the input.

Thus, the steady-state error to a step input is equal to zero.

Let us compute the relationship, in frequency, between input,initial conditions, and output:

Y (s) = (C(sI − A)−1B +D)U(s) + Ce

At(X(0)− (sI − A)

−1B).

Notice that the effect of the non-zero state X(0) vanishes since

the state matrix A is stable.

The steady-state to a state response can be thus computed

through

[1 0

] [ s+ 1 −3

0 s+ 4

]−1 [1

0

]+ 0 =

1

s+ 1,



so the steady-state is computed by setting s = 0, which gives 1.

This means that the steady-state error to a step input is zero.

2



Loss of Observability and/or Controllability

Recall a topic developed in Lecture 4. Consider system in

series:

Stabilizable, detectable 117

Controllable and observable eigenvalues can be relocated by feedback.

In some cases, it is su!cient that the unstable eigenvalues are

controllable and observable, leading to the following weaker concepts

Definition

(A, B) is stabilisable if all!!i(A)

""" Re!i(A) ! 0#

are controllable

(A, C) is detectable if all!!i(A)

""" Re!i(A) ! 0#

are observable

Note that the eigenvalues on the imaginary axis are considered as

unstable.

Controllability/observability 9: pole-zero cancellation 118

s + 3

s + 1! s + 1

s + 2! !

u(s) z(s) y(s)

Simulation diagram (time domain):

$!

$!"1!

"1 " "2 "

1! 1!

!#

! !#

!!#

! !#

!2! !u(t) z(t) y(t)"1 "2


Diagram uniquely defines (A, B, C,D):

%A B

C D

&=

'()

"1 0 2

"1 "2 "1

1 1 1

*+,

-B AB

.=

%2 "2

"1 0

&rank is 2

%C

CA

&=

%1 1

"2 "2

&rank is 1

Controllable, not completely observable.


Eigenvectors of !1 = "1, !2 = "2:

m1 =

%1

"1

&m2 =

%0

1

&

%C

CA

&m1 = 0

/ 01 2unobservable

%C

CA

&m2 #= 0

/ 01 2observable

Unobservable eigenvalue !2 = "1 is cancelled by forming transfer

function

C(sI " A)!1B + D =

3s + 3

s + 2

4

Show that it is controllable, though not observable. Show

that this corresponds to zero/pole cancellation in the Transfer

Function. Now consider system in series:

Controllability/observability 9: zero-pole cancellation 121

s + 1

s + 2! s + 3

s + 1! !

v(s) x(s) w(s)

Simulation diagram (time domain)

!!

!!2!

!2 " !1 "

1! 1!

!#

! !#

!!#

! !#

!!1! !v(t) x(t) w(t)!2 !1


Diagram uniquely defines (A, B, C,D):

"A B

C D

#=

$%&

!1 2 2

0 !2 !1

1 1 1

'()

*B AB

+=

"2 !4

!1 2

#rank=1

"C

CA

#=

"1 1

!1 0

#rank=2

Observable, not completely controllable


Eigenvectors of "1 = !1, "2 = !2:

m1 =

"1

0

#m2 =

"2

!1

#

m1 /" range*

B AB+

, -. /uncontrollable

m2 " range*

B AB+

, -. /controllable

Uncontrollable eigenvalue " = !1 is cancelled by forming transfer

function

C(sI ! A)!1B + D =

0s + 3

s + 2

1

Minimal order 124

Result

Let R be controllable subspace of (A, B). Let Tc bring (A, B)

to form of (94), then columns 1, 2, . . . , nc of T!1c span R

Let N be unobservable subspace of (A, C). Let To bring (A, C)

to form of (98), then columns no + 1, . . . , n of T!1o span N

In general, Tc and To will be di!erent

Definition

(A, B, C) is of minimal order if (A, B) is controllable and (A, C)

is observable. Then (A, B, C) is called a minimal realisation

Show that it is observable, though not controllable. Show

that this corresponds to zero/pole cancellation in the Transfer

Function.

Solution:For both instances, let us write out the simple two dimensional



state-space representations of the two models, which are both

observable and controllable, interconnect them and show the lack

of either property.

Let us then introduce the transfer functions of the system

components (rather than of the 2-d compositions), and show

the cancellations.

Introducing state variable [ξ1 ξ2]T , input u and output y,

we have:

A =

[−1 0

−1 −2

], B =

[2

−1

],

C =[

1 1], D = [1].

The Kalman controllability matrix is:[2 −2

−1 0

],

which is full rank. Thus, the composition is controllable. However

the observability one is: [1 1

−2 −2

],

which is not full rank. Thus, the composition is not observable.

Given a state-space representation, a transfer function can be



obtained as C(sI − A)−1B + D. The first module has the

following transfer functions:

Gf(s) =s+ 3

s+ 1,

whereas the second

Gs(s) =s+ 1

s+ 2.

You can see that the pole of the first TF cancels with the zero of

the second TF.

Similarly, for the second model, introducing state variable

[ξ1 ξ2]T , input v and output w, we have:

A =

[−1 2

0 −2

], B =

[2

−1

],

C =[

1 1], D = [1].

The Kalman controllability matrix is:[2 −4

−1 2

],

which is not full rank. Thus, the composition is not controllable.



The observability one is: [1 1

−1 0

],

which is full rank. Thus, the composition is observable.

Now, the first module has the following transfer functions:

Gf(s) =s+ 1

s+ 2,

whereas the second

Gs(s) =s+ 3

s+ 1.

You can see that the zero of the first TF cancels with the pole of

the second TF.

2



Controllable Canonical Form from polynomials ofTransfer Function

Consider the following transfer function:

Y (s)

U(s)=

(s+ 10)(s2 + s+ 25)

s2(s+ 3)(s2 + s+ 36).

Give a state-space description of the system in its canonical

controllable form.

Solution:Let us recall some theory to begin with.

Consider a transfer function as a ratio between two polynomials,

as follows:

Y (s)

U(s)=

b1sn−1 + b2s

n−2 + . . .+ bn

sn + a1sn−1 + a2sn−2 + . . .+ an.

There exists a state-space representation of this transfer function

(a.k.a., a realization), which takes the form of the controllable



canonical form, where:

A =

−a1 −a3 . . . −an

1 0 . . . 0...

0 ...

0 . . . 0 1 0

, B =

1

0...

0

0

,C =

[b1 b2 . . . . . . bn

], D = [0].

In MATLAB, this realization is obtained with the command

tf2ss . Notice that in general it is not always possible to

obtain such a realization. (There are sufficient conditions on the

rational functions for that, but we do not discuss them right now.)

Also, notice that this realization is not unique. In particular, we

may be interested to obtain the observable canonical form.

In our instance, we have:

Y (s)

U(s)=

(s+ 10)(s2 + s+ 25)

s2(s+ 3)(s2 + s+ 36)

=s3 + 11s2 + 35s+ 250

s5 + 4s4 + 39s3 + 108s2

=b(s)

a(s).

From this expression, we have that b1 = 0, b2 = 1, b3 =11, b4 = 35, b5 = 250; a1 = 4, a2 = 39, a3 = 108, a4 =



0, a5 = 0. The controllable canonical form is:

d

dt

x1

x2

x3

x4

x5

=

−4 −39 −108 0 01 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 0

x1

x2

x3

x4

x5

+

10000

uy =

[0 1 11 35 250

]x+ [0]u.

As a side note, let us remark that the controllable canonical

form is not unique. This should be clear if, within the block

diagram representation of a model, we re-order the state variable

in the opposite sense. The corresponding canonical form would

be:

A =

0 1 0 . . . 0

0 0 0 0... 1 0

0 0 0 0 1

−an . . . −a3 −a2 −a1

, B =

0

0...

0

1

,C =

[bn bn−1 . . . b2 b1

], D = [0].

Both canonical forms display analogous features, and are

equivalent in terms of their controllability properties. A similar

thing happens for the case of the observable canonical form. 2



Observer Design for DC Servo

Problem 2: Observer design.

x2 x1

Observer

a2su

a1s

u 1

z2

z

Figure 2: Simple model of a DC Servo system.

Figure 2 shows a block diagram representation of a simple model of a DC servo system: x1 is a voltage signalproportional to the output angular velocity x2 (ie. a tacho signal is not available).

(a) Design a full order observer, with observer gain matrix T given by

T =

!

T1

T2

"

, (3)

for x1 and x2 so that the characteristic polynomial associated with the error dynamics is given by:

!e(s) = s2 + 2!e"es + "2

e (4)

(“Design” means write the equations for the observer, with expressions for gains T1 and T2.)

(b) Now, the observer is a system with inputs u and x1, and outputs z1 and z2. Thus, there are four possibletransfer functions between inputs and outputs – these may be included as elements in a 2!2 matrix. Evaluatethe following matrix of transfer functions M(s) between the inputs to the observer u and x1, and its outputsz1 and z2:

M(s) =

!

z1(s)/u(s) z1(s)/x1(s)z2(s)/u(s) z2(s)/x1(s)

"

(5)

as a function of gains T1 and T2, as well as system parameters a1 and a2.

(c) Now determine M(s) as T2 " #. Discuss the meaning of the result.


x1 x2

z1

uinput

velocityobserved variable

observeroutput

observer

1s 2+s

2 s

Figure 3: Velocity Observation System.

Figure 3 shows a velocity observation system where x1 is the velocity to be observed. An observer is to beconstructed to track x1, using u and x2 as inputs. The variable x2 is obtained from x1 through a sensor havingthe known transfer function

2 $ s

2 + s(6)

2

The Figure above shows a block diagram representation of a

simple model of a DC servo system. The variable x1 is a voltage

signal that corresponds to the output y, whereas x2 is the angular

velocity.

1. Design an observer, with observer gain matrix L = [l1 l2]T ,

for the variables x1, x2, and such that the characteristic

polynomial associated with the error dynamics is given by:

∆(s) = s2

+ 2ζωs+ ω2.

2. The observer is a system with inputs u and x1, and with

outputs z1 and z2. Thus, there are four possible transfer

functions between inputs and outputs these may be included

as elements in a 2× 2 matrix. Evaluate the following matrix



of transfer functions M(s) between the inputs to the observer

u and x1, and its outputs z1 and z2

M(s) =

[Z1(s)

U(s)

Z1(s)

X1(s)Z2(s)

U(s)

Z2(s)

X1(s)

],

as a function of gains l1 and l2, as well as of the system

parameters a1, a2.

3. Determine M(s) as l2 → ∞. Discuss the meaning of the

result.

Solution:We have:

X1(s) =a2

sX2(s) ⇒ x1(t) = a2x2(t)

X2(s) =a1

sU(s) ⇒ x2(t) = a1u(t)

y(t) = x1(t)

The state-space model follows:

d

dt

[x1

x2

]=

[0 a2

0 0

] [x1

x2

]+

[0

a1

]u

y =[

1 0] [ x1

x2

]+ [0]u.



Let us introduce an observer matrix L =

[l1l2

]. The closed-

loop matrix for the estimator is

A− LC =

[−l1 a2

−l2 0

].

Let us compute the characteristic polynomial:

det(sI − (A− LC)) = s(s+ l1) + a2l2,

which is then equated to the desired characteristic polynomial,

thus obtaining:

l1 = 2ζω, l2 =ω2

a2

.

Let us now look closer at the observer, with variable z =

[z1 z2]T : ˙z = Az + Bu + L(y − y) = Az + Bu + L(y −

Cz) = (A−LC)z+[B L][u y]T , where y = x1. We obtain:

d

dt

[z1

z2

]=

[−l1 a2

−l2 0

] [z1

z2

]+

[0 l1a1 l2

] [u

y

]y =

[1 0

0 1

] [z1

z2

].



The transfer function can be directly found as follows:

M(s) = C(sI − A)−1B +D

=

[1 0

0 1

] [s+ l1 −a2

l2 s

]−1 [0 l1a1 l2

]=

1

s2 + l1s+ a2l2

[a1a2 sl1 + a2l2

a1(s+ l1) sl2

].

If l2 →∞, we have:

liml2→∞

M(s) =

[0 1

0 s/a2

].

Since l2 = ω2/a2, whenever l2 → ∞, then ω2 → ∞. The

speed of the estimator thus increases in speed and its error

converges to zero infinitely fast at the limit. In the long run, we

expect z1(t) → x1(t), z2(t) → x2(t) =x1(t)

a2, and that the

effect of the input on the estimate is null.

2



Heading Controller for Aircraft Ground Control

EECS 128 Introduction to Control Design Techniques

Problem Set 3

Professor C. TomlinDepartment of Electrical Engineering and Computer Sciences

University of California at Berkeley, Fall 2008Issued 10/2; Due 10/9

Problem 1: Heading controllers for ground control of aircraft. A top view of the tricycle landing gear

!a

"

L

V

Figure 1: Tricycle landing gear top view.

for a small unmanned aerial vehicle is shown in Figure 1. Consider the case in which the vehicle is movingwith constant forward velocity V (achieved by an input motor thrust, not shown here). The only input wewould like to consider here is the heading actuator !a, which a!ects the heading ". The dynamics betweenheading actuator and heading, for small angle changes, can be modeled as:

"(s) =V

L

1

s(#s + 1)!a(s)

(a) Design a controller for this system, so that a given step reference heading "ref = 1 is achieved with nosteady state error.

(b) Design a controller for this system, so that a given step reference heading rate $ref = "ref = 1 (corre-sponding to a changing heading) is achieved with no steady state error.

Problem 2. Root locus design.

G(s)R YK(s)

Figure 2: Unity Feedback System with proportional controller K(s), plant G(s)

Consider the unity feedback system of Figure 2 with G(s) = 1s(s+1) . You would like to have closed loop poles

1

The above figure displays a top view of a tricycle landing

gear of an aircraft. Consider the case in which the vehicle is

moving with constant forward velocity V (achieved by an input

motor thrust, which is not shown here). The only input we would

like to consider here is the heading actuator δa, which affects the

heading ψ. The dynamics between heading actuator and heading,

for small angle changes, can be modeled as

Ψ(s) =V

L

1

s(τs+ 1)∆a(s),



where in the formula we have denoted with capitals letters the

respective Laplace transforms, and where V, L, τ > 0.

1. Write a block diagram representation of the feedback loop

that presents a controller (call it C(s)) and the system block

with input δa and output ψ. The block diagram should

also encompass a reference signal. (If necessary, you can get

inspiration from the first Section of Chapter 9 in the A.M.)

2. Based on this representation, design a controller (the simplest

possible) for this system, so that a given step reference heading

ψref = 1 is achieved with no steady state error. (In other

words, apply a step input at the reference and make sure to

get zero steady state error.)

3. Furthermore, design a controller for this system (again, the

simplest possible), so that a given step reference heading rate

ωref = ψref = 1 (corresponding to a change in the heading)

is achieved with no steady state error.

Solution:The block diagram appears on page 268 of the A.M. book, in

Figure 9.1 a), where P (s) = VL

1s(τs+1) and r = ψref is the

reference signal.

Let us call G(s) = Ψ(s)/∆a(s) = VL

1s(τs+1). Given a

controller C(s), the relationship between ψ and ψref can be



obtained as follows:

Ψ(s)

Ψref(s)=

C(s)G(s)

1 + C(s)G(s).

The steady-state of this transfer function is C(0)G(0)1+C(0)G(0), thus its

error (w.r.t. ψref = 1) is

1−C(0)G(0)

1 + C(0)G(0)=

1

1 + C(0)G(0)

=1

1 + C(s)VL1

s(τs+1)

∣∣∣∣∣s=0

The requirement is thus

1

1 + C(s)VL1

s(τs+1)

∣∣∣∣∣s=0

= 0,

which can be simply achieved by choosing C(s) = C, where C

is a non-zero gain.

The second instance can be obtained by replicating the block

diagram above, and adding a derivative at the output of the block

described by G(s), thus relating Ψ(s) with its derivative sΨ(s).

The transfer function relating ∆a(s) to the derivative of Ψ(s) is

thus sG(s) = VL

1(τs+1). Reasoning along the previous lines, we



get to the following requirement

1

1 + C(s)VL1

(τs+1)

∣∣∣∣∣s=0

= 0,

which can be simply achieved by choosing C(s) = C/s, which

is a simple integrator with gain C.

2



Observer Design Problem


x2 x1

Observer

a2su

a1s

u 1

z2

z

Figure 2: Simple model of a DC Servo system.

Figure 2 shows a block diagram representation of a simple model of a DC servo system: x1 is a voltage signalproportional to the output angular velocity x2 (ie. a tacho signal is not available).

(a) Design a full order observer, with observer gain matrix T given by

T =

!

T1

T2

"

, (3)

for x1 and x2 so that the characteristic polynomial associated with the error dynamics is given by:

!e(s) = s2 + 2!e"es + "2

e (4)

(“Design” means write the equations for the observer, with expressions for gains T1 and T2.)

(b) Now, the observer is a system with inputs u and x1, and outputs z1 and z2. Thus, there are four possibletransfer functions between inputs and outputs – these may be included as elements in a 2!2 matrix. Evaluatethe following matrix of transfer functions M(s) between the inputs to the observer u and x1, and its outputsz1 and z2:

M(s) =

!

z1(s)/u(s) z1(s)/x1(s)z2(s)/u(s) z2(s)/x1(s)

"

(5)

as a function of gains T1 and T2, as well as system parameters a1 and a2.

(c) Now determine M(s) as T2 " #. Discuss the meaning of the result.


x1 x2

z1

uinput

velocityobserved variable

observeroutput

observer

1s 2+s

2 s

Figure 3: Velocity Observation System.

Figure 3 shows a velocity observation system where x1 is the velocity to be observed. An observer is to beconstructed to track x1, using u and x2 as inputs. The variable x2 is obtained from x1 through a sensor havingthe known transfer function

2 $ s

2 + s(6)

2

(This exercise is to be solved after the Lecture of Monday,

October 24.) The Figure above shows a velocity observation

system, where x1 is the velocity to be observed. An observer is

to be constructed to track x1, using the control u and the output

x2 as its inputs. The variable x2 is obtained from x1 through a

sensor having the following transfer function

G(s) =2− s2 + s

.

Answer to the following points:

1. Derive a set of state-space equations for the system, with state

variables x1 and x2 , input u and output x2. (To make sure

you’ve done it right, check your result by calculating back the

transfer function of the obtained model.)



Plot the poles and the zeros of the transfer function in

MATLAB. Also, represent its Bode plot with MATLAB.

2. Design an observer with states z1 and z2 to track respectively

x1 and x2. Choose both observer eigenvalues to be at −4.

Write out the state space equations for the observer.

3. Derive the combined state-space equations for the system

composed of the model and of the observer. To do so, consider

four state variables x1, x2, e1 = x1 − z1, e2 = x2 − z2.

Select u as the input and z1 as the output. Is the composed

model controllable and/or observable? If meaningful, try to

give physical reasons for any of the states being uncontrollable

or unobservable.

4. What is the transfer function relating u to z1? (Explain your

result.)

Solution:We have:

X1(s) =1

sU(s)⇒x1(t) = u(t)

X2(s) =2− s2 + s

X1(s)⇒(2 + s)X2(s) = (2− s)X1(s)

⇒x2(t) + 2x2(t) = −x1(t) + 2x1(t)

⇒x2(t) = 2x1(t)− 2x2(t)− u(t)

y(t) = x2(t)



The state-space model follows:

d

dt

[x1

x2

]=

[0 0

2 −2

] [x1

x2

]+

[1

−1

]u

y =[

0 1]x+ [0]u.

The transfer function is:

C(sI − A)−1B +D =

[0 1

] [ s 0

−2 s+ 2

]−1 [1

−1

]+ [0] =

1

s(s+ 2)

[0 1

] [ s+ 2 0

2 s

] [1

−1

]=

2− ss(s+ 2)

.

The pole-zero map and the Bode diagram of the transfer function

can be generated by the following instructions:

H = tf([-1 2],[1 2 0]);

pzmap(H)

bode(H)

The pole-zero map is displayed in following Figure, where poles

are denoted by crosses, zeros by circles.



−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Pole−Zero Map

Real Axis

Imag

inar

y Ax

is

The Bode diagram is displayed in the following Figure:

−40

−20

0

20

40

60

Mag

nitu

de (d

B)

10−2 10−1 100 101 10290

135

180

225

270

Phas

e (d

eg)

Bode Diagram

Frequency (rad/sec)

Let us now introduce an observer gain as L =

[l1l2

]. The



closed-loop matrix for the observer is

A− LC =

[0 −l12 −(l2 + 2)

].

Let us compute the characteristic polynomial:

det(sI − (A− LC)) = s2

+ s(l2 + 2) + 2l1.

We can assign the eigenvalues of the observer to the point −4

by choosing

l1 = 8, l2 = 6.

The state space equations for the observer, with variable z =

[z1 z2]T , are:

z = (A− LC)z + [B|L]

[u

y

], y = z2;

d

dt

[z1

z2

]=

[0 −8

2 −8

] [z1

z2

]+

[1 8

−1 6

] [u

y

]y = [0 1]

[z1

z2

].

Let us introduce the error e = x − z, and its dynamics

e = (A − LC)e, which are to be considered along with the



open-loop dynamics x = Ax + Bu and with the observation

y = z1 = x1 − e1. The overall dynamics can be written out as

follows:

d

dt

[x

e

]=

[A 0

0 A− LC

] [x

e

]+

[B

0

]u

y =[

1 0 −1 0] [ x

e

],

or explicitly

d

dt

x1

x2

e1

e2

=

0 0 0 0

2 −2 0 0

0 0 0 −8

0 0 2 −8

x1

x2

e1

e2

+

1

−1

0

0

u

y =[

1 0 −1 0]

x1

x2

e1

e2

.The controllability matrix follows:

C =

1 0 0 0

−1 4 −8 16

0 0 0 0

0 0 0 0

,



so that

rank(C) = 2 < 4,

which enables us to conclude that the model is uncontrollable.

More precisely, it is so over the error states e1, e2, since they are

bound to decay to zero, regardless of the control effort.

The observability matrix follows:

O =

1 0 −1 0

0 0 0 8

0 0 16 −64

0 0 −128 384

so that

rank(O) = 3 < 4,

which enables us to conclude that the model is unobservable. In

fact, the observer is only estimating values of x1, not of x2.

The transfer function relating u to z1 can be directly derived



as:

C(sI − A)−1B +D

=[

1 0 −1 0]

s 0 0 0−2 s+ 2 0 00 0 s 80 0 −2 s+ 8

−1

1−100

=

1

s.

In fact, we expect that z1 → x1 = 1su.

2


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sc4026 Exercise Session 7 (with solutions) - TU Delftaabate/2012sc4026/2012sc4026_ex7_soln.pdf · sc4026 Exercise Session 7 (with solutions) Alessandro Abate [email protected] Marco