Stability Analysis - Yolafadhil.yolasite.com/resources/Chapter Six Stability.pdf · Find the range of fain, K, for the following characteristic equation, that will cause the system

Stability Analysis

Fourth Academic Year

Electrical Engineering Department

College of Engineering

Salahaddin University

January 2016

Stability of a System

Stability is the most important system specification.

If a system is unstable, transient response and steady-state errors are moot

points.

An unstable system cannot be designed for a specific transient response or

steady-state error requirement.

Unstable closed-loop, feedback, systems are useless.

Instability causes damage, break down and/or burn out, to the systems.

Instability malfunctions to the system normal operations.

Dr. Fadhil Aula EED at University of Slahaddin 2 of 33

Types of Stability: According to Natural Response

The total time response of a system is given by

𝒄 𝒕 = 𝒄𝒇𝒐𝒓𝒄𝒆𝒅(𝒕) + 𝒄𝒏𝒂𝒕𝒖𝒓𝒂𝒍(𝒕)

The definition of Stability implies that only the Forced Response remains as the

Natural Response approaches zero.

The types of the stability of a system according to the natural response are

1. Stable System (Asymptotically Stable): a system is STABLE if the natural

response approaches zero as time approaches infinity. Thus, the system should

has, on s-plane, negative real poles and/or complex conjugate: Re{pi}<0 the term

decays exponentially.

2. Unstable System: a system is UNSABLE if the natural response approaches

infinity as time approaches infinity. Unstable system, on s-plane, has positive

real poles or repeated jw axis poles.

3. Marginally Stable: a system is marginally stable if the natural response neither

decay nor grows but remains constant or oscillates. Marginally stable has

simple(not repeated) poles on the jw axis.


Types of Stability: Detecting Stable System

Considering on the natural response definition of the stability:

1. Stable systems have closed-loop transfer functions with poles only in the left half

of the s-plane (LHP)

2. If the closed-loop systems poles are in the right half of the s-plane (RHP), or a

positive real part of complex poles, the system is unstable.


Types of Stability


Types of Stability Cont.


Stability Analysis

Feedback is often used to stabilize a system (closed-loop control system).

In some cases, a sufficient high gain can actually destabilize a system.


Stability Analysis: Examples

Example 6.1: What is the type of stability of each of the following open-loop

systems?


Routh-Hurwitz Criterion

Routh-Hurwitz (Routh, 1905) criterion is used to know how many closed-loop

system poles are in the LHP, in the RHP, and on the jw-axis.

Routh criterion technique gives the number of poles in each section of the s-

plane, but does not give their coordinates.

Re

Im

Unstable RegionRHP

Stable Region

LHP

Marginally Stable Region


Routh-Hurwitz Criterion

This technique requires two steps:

1. Generate a data table called a Routh table, and

2. Interpret the Routh table to tell how many closed-loop system poles are in

the LHP, RHP, and on the jw-axis.

The power of the Routh criterion technique lies in design rather than analysis.

For example, if there is unknown parameters in the denominator of a transfer

function, it is difficult to determine via a calculator the range of this parameter to

yield stability.

Main idea of this technique is the number of roots of the characteristic equation

in the RHP is equal to the number of sign change in the first column of the Routh

table.


Routh-Hurwitz Criterion: Generating a Basic Routh Table

Consider the following closed-loop transfer function

Since in Routh-Hurwitz criterion only system poles are interested, therefore, the

denominator is taken into account.

In order the characteristic equation of the above system does not have roots in RHP,

it is necessary but not sufficient that the following hold:

1. All the coefficient of the polynomial of the characteristic equation have the same

sign.

2. None of the coefficient vanishes.


Routh-Hurwitz Criterion: Generating a Basic Routh Table

First, creating the Routh table:


Routh-Hurwitz Criterion: Generating a Basic Routh Table Cont.

Only first two rows of the array are obtained from

the characteristic equation the remaining are

calculated as follows:


Routh-Hurwitz Criterion: Examples

Example 6.2 Find the stability of the following characteristic equation:

Answer:



Example 6.3 Find the stability of the following closed-loop system:

Answer:

There are two sign changes in the first column, thus the system is unstable with

two poles in RHP.



Example 6.4 Find the stability of the following characteristic equation:

𝟐𝒔𝟒+ 𝒔𝟑+ 𝟑𝒔𝟐+ 𝟓𝒔 + 𝟏𝟎 = 𝟎

Answer:

Because the equation has no missing terms and the coefficients are all of the same sign,

it satisfies the necessary conditions.

The Routh table is

The system is unstable because there are two sign changes in the first column, the

equation has roots in the RHP.


Routh-Hurwitz Criterion: Special Cases

There are two special cases may occur:

1. The Routh table may have a zero only in the first column of a row, or

2. The Routh table may have an entire row that consists of zeros.


Routh-Hurwitz Criterion: Special Case: Zero only the First Column

If the first element of a row is zero, a division by zero would be required to form the

next row.

To avoid this phenomenon, an epsilon, , is assigned to replace the zero in the first

column.

The values of is then allowed to approach zero from either the positive or the

negative side, after which the signs of the entries in the first column can be

determined.


Routh-Hurwitz Criterion: Special Case: Zero only the First Column-Example

Example 6.5 Find the stability of the following closed-loop transfer function:

Answer

If is chosen positive, therefore a sign change from the s3 row to the s2 row, and there will be

another sign change from the s2 row to the s1 row. Hence, the system is unstable and has two poles

in RHP.

Alternatively, we could choose negative. There would a sign change from the s4 row to the s3 row.

Another sign change would occur from the s3 row to the s2 row. Our result would be exactly the

same as that for a positive choice for . Thus, the system is unstable, with two poles in the RHP.Dr. Fadhil Aula EED at University of Slahaddin 19 of 33

Routh-Hurwitz Criterion: Special Case: Entire Row is Zero

When the entire row is zero, this indicates that one or more of the following exist:

1. The roots are symmetrical and real,

2. The roots are symmetrical and imaginary, or

3. The roots are quadrantal


Routh-Hurwitz Criterion: Special Case: Entire Row is Zero

To continue with Routh table when a row of zero appears, the following steps are

taken:

1. From the auxiliary equation P(s) = 0 by using the coefficients from the row just

preceding the row of zeros.

2. Take the derivative of the auxiliary equation with respect to s; this gives dP(s)/ds = 0

3. Replace the row of zeros with the coefficients of dP(s)/ds.

4. Continue with Routh table in the usual manner with the newly formed row of the

coefficients replacing the row of zeros.

5. Interpret the change of signs, if any, of the coefficients in the first column of the

Routh table in the usual manner.


Routh-Hurwitz Criterion: Special Case: Entire Row is Zero – Example

Example 6.6

Determine the stability of the following closed-loop transfer function

Answer


Routh-Hurwitz Criterion: Practical Examples

Example 6.7

Find the stability of the following system

Answer

First step is to find the closed-loop transfer function as

Since the system has two RHP poles

and two LHP poles.



Example 6.8


Answer


There are two sign changes, therefore the

system is unstable with two poles in RHP.

The remaining poles are in the LHP.



Example 6.9


Answer


The system is unstable!



Example 6.10

Find the range of fain, K, for the following system, that will cause the system to be

stable, unstable, and marginally stable, assume K> 0.

Answer


a) For stable system 0 < K < 1386

b) For unstable system K > 1386

c) For marginally stable system

K = 1386



Example 6.11

Find the range of fain, K, for the following characteristic equation, that will cause the

system to be stable.

Answer

From s2 row, the condition of stability is K > 0, and from the s1 row, the condition of

stability is


Routh-Hurwitz Criterion: Assignments

HW 6.1

Find the range of K to make the following system stable.

𝒔𝟒 + 𝟓𝒔𝟑 + 𝟓𝒔𝟐 + 𝟒𝒔 + 𝑲 = 𝟎

HW 6.2

Find the range of gain, K, for the following system, that will cause the system to be

stable.


State-Space Stability

Stability of State Space depends on eigenvalues of matrix A.

Because the values of the system’s poles are equal to the eigenvalues of the system

matrix A.

Eigenvalues of matrix A is found by det(λI –A) = 0.


State-Space Stability – Example

Example 6.12

Consider the following state-space model of a control system

Find how many poles are in the LHP, in the RHP, and on the jw-axis.

Answer

Let first form λI-A


State-Space Stability – Example

Example 6.12

Consider the following state-space model of a control system

Find how many poles are in the LHP, in the RHP, and on the jw-axis.

Answer

Second, find the det.(λI –A) :

Since there is only one sign change

in the first column, therefore, the

system has one poles in the RHP

and two poles in the LHP. The

system is, therefore, unstable.


State-Space Stability –Assignments

HW 6.3

The following system in state space represents the forward path of a unity feedback

system. Use the Routh Hurwitz criterion to determine if the closed-loop system is stable.

𝒙 =𝟎 𝟏 𝟎𝟎 𝟏 𝟑−𝟑 −𝟒 −𝟓

𝒙 +𝟎𝟎𝟏𝒖

𝒚 = 𝟎 𝟏 𝟏 𝒙


End of Chapter Six!


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Stability Analysis - Yolafadhil.yolasite.com/resources/Chapter Six Stability.pdf · Find the range of fain, K, for the following characteristic equation, that will cause the system