Lecture11-SystemStability

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Page : 1 EE406 Control Systems Lecture 11 : System Stability

UCSI University Faculty of Engineering

Kuala Lumpur, Malaysia Department of Mechatronics

Lecture 11

System Stability

Mohd Sulhi bin Azman Lecturer

Department of Mechatronics UCSI University

[email protected]

1 August 2011

Contents

• Poles, Zeros, Gain and Pole-Zero Map

• Applications in MATLAB

• Stable, Marginally Unstable & Unstable

• Necessary condition for stability

• Routh stability criterion

• Hurwitz stability criterion

• Routh-Hurwitz Criterion

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Definitions

• Poles

– Poles are the roots of D(s) (the denominator of the transfer function), obtained by setting D(s) = 0 and solving for s.

• Zeros

– The roots of the numerator of the transfer function obtained by setting N(s) = 0 and solving for s.

• Gain

• Pole-Zero Map

– A plot of poles and zeros.

Tease Your Brain

• Consider the following function. Determine the

poles and zeros and plot them on p-z map.

(

)

2

3 4 ( )

2 5

s G s

s s + =

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Definitions

• A stable system has a closed-loop transfer

functions with poles only in the left half plane.

• An unstable system has a closed-loop transfer

functions with at least one pole in the right half

plane and/or poles of multiplicity greater than

one on the imaginary axis.

• A marginally stable system has closed-loop

transfer functions with only imaginary axis poles

of multiplicity 1 and poles in the left half plane.

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Example 1

• Test the stability for the following system:

(

)(

1

)

( )

G s

s a s b =

+ +

Solution to Example 1

• Taking the inverse Laplace transforms:

(

)(

)

1 1 1 2

1 2

1 k k at bt

k e k e s a s b s a s b

− _ _₌ − _ ₊ _₌ − ₊ −

+ +  + + 

 

 

L L

0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1 1.2

1.4 Step Response

Time (sec) A m p lit u d e Re -a -b Im

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Example 2

• Test the stability for the following system:

(

)(

1

)

( )

F s

s a s b =

− −

• Taking the inverse Laplace transform:

Poles in the RHP System Response

Re

a b

Im

0 5 10 15 20 25 30

- 20 - 15 - 10 -5 0 5x 10

5

Step Res ponse

Time ( sec )

Am

plit

ude

(

)(

)

1 1 1 2

1 2

1 k k at bt

k e k e s a s b s a s b

− _ _₌ − _ ₊ _₌ ₊

− −  − − 

 

 

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Bounded Solution

• A system is said to be stable if every bounded input produced a bounded output. The following system shows a stable system:

• The following is unstable system:

Bounded Signals

• The following diagram shows a BIBO stable system:

bounded signal (within limits)

bounded

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Unbounded Signals

• The following shows unbounded signals:

4

x

Unbounded signal Unbounded

signal 3

x

Conditions for Stability

• There are two conditions in order for a system

to be stable. The first condition is the

“sufficient condition” and the second condition

is the “necessary condition”.

• The sufficient condition is the minimum

requirement in order for a system to be stable.

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Necessary Condition for Stability

• Consider the following characteristic equation:

• The stability of this system may be determined by

solving this characteristic equation. Solving this equation will yield roots of the characteristic equation.

• The necessary but insufficient condition for the stability of a system is that all coefficients of the above characteristic equation be real and have the same signs. Furthermore, none of the coefficients should be zero.

1 2

0 1 2 1

( ) n n n 0

n n

q s =a s +a s − +a s − + +_⋯ a₋s+a =

Necessary Condition for Stability

• We also take note of the following results:

1. If all the roots of the characteristic equation have negative real parts, then the system is stable.

2. If any root of the characteristic equation has a positive real part of if there is a repeated root on the jw-axis, then the system is unstable (because it lies on the right-half plane).

3. If some of the coefficients are zero or negative, then it can be concluded that the system is unstable.

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Necessary Condition for Stability

• Theorem:

1. The positive-ness of the coefficients of characteristic equation is necessary as well as sufficient condition for stability of system of a first and second order.

2. The positive-ness of the coefficients of the characteristic equation ensures that the negative-ness of real roots, but it does not ensure the negative-ness of the real parts of the complex roots for third and higher order systems.

• In sum, a necessary and sufficient condition for a feedback system to be stable is that all of the poles of the system transfer function must have negative real parts.

Theorem 1 : Proof

• Consider the first order characteristic

equation:

• Then:

• Your turn: prove Theorem 1 for

0 1

( ) 0

q s =a s+ =a

1 0 a s a = − 2

0 1 2

( ) 0

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Theorem 2 : Proof

• Your turn – prove Theorem 2 by using the following equation:

• You shall see that the real part of the complex roots is positive, thus indicating the instability if the system even though all of the coefficients of the characteristic equation are positive.

• Therefore, for systems of three and higher order, we need another method/way of examining the stability of the system. And this is shown in the next slide.

3 2

2

8

0

s

+ + + =

s

Routh & Hurwitz Stability Criterion

• The Routh and Hurwitz criterion is used to

further examine the stability of a system.

• The Hurwitz criterion is in terms of

determinants and the Routh criterion is in terms

of array formulation, which is more convenient

to handle.

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Stability in s-domain

• Besides sketching the pole-zero map, we can

also determine the stability of the system in

s-domain by using the Routh-Hurwitz Criterion.

• The method tells how many (not where) poles

are there in each section of the s-plane : the

right half-plane (RHP), left half-plane (LHP) and

on the

j

ω

-axis.

• We need to create a Routh table first, then only

can we test the stability of the system.

Routh Table

• Given the following polynomial:

• The Routh table is given as:

1 2 2

1 2 2 1 0

( )

_n n _n n _n n

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Routh Table

• Where the computation of “b” and “c” is given as:

• We note that a sign change in the first column indicate that the system is unstable. And the number of changes of sign equals the number of roots with positive real parts. And this is known as the Routh-Hurwitz stability criterion.

1 2 2 1 1

n n i n n i i

n

a

a a

b

a

− − − − −

−

=

1 2 1 1 1 1

n i i n i

c a

b a

c

b

− −

−

+ −

=

Steps to Create Routh Table

• There are essentially two (2) steps that needs

to be followed in creating a Routh table.

• Step 1

– Generate a data table called the Routh table.

• Step 2

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Creating Routh Table : Step 1

• Step 1a:

– Begin by labeling the rows with powers of “s” from the highest power of the denominator of the closed loop transfer function until s0_.

• Step 1b:

– Next, start with the coefficient of the highest power of s in the denominator and list, horizontally in the first row every other coefficient.

Creating Routh Table : Step 1

• Step 1c:

– In the second row, list horizontally starting with the next highest power of s, every coefficient that was skipped in the first row.

• Step 1d:

– Each of the remaining entry is a negative determinant of entries in the previous two rows, divided by the entry in the first column directly above the calculated row.

– Two left hand column of the determinant is always the first column of the previous two rows; and, the right hand column is the elements of the column above and to the right.

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Creating Routh Table : Step 2

• Step 2 – interpret your results.

– If there are no sign changes (i.e. from positive to negative) in the first column, then the system is stable.

– If there are sign changes in the first column, then the system is unstable.

Example 3

• Generate the Routh Hurwitz table for:

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Solution to Example 3

• Step 1 : Generate the Routh table:

• Step 2 : Interpret your result.

– The number of roots of the polynomial that are in the right-half plane is equal to the number of sign

changes in the first column.

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Special Case

• There are two special cases which can sometimes occur:

1. A zero in the first column of the row.

• In this case, we need to use the “epsilon method” where if only one element in the array is zero, it may then be replaced with a small positive number ϵ, that is allowed to approach zero after completing the array. This will be shown in Example 4.

2. Entire row consisting of zeros.

• In this case, we need to establish an auxiliary (helping) polynomials which immediately precedes the zero entry in the Routh array. The order of the auxiliary polynomial is always even and indicates the number of symmetrical roots pairs of the characteristic equation. This is further discussed in Example 5.

Example 4

• In this example, you shall see that the entire

first column is zero. Therefore, we need to use

the epsilon method.

• Generate a Routh table for the following

system:

5 4 3 2

10

( )

2

3

6

5

3

T s

s

=

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• Step 1 : Generate the Routh table.

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• Step 2 : interpret your result.

– The system is unstable, regardless whether we choose a positive or negative epsilon.

Summary of Steps When First Column is Zero

• If the first element of a row is zero, then to

form the next row will involve with division by

zero.

• To avoid this, an epsilon, ε, is used to replace

zero in the first column.

• Then the value is allowed to approach zero

either from the positive or negative side, after

which the signs of the entries in the first

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Example 5

• In this example, you shall see that the entire row is zero. We’re going to form auxiliary equation and use differentiation to compute the values of the next row. • Note that an entire row of zeros will appear in the

Routh table when a purely even or purely odd power of polynomial is a factor of the original polynomial.

• Generate a Routh table for the following system.

5 4 3 2

10

( )

7

6

42

8

56

T s

s

=

+

+ +

Solution to Example 5

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• The row s

3

is now zero. So, at row s

4

, we form an

auxiliary (additional) polynomial :

• Next, we differentiate and we get:

• Then, use the coefficients of the

differentiated function and then place it in our

Routh table; and proceed as usual.

4 2

( )

6

8

P s

= +

s

+

3

( )

4

12

0

P s

′

=

s

+

s

+

Summary of Steps When Entire Row is Zero

• When there is an entire row consists of zeros, there is an even polynomial that is a factor of the original polynomial. This case is handled differently from the first two cases. The steps are:

1. Start by forming the Routh table for the denominator. When an entire row of zero occurs, then proceed to step 2.

2. Return to the row immediately above the row of zeros and form an auxiliary polynomial using the entries in that row as coefficients. The polynomial will start with the power of “s” in the label column and continue by skipping every other one and diminishing in power.

3. Next, differentiate the polynomial with respect to “s”. 4. Then, use the coefficients to replace the row of zeros. 5. Then, the remainder of the table is formed in a straight

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Example 6

• Example :

– Find the value of K, for the following system that will cause the system to be stable, unstable of marginally stable.

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• Make the following assumption:

• For K < 1386, the system is stable.

• For K > 1386, the system is unstable.

• For K = 1386, use the epsilon method, and therefore, there are no poles on the right half-plane, which gives the system to be marginally stable.

1386,

1386

K

=

K

>

K

<

Recap

• What, then again, is the Routh stability criterion?

– It states that the number of roots of the characteristic equation with positive real parts is equal to the number of changes in sign of the coefficients of the first column of the Routh-Hurwitz array.

– It should also be noted that the exact values of the terms in the first column need not be known; instead, only the signs are needed.

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Stability in Time Domain

• The Routh-Hurwitz criterion is often used in analyzing the stability of a system represented in transfer

function. It is the necessary and sufficient condition for stability of a system.

• Now, how about analyzing the stability of state space systems? That is easy. We only need to determine the eigenvalue(s) of the system.

• The stability condition states that if the eigenvalues:

– Have positive real parts, then it is an unstable system. – Have negative real parts, then it is a stable system.

Next Step

• Textbook reference : Chapter 6.

• Homework 11 has been posted on the course

website. Attempt them. You do not have to

submit Homework 11 as it will not be graded.