Chapter (5)

STABILITY

1
Learning Outcomes
After completing this chapter the student will be
able to:
1) Known the definition of stability.
2) Determine the stability conditions by evaluating
poles of transfer function.
3) Make and interpret a basic Routh table to determine
the stability of a system.
4) Make and interpret a basic Routh table where either
the first element of a row is zero or an entire row is
zero. 2
1. Introduction

3
1. Introduction

The stability of a cone.

4
1. Introduction
• Requirements for design of a control system
– Transient Response
– Stability
– Steady State Errors
• Stability – most important parameter for
design
• Total response of a system is = forced
response + natural response, or
c (t ) = c forced (t ) + c natural (t )
5
System Stability Definition
Types of stability based on Natural response definition:
1. A system is STABLE if the natural response approaches zero as time
approaches infinity
2. A system is UNSTABLE if the natural response approaches infinity as
time approaches infinity
3. A system is MARGINALLY STABLE if the natural response neither
decays nor grows but remains constant or oscillates.

Therefore,
1. A system is stable if every bounded input yields a bounded output
2. A system is unstable if any bounded input yields an unbounded
output 6
1. Introduction

Stability in the s-plane.

7
How to define stability
R(s) C(s)
+
- G(s)

H(s)
Stability with respect to G(s)? Stability with respect to
All poles in the left half plane 𝑮(𝒔)
?
𝟏+𝑮 𝒔 𝑯(𝒔)
Poles of 1+G(s)H(s) in the
left half.

8
❖ How do we determine if a system is stable?

If ALL poles of the closed-loop system are in the left half of

the plane and hence, have a negative real part, the system is
stable.

unstable systems have closed loop transfer functions with at

least one pole in the right half-plane and/or poles of
multiplicity greater than 1 on the imaginary axis.

marginally stable systems have closed-loop transfer

functions with only imaginary axis poles of multiplicity 1
and poles in the left half-plane
9
 ❖ System Stability Definition – Stable System

Bounded input
yields bounded
output
As time approaches
infinity , the natural
response approaches
zero

Stable system
have poles only
in the left hand
plane
10
 ❖ System Stability Definition – Unstable System
Bounded input
Time approaches
yields an
infinity the natural
unbounded
response
output
approaches
infinity

Unstable
system have
at least one
pole in the And/or poles of multiplicity greater
right hand than one on imaginary axis
plane 11
 System Stability Definition
Unstable system – closed
loop transfer function
poles with at least one
j pole in the right half
and/or poles of
Stable system –
multiplicity greater than 1
closed loop   on the imaginary axis
transfer function  𝑨𝒕𝒏 𝒄𝒐𝒔(𝝎𝒕 + ∅)
poles only in the 1
left half plane 
    
 Marginally stable –
closed loop transfer
 -1 function with only
   imaginary axis
poles of multiplicity
1 and poles in the
left half plane.

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❖ How do we determine if a system is stable?
It is not always a simple matter to determine if a feedback control
system is stable.
Consider the system below

We need a method to test for stability without having to

solve for the roots of the denominator!! 13
❖ Routh-Hurwitz Method
➢ A method that yields stability information without the need to
solve for the closed-loop system poles.
➢ Using this method, we can tell how many closed-loop system
poles are in the left half-plane, in the right half-plane, and on
the jw-axis (imaginary axis).
➢ Notice that we say how many, not where. We can find the
number of poles in each section of the s-plane, but we cannot
find their coordinates.
➢ The method requires two steps:
(1) Generate a data table called a Routh table
(2) Interpret the Routh table to tell how many closed-
loop system poles are in the left half-plane, the
right half-plane, and on the jw-axis.
14
❖ Routh-Hurwitz Method
Generating of Routh table
Consider the equivalent closed-loop transfer function below. Since we are
interested in the system poles, we focus our attention on the denominator

15
Generating of Routh table

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 Generating of Routh table
Example:

17
 Generating of Routh table
Example:

18
❖ Routh-Hurwitz Method
Interpretation of Routh table

Routh-Hurwitz criterion declares that 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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 ❖ Routh-Hurwitz Method
Interpretation of Routh table

Two sign changes in the first column!

Two sign changes = two right half plane poles, therefore unstable system 20
❖ Routh-Hurwitz Stability Criterion –Example

How many roots are in the right-half plane

and in the left-half plane?

P(s) = 3s 7 + 9s 6 + 6s 5 + 4s 4 + 7s 3 + 8s 2 + 2s + 6

Solution
4 in RHS & 3 in LHS

21
❖ Routh-Hurwitz Stability Criterion –Example
Example:

Find the range of gain, K, for the system that

will cause the system to be stable.

22
❖ Routh-Hurwitz Stability Criterion –Example
Example:
First find the closed-loop transfer function as

Next form the Routh table shown as

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❖ Routh-Hurwitz Stability Criterion –Example
Example:

➢ Since K is assumed positive, we see that all elements in the first column
are always positive except the s1 row.
➢ This entry can be positive, zero, or negative, depending upon the value
of K.
➢ If K < 1386, all terms in the first column will be positive, and since there
are no sign changes, the system will have three poles in the left half-
plane and be stable. 24
❖ Routh-Hurwitz Stability Criterion –Example
Example:

K < 1386, The system is stable

K > 1386, The system is unstable

K = 1386, the system is marginally stable
25
❖ Routh-Hurwitz Stability Criterion –Example

Determine the value of gain K to make the

system stable

26
❖ Routh-Hurwitz Stability Criterion –Example
For a unity feedback system with the forward
transfer function

Find the range of K to make the system stable.

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❖ Routh-Hurwitz Stability Criterion

Special cases:
A. Zero in the first column
B. Zero in the entire row

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 (a) Zero Only in the First Column
If the first element of a row is zero, division by zero would be
required to form the next row. To avoid this phenomenon, will
do the following:
(1) An epsilon, , is assigned to replace the zero in the first
column.
(2) The value 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.
29
❖ Routh-Hurwitz Method : Special Cases
(a) Zero Only in the First Column
Example: 10
T (s) = 5
s + 2s4 + 3s3 + 6s2 + 5s + 3

How many poles?

Five poles

30
❖ Routh-Hurwitz Method : Special Cases
(a) Zero Only in the First Column

How many sign

changes?

Two sign changes

Two poles are on

the right half
plane

The system is
unstable

31
❖ Routh-Hurwitz Method : Special Cases
(b) Entire Row is Zero
This case must be handled differently from the case of a
zero in only the first column of a row. The procedure is as
follows:
1. we 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 power of s.
2. differentiate the polynomial with respect to s
3. use the coefficients of the resulted polynomial to replace
the row of zeros.

32
❖ Routh-Hurwitz Method : Special Cases
(b) Entire Row is Zero
10
Example: T (s) = 5
s + 7s4 + 6s3 + 42s2 + 8s + 56

0 0 0

P( s ) = s 4 + 6 s 2 + 8 dP( s)
= 4s 3 + 12s
What to do? ds

4 12 0
33
❖ Routh-Hurwitz Method : Special Cases
(b) Entire Row is Zero
10
Example: T (s) = 5
s + 7s4 + 6s3 + 42s2 + 8s + 56

dP
P(s) = s + 6 s + 8 
4 2
= 4 s 3 + 12s
ds

How many sign changes?

No sign changes 34
❖ Routh-Hurwitz Method : Special Cases
(b) Entire Row is Zero
What can we learn when the entire
row is zero?
An entire row of zero will appear in
the Routh Table when a purely
even or a purely odd polynomial is
a factor of original polynomial

Even polynomial only has

roots symmetry about the
origin

If we do not have row of

zeros, we don’t have roots on
imaginary axis
35
❖ Routh-Hurwitz Method : Special Cases
s 8 + s 7 + 12s 6 + 22s 5 + 39s 4 + 59s 3 + 48s 2 + 38s + 20

0 0 0 0

P(s) = s 4 + 3s 2 + 2 dP ( s )
= 4s 3 + 6s
ds
4 6 0 0

36
❖ Routh-Hurwitz Method : Special Cases
s 8 + s 7 + 12s 6 + 22s 5 + 39s 4 + 59s 3 + 48s 2 + 38s + 20

Apply to
original
polynomial

Apply only
to even
polynomial

37
❖ Routh-Hurwitz Method
Example:

38
❖ Routh-Hurwitz Method
Example:
Find the range of gain, K, to ensure stability in the unity feedback system

Solution:

39
❖ Routh-Hurwitz Method
Example:

Conditions state that K < -12, K > 2, and K > -33.

These conditions cannot be met simultaneously.
So, the system is not stable for any value of K.
40