Stability

T.Kaputu
 Stability

An LTI system is said to be stable if it satisfies the following conditions.

a) Even after excitation by a bounded input, output must be bounded

b) in the absence of input, output must be zero irrespective of initial
conditions.
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
Effect of Poles Location
 Effect of Poles Location

(j) Complex conjugate poles at s = a ± j b

 Effect of Poles Location

From the definition of stability, the impulse response must tend to zero as time approaches infinity.
Therefore, the following conclusions can be drawn:

i. For all the poles located in the left half of s-plane, the impulse response tends to zero and hence
the system is stable.
ii. For poles on the jw-axis and simple, the system is marginally stable or limitedly stable.
iii. For all poles located in the right half of s-plane, the impulse response does not tend to zero and it
increase with time. Therefore, the output becomes unbounded. Hence the system is unstable.

Therefore, if the roots of the characteristic equation

i. have a negative real part, system is stable.
ii. have a zero real part and the pole is not repeated, the system is marginally or limitedly stable.
iii. have a positive real part, the system is unstable.
 Stability

𝐶 𝑠 𝐺 𝑠
=
𝑅 𝑠 1 ± 𝐺 𝑠 𝐻(𝑠)
 Stability

The transfer function of any closed-loop system is given by

𝐶 𝑠 𝑏m 𝑠 𝑚 + 𝑏m−1 𝑠 𝑚−1 + ⋯ + 𝑏0
=
𝑅 𝑠 𝑎n 𝑠 𝑛 + 𝑎n−1 𝑠 𝑛−1 + ⋯ + 𝑎0

𝒂 and 𝒃 are constants. To find the poles of this closed-loop system, let us take

𝑎n 𝑠 𝑛 + 𝑎n−1 𝑠 𝑛−1 + ⋯ + 𝑎0 = 0

The roots of this characteristic equation represent the closed-loop poles. The stability of
the system depends on these poles.
 Routh-Hurwitz Criterion
▪ yields stability information without the need to solve for the closed-loop system poles

▪ Determine whether a system is stable.

▪ An easy way to make sure feedback isn't destabilizing
▪ Construct the Routh Table

Three steps:
1) Write the characteristics equation with all the powers of s (in descending order)
2) Generate a data table called a Routh table
3) Interpret the Routh table to tell how many closed-loop system poles are in the left
half-plane and in the right half-plane.
Generating a Basic Routh Table
 Generating a Basic Routh Table

Note: any row of the Routh table can be multiplied by a positive constant without changing the values of the
rows below.
 Example 1
Make the Routh table for the system shown below

The first step is to find the equivalent closed-loop system because we want to test the
denominator of this function, not the given forward transfer function, for pole location.

Using the feedback formula, we obtain the equivalent system

The Routh-Hurwitz criterion will be applied to this denominator.

Example 1
 Example 2
Determine the stability of the closed-loop transfer function

𝒔 + 𝟐𝟔
𝒔𝟐 + 𝟓𝒔𝟐 + 𝟏𝟎𝒔 + 𝟑
 Routh-Hurwitz Criterion: Special Cases

Two special cases can occur:

a) The Routh table sometimes will have a zero only in the first column of a row, or
b) the Routh table sometimes will have an entire row that consists of zeros.

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, an epsilon, 𝝐, is assigned to replace the zero in the first column.
The value 𝝐 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.
 Examaple 3
Stability via Epsilon Method
Determine the stability of the closed-loop transfer function
10
𝑇 𝑠 = 5
𝑠 + 2𝑠 4 + 3𝑠 3 + 6𝑠 2 + 5𝑠 + 3
Example 3
 Example 4
𝐶 𝑠 𝑠 2 + 12𝑠 + 15
= 5
𝑅 𝑠 𝑠 + 2𝑠 4 + 4𝑠 3 + 8𝑠 2 + 3𝑠 + 1
 Stability via Reverse Coefficients
Determine the stability of the closed-loop transfer function
10
𝑇 𝑠 = 5
𝑠 + 2𝑠 4 + 3𝑠 3 + 6𝑠 2 + 5𝑠 + 3

First write a polynomial that has the reciprocal roots of the denominator
𝐷 𝑠 = 3𝑠 5 + 5𝑠 4 + 6𝑠 3 + 3𝑠 2 + 2𝑠 + 1

Since there are two sign changes, the

system is unstable and has two right-
half-plane poles.
 Entire Row is Zero
• We now look at the second special case.
• Sometimes while making a Routh table, we find that an entire row consists of zeros
because there is an even polynomial that is a factor of the original polynomial.
• This case must be handled differently from the case of a zero in only the first column
of a row.
• Let us look at an example that demonstrates how to construct and interpret the Routh
table when an entire row of zeros is present.
 Example 5
Stability via Routh Table with Row of Zeros
Determine the number of right-half-plane poles in the closed-loop transfer function

10
𝑇 𝑠 = 5
𝑠 + 7𝑠 4 + 6𝑠 3 + 42𝑠 2 + 8𝑠 + 56
 Example 5
10
𝑇 𝑠 = 5
𝑠 + 7𝑠 4 + 6𝑠 3 + 42𝑠 2 + 8𝑠 + 56

𝑠5 1 6 8
𝑠4 7 42 56
𝑠3 0 0 0
𝑠2
𝑠1
𝑠0

Now we are faced with the problem of zeros in the third row.
 Example 5
1. Form a new polynomial using the entries in the row above zeros.
The polynomial will start with power of s in that row, and continue by skipping every
other power of s, i.e.

𝑃 𝑠 = 𝑠 4 + 6𝑠 2 + 8 4.1

2. Next we differentiate the polynomial with respect to s and obtain

𝑑𝑃(𝑠)
= 4𝑠 3 + 12𝑠 + 0
𝑑𝑠 4.2
 Example 5
3. Finally the row with all zeros in the Routh table is replaced with the coefficients in
Eq.(4.2), and continue the table.

𝑠5 1 6 8
𝑠4 1 1 8
𝑠3 4 12 0
𝑠2 3 8 0
𝑠1 4/3 0 0
𝑠0 8 0 0

• From the table it can seen that all entries in the first column are positive. Hence,
there are no right-half-plane poles
 Example 6
Find the range of gain, 𝐾, for the system of Figure 11.1 that will cause the system to be
stable, unstable, and marginally stable. Assume 𝐾 > 0.

𝑹 𝒔 + 𝐾 𝑪 𝒔
𝑠(𝑠 + 7)(𝑠 + 11)
-

FIGURE 11.1 Feedback control system

 Example 6
First find the closed-loop transfer function as
BIBO Stable
𝐾
𝑇 𝑠 = 3 0 < 𝐾 < 1386
𝑠 + 18𝑠 2 + 77𝑠 + 𝐾
Unstable
Next form the Routh table
𝐾 > 1386
𝑠3 1 77
(2 sign changes)
𝑠2 18 𝐾
𝑠1 1386 − 𝐾
18
𝑠0 𝐾
 Example 6
• Since K is assumed positive, we see that all elements in the first column are always
positive except the 𝑠1 row.

• 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.

• If 𝐾 > 1386, the 𝑠1 term in the first column is negative

• There are two sign changes, indicating that the system has two right-half-plane poles
and one left half-plane pole, which makes the system unstable.

• If 𝐾 = 1386, we have an entire row of zeros, which could signify 𝑗𝜔 poles.

 Example 6
• Returning to the 𝑠 2 row and replacing 𝐾 with 1386, we form the even polynomial

𝑃 𝑠 = 18𝑠 2 + 1386

• Differentiating with respect to s, we have

𝑑𝑃(𝑠)
= 36𝑠 + 0
𝑑𝑠

𝑠3 1 77 For the case of 𝐾 = 1386, system

𝑠2 18 1386 is marginally stable
𝑠1 36
𝑠0 1386
 Example 7
𝐾 𝑠+4
Given the unity-feedback system of Figure 3.1 with 𝐺 𝑠 =
𝑠 𝑠+1 𝑠+2
do the following:

i. Construct the Routh-Hurwitz table

ii. Find the range of K that keeps the system stable
iii. Find the Value of K that makes the system oscillate
iv. find the frequency of oscillation when K is set to the value that makes the
system oscillate

Figure 3.1
Example 7
 Stability in State Space
• we can determine the stability of a system represented in state space by finding the
eigenvalues (poles) of the system matrix, A, and determining their locations on the s-
plane

we can find the poles using equation:

det 𝜆𝑰 − 𝑨 = 0
 Example
Given the system

0 3 1 10
𝒙ሶ = 2 8 1 𝒙+ 0 𝑢 𝑦= 1 0 0𝑥
−10 −5 −2 0

find out how many poles are in the left half-plane, in the right half-plane, and on the
jω-axis.
Example Sol
Questions?