Routh’s Stability Criterion

Prof. Dr. Md. Faruk Hossain

Department of Electrical & Electronic Engineering
Rajshahi University of Engineering & Technology
Rajshahi-6204, Bangladesh
Why it is necessary to determine the poles of
a linear systems?
The most important problem in linear control systems
concerns stability. A control system is stable if and only if all
closed loop poles lie in the left half s-plane. If any pole lies in
the right hand side of the y-axis i.e. right half s-plane, then the
system will be unstable. For knowing the system stability, it is
necessary to determine the poles of a linear system.

There are two methods to perform this job such as:

 Routh’s stability criterion.
 Hurwitz criterion
 Hurwitz Criterion
With the help of characteristic equation, we will make a
number of Hurwitz determinants in order to find out the
stability of the system. We define characteristic equation of
the system as

Now there are n determinants for nth order characteristic

equation.
Let us see how we can write determinants from the
coefficients of the characteristic equation. The step by step
procedure for kth order characteristic equation is written
below:
 Hurwitz Criterion
The necessary and sufficient condition that all roots of the
characteristics equation lie in the left half plane is that the
equation’s Hurwitz determinant, Dk , k=1, 2, 3, ……. n, must
all be positive.
𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + 𝑎𝑛−2 𝑠 𝑛−2 + − − − 𝑎1 𝑠+𝑎0 = 0

Now there are n determinants for nth order characteristic

equation.
Let us see how we can write determinants from the
coefficients of the characteristic equation. The step by step
procedure for kth order characteristic equation is written
below:
 Hurwitz Criterion
𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + 𝑎𝑛−2 𝑠 𝑛−2 + − − − 𝑎1 𝑠+𝑎0 = 0
𝑎𝑛−1 𝑎𝑛−3 𝑎𝑛−5
𝑎𝑛−1 𝑎𝑛−3
𝐷𝟏 = |𝑎𝑛−1 | 𝐷2 = 𝑎 𝐷 3 = 𝑎𝑛 𝑎𝑛−2 𝑎𝑛−4
𝑛 𝑎𝑛−2
0 𝑎𝑛−1 𝑎𝑛−3
𝑎𝑛−1 𝑎𝑛−3 𝑎𝑛−5 𝑎𝑛−7
𝑎𝑛 𝑎𝑛−2 𝑎𝑛−4 𝑎𝑛−6 When the coefficients with indices
𝐷4 = 0 𝑎𝑛−1 𝑎𝑛−3 𝑎𝑛−5 than n or negative should be replaced
with zeros.
0 𝑎𝑛 𝑎𝑛−2 𝑎𝑛−4

𝑎𝑛−1 𝑎𝑛−3 𝑎𝑛−5 𝑎𝑛−7 .. .. 0

𝑎𝑛 𝑎𝑛−2 𝑎𝑛−4 𝑎𝑛−6 .. .. 0
If all determinants have
0 𝑎𝑛−1 𝑎𝑛−3 𝑎𝑛−5 .. .. 0 positive (+ve) value,
𝐷𝑛 = 0 𝑎𝑛 𝑎𝑛−2 𝑎𝑛−4 .. .. 0 then this system is in
.. .. .. .. .. .. .. stable conditions.

.. .. .. .. .. .. ..
0 0 0 0 .. . . 𝑎0
 Routh-Hurwitz criterion
• The criterion is an algebraic method that provides information on the
absolute stability of a linear time invariant system that has a characteristics
equation with constant coefficients.
• The criterion represents a method of determining the location of zeros of a
polynomial with constant real coefficient with respect to the left and right
half of the s-plane without actually solving for the zeros.
• This criterion tests whether any of the roots of characteristic equation lies
in the right half in s-plane.
• The number of roots that lie on 𝒋𝝎-axis and in the right half plane are also
indicated.
• The Routh-Hurwitz criterion represents a method of determining the
location of zeros of a polynomial with constant real coefficients with
respect to the left half and right half of the s-plane without actually solving
for the zeros.
 State the Routh’s Stability Criterion
• Routh’s criterion state that “the number of roots of characteristics
equation with positive real parts is equal to the number of changes of sigh
of the coefficients in the first column”

• Therefore, the system is stable if all terms in first column have the same
sign. It should be noted that exact values of the terms in the first column is
not required to be known. Only the sign is the requirement.

𝐶(𝑠) 𝑏0 𝑠 𝑚 + 𝑏1 𝑠 𝑚−1 + … … … … … … … + 𝑏𝑚−1 𝑠 + 𝑏𝑚 𝐴(𝑠)

= =
𝑅(𝑠) 𝑎0 𝑠 𝑛 + 𝑎𝑠 𝑛−1 + … … … … … … … + 𝑎𝑛−1 𝑠 + 𝑎𝑛 𝐵(𝑠)

• Where, a’s and b’s are constant and 𝒎 ≤ 𝒏. By using Routh’s criterion to find the
roots of the polynomial A(s). i.e. poles of the linear system.

[1] K. Ogata, “Modern Control Engineering”, Third Edition, Prentice-Hall of India Private Limited.
 Conditions of Routh’s stability criterion

The linear system by Routh’s stability criterion types of

conditions are they are---

1. Sufficient condition:
Only the sign and the change of sign in the first column is required to
be known, not the exact value of them.

2. Necessary condition:
a. None of the coefficient of the characteristics equation should be
zero or missing.
b. Every coefficient should be real and should be the same sign.
 Limitation for Routh’s stability criterion
Routh’s stability criterion has a number of limitation. They are
given below:

1. This criterion can be used if only the equation is an

algebraic one and has real coefficients and none of them
being zero.

2. It is valid only for determining the roots of the

characteristic equation with respect to the left half or
right half of the complex s-plane.

3. The stability boundary is only for the 𝒋𝝎-axis of the s-

plane. It can not be used for any other plane.
Describe the procedure of Routh’s stability criterion
The nth order differential equation of the general form is as follows:
𝑑 𝑛 𝑥(𝑡) 𝑑 𝑛−1 𝑥(𝑡) 𝑑 𝑛−2 𝑥(𝑡) 𝑑𝑥 𝑡
𝑎𝑛 + 𝑎𝑛−1 +𝑎𝑛−2 + ⋯ … … … … … + 𝑎1 +𝑎0 𝑥 𝑡 = 𝑦(𝑡)
𝑑𝑡 𝑛 𝑑𝑡 𝑛−1 𝑑𝑡 𝑛−2 𝑑𝑡

Taking Laplace transform on both sides with zero initial condition

𝑋 𝑠 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + 𝑎𝑛−2 𝑠 𝑛−2 + − − − 𝑎1 𝑠 +𝑎0 = 𝑌(𝑠)

𝑌(𝑠)
𝑄 𝑠 = = 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + 𝑎𝑛−2 𝑠 𝑛−2 + − − − 𝑎1 𝑠+𝑎0 = 0
𝑋(𝑠)

This is the characteristics equation of the system. The coefficient of the

characteristics equation are arranged in the pattern shown in Routhian Array.

This coefficients are used to evaluate the rest of coefficients to complete the array
Describe the procedure of Routh’s stability criterion
𝑌(𝑠)
𝑄 𝑠 = = 𝑎𝑛 𝑠 𝑛 + 𝑎𝑛−1 𝑠 𝑛−1 + 𝑎𝑛−2 𝑠 𝑛−2 + − − − 𝑎1 𝑠+𝑎0 = 0
𝑋(𝑠)
The coefficient of the characteristics equation are arranged in the pattern shown in
Routhian Array.

Routhian Array 𝑎𝑛−1 ∙ 𝑎𝑛−2 − 𝑎𝑛 ∙ 𝑎𝑛−3

𝑏1 =
𝒔𝒏 𝑎𝑛 𝑎𝑛−2 𝑎𝑛−4 𝑎𝑛−6 … … … 𝑎𝑛−1
𝒔𝒏−𝟏 𝑎𝑛−1 𝑎𝑛−3 𝑎𝑛−5 𝑎𝑛−7 … … … 𝑎𝑛−1 ∙ 𝑎𝑛−4 − 𝑎𝑛 ∙ 𝑎𝑛−5
𝑏2 =
𝒔𝒏−𝟐 𝑏1 𝑏2 𝑏3 𝑏4 … … … 𝑎𝑛−1
𝒔𝒏−𝟑 𝑐1 𝑐2 𝑐3 𝑐4 … … … And so on

𝒔𝒏−𝟒 𝑑1 𝑑2 𝑑3 𝑑4 … … … 𝑏1 ∙ 𝑎𝑛−3 − 𝑎𝑛−1 ∙ 𝑏2

𝑐1 =
… … … … … … … … 𝑏1
𝑏1 ∙ 𝑎𝑛−5 − 𝑎𝑛−1 ∙ 𝑏3
𝒔𝟏 𝑐2 =
𝑏1
𝒔𝟎
And so on

𝑐1 ∙ 𝑏2 − 𝑏1 ∙ 𝑐2 𝑐1 ∙ 𝑏3 − 𝑏1 ∙ 𝑐3
𝑑1 = 𝑑2 = And so on
𝑐1 𝑏1
Theorems for Routh's Stability criterion
Theorem I: Division of a row
The coefficients of any row can be multiplied or divided by any positive number
without changing the signs of the first column.

Example: 𝑄 𝑠 = 𝑠 6 + 3𝑠 5 + 2𝑠 4 + 9𝑠 3 + 5𝑠 2 + 12𝑠 + 20 = 0

Routhian Array
𝒔𝟔 1 2 5 20

𝒔𝟓 3 9 12 (After divided by 3)
1 3 4
𝒔𝟒 -1 1 20
𝒔𝟑 4 24 (After divided by 4)
1 6
𝒔𝟐 7 20
22 (After multiplied by 7)
𝒔𝟏
7 There are two changes of signs in the first column.
22 Hence there are two roots with positive real parts i.e.
𝒔𝟎 20 the right half of s-plane. Hence, the system is unstable.
Theorems for Routh's Stability criterion
Theorem II: A zero coefficient in the first column
When the first term in a row is zero but not all other terms are zero, then either of the
following methods are used.
(a) Method I: Substitute a small positive number for the zero and produced to
evaluate the rest of the Routhian Array
Example: 𝑄 𝑠 = 𝑠 5 + 𝑠 4 + 2𝑠 3 + 2𝑠 2 + 3𝑠 + 15 = 0

Routhian Array Routhian Array

𝒔𝟓 1 2 3 𝒔𝟓 1 2 3
𝒔𝟒 1 2 5 𝒔𝟒 1 2 5
𝒔𝟑 0 -12 𝒔𝟑  -12
𝒔𝟐 ∞ 𝒔𝟐 2 + 12 15 (After replacing 0 by )
𝒔𝟏 
𝒔𝟏 A1 Where,
𝒔𝟎
−12 2𝛿 + 12 − 15𝛿 2
𝒔𝟎 15 𝐴1 =
2𝛿 + 12
= −12
There are two changes of signs in the first column. Hence there are two roots with positive
real parts i.e. the right half of s-plane. Hence, the system is unstable.
Theorems for Routh's Stability criterion
Theorem II: A zero coefficient in the first column
𝟏
(b) Method II: The original characteristics equation can be replaced by taking 𝒔 = 𝒙
and then calculating for 𝒙.
Example: 𝑄 𝑠 = 𝑠 5 + 𝑠 4 + 2𝑠 3 + 2𝑠 2 + 3𝑠 + 15 = 0
1 1 2 2 3
Routhian Array 𝑄 𝑠 = 5 + 4 + 3 + 2 + + 15 = 0
𝑥 𝑥 𝑥 𝑥 𝑥
𝒔𝟓 1 2 3
15𝑥 5 + 3𝑥 4 + 2𝑥 3 + 2𝑥 2 + 𝑥 + 1 = 0
𝒔𝟒 1 2 5
𝒔𝟑 0 -12 Routhian Array
𝒔𝟐 ∞ 𝒙𝟓 15 2 1
𝒔𝟏 𝟏 𝒙𝟒 3 2 1
Replacing by 𝒔 =
𝒔𝟎 𝒙 𝒙𝟑 -8 -4 (After divided by 4)
-2 -1
There are two changes of signs in the 𝒙𝟐 0.5 1 (After multiplied by 2)
first column. Hence there are two 1 2
roots with positive real parts i.e. the 𝒙𝟏 3
right half of s-plane. Hence, the
system is unstable. 𝒙𝟎 2
Theorems for Routh's Stability criterion
Theorem III: A zero row : When all the coefficient of one row are zero:
1. The auxiliary equation can be formed from the processing row.
2. The Routhian array can be completed by replacing the all zero by the coefficients
obtained by differentiating the auxiliary equation.
3. The roots of the auxiliary equation are also be the roots of original equation.
Example: 𝑄 𝑠 = 𝑠 4 + 2𝑠 3 + 11𝑠 2 + 18𝑠 + 18 = 0
Routhian Array The auxiliary equation, 𝒔𝟐 + 𝟗 = 𝟎 ∴ 𝒔 = ±𝒋𝟑
𝒔𝟒 1 11 18 Differentiating this equation, 𝟐𝒔 + 𝟎 = 𝟎

𝒔𝟑 2 18 (After divided by 9) Routhian Array

1 9
𝒔𝟓 1 11 18
𝒔𝟐 2 18 (After divided by 9)
1 9 𝒔𝟒 2 18
1 9
𝒔𝟏 0
𝒔𝟑 2 18
𝒔𝟎 1 9
𝒔𝟐 1 9
There are no changes of sign in the first column.
Hence there are no roots in the right half of s-plane. 𝒔𝟏 2 0
Hence, the system is stable. 𝒔𝟎 9
Assignments (All Examples in Books of B.C. Kuo & K. Ogata)
1. 𝑠 5 + 𝑠 4 + 10𝑠 3 + 72𝑠 2 + 172𝑠 + 240 = 0
2. 𝑠 5 + 𝑠 4 + 2𝑠 3 + 2𝑠 2 + 3𝑠 + 5 = 0
3. 𝑠 5 + 2𝑠 4 + 3𝑠 3 + 6𝑠 2 + 10𝑠 + 15 = 0
4. 𝑠 6 + 4𝑠 5 + 12𝑠 4 + 16𝑠 3 + 41𝑠 2 + 36𝑠 + 72 = 0
5. 𝑠 5 + 6𝑠 4 + 15𝑠 3 + 30𝑠 2 + 44𝑠 + 5 = 0
6. 𝑠 5 + 6𝑠 4 + 15𝑠 3 + 30𝑠 2 + 44𝑠 + 5 = 0
B.C. Kuo Example: 6-3 to 6-7, Problems: 6-1 to 6-5
Find the range of k for stability of the following
system using Routh’s/Root Hurwitz Criterion
R(s) 𝑘(𝑠 + 2) C(s)
𝑘(𝑠 + 2)
𝐺 𝑠 = +_
𝑠(𝑠 + 1)(𝑠 + 3)(𝑠 + 5) 𝑠(𝑠 + 1)(𝑠 + 3)(𝑠 + 5)
The characteristic equation
𝑄 𝑠 = 1 + 𝐺 𝑠 𝐻(𝑠) = 0 𝑘(𝑠+2)
 1 + 𝑠(𝑠+1)(𝑠+3)(𝑠+5) = 0
𝑠 𝑠 + 1 𝑠 + 3 𝑠 + 5 + 𝑘 𝑠 + 2 = 0 𝑠 4 + 9𝑠 3 + 23𝑠 2 + 15 + 𝑘 𝑠 + 2𝑘 = 0
𝟏𝟗𝟐−𝒌 𝟏𝟓+𝒌 −𝟏𝟔𝟐𝒌
where,
𝟏𝟗𝟐−𝒌
Routhian Array
𝟏𝟗𝟐−𝒌 𝒌 ≤ 𝟏𝟗𝟐
𝟒 For 𝒔𝟐 row, ≥𝟎 𝟏𝟗𝟐 − 𝒌 ≥ 𝟎
𝒔 1 23 2𝑘 𝟗

𝟏𝟗𝟐−𝒌 𝟏𝟓+𝒌 −𝟏𝟔𝟐𝒌

For 𝒔𝟏 row, ≥𝟎
𝒔𝟑 9 15 + 𝑘 𝟏𝟗𝟐−𝒌

192 − 𝑘  𝟏𝟗𝟐 − 𝒌 𝟏𝟓 + 𝒌 − 𝟏𝟔𝟐𝒌 ≥ 𝟎

𝒔𝟐 2𝑘
9  − (𝒌𝟐 − 𝟏𝟕𝟕𝒌 − 𝟐𝟖𝟖𝟎 + 𝟏𝟔𝟐𝒌) ≥ 𝟎

𝒔𝟏 𝐴 (𝒌 − 𝟔𝟏. 𝟗𝟒)(𝒌 + 𝟒𝟔. 𝟓) ≤ 𝟎  𝒌 − 𝟔𝟏. 𝟗𝟒 < 𝟎

𝒔𝟎 2𝑘  𝒌 + 𝟒𝟔. 𝟓 > 𝟎 𝒌 > −𝟒𝟔. 𝟓) 𝒌 < 𝟔𝟏. 𝟗𝟒

For 𝒔𝟎 row, 𝟐𝒌 ≥ 𝟎 𝒌>𝟎 Hence, for the stable condition, 6𝟏. 𝟗𝟒 > 𝒌 > 𝟎
Hence, just oscillate condition, 𝒌 = 𝟏𝟗𝟐 Hence, unstable condition, 6𝟏. 𝟗𝟒 < 𝒌 > 𝟏𝟗𝟐
Problem: Closed loop transfer function of a control system is-
𝐶(𝑠) 𝑘
=
𝑅(𝑠) 𝑠 5 + 𝑠 4 + 3𝑠 3 + 𝑠 2 + 1 + 𝑘 𝑠 + 𝑘
a) Find the range of value of k for which the time response is stable.
b) Select the values of k for which will produce imaginary roots.

The characteristic equation 𝑠 5 + 𝑠 4 + 3𝑠 3 + 𝑠 2 + 1 + 𝑘 𝑠 + 𝑘 = 0

Routhian Array 𝟏−𝟐𝒌 𝟐+𝒌 −𝟐𝟓𝒌
where, 𝐀=
𝒔𝟓 1 3 1+𝑘 𝟏−𝟐𝒌

𝒔𝟒 2 1 k
𝒔𝟑 5 2+𝑘 For 𝒔𝟐 row, (𝟏 − 𝟐𝒌) > 𝟎 𝒌 <0.5
𝟐𝒌 − 𝟏 < 𝟎
2 2
5 2+k For 𝒔𝟏 row,
𝟏−𝟐𝒌 𝟐+𝒌 −𝟐𝟓𝒌
>𝟎
𝟏−𝟐𝒌
𝒔𝟐 1 − 2𝑘 k  𝟏 − 𝟐𝒌 𝟐 + 𝒌 − 𝟐𝟓𝒌 > 𝟎
5  − (𝟐𝒌𝟐 + 𝟐𝟖𝒌 − 𝟐) > 𝟎
1-2k 5k
(𝒌 + 𝟏𝟒. 𝟎𝟕𝟏)(𝒌 − 𝟎. 𝟎𝟕𝟏) > 𝟎  𝒌 − 𝟎. 𝟎𝟕𝟏 < 𝟎
𝒔𝟏 𝐴
 𝒌 + 𝟏𝟒. 𝟎𝟕𝟏 > 𝟎 𝒌 > −𝟏𝟒. 𝟎𝟕𝟏 𝒌 < 𝟎. 𝟎𝟕𝟏
𝒔𝟎 𝑘
For 𝒔𝟎 row, 𝒌>𝟎 𝒌>𝟎 Hence, for the stable condition, 𝟎. 𝟎𝟕𝟏 > 𝒌 > 𝟎
Hence, just oscillate condition, 𝒌 = 𝟎. 𝟎𝟕𝟏
Problem: The characteristics equation of a polynomial is
𝑄 𝑠 = 𝑠 4 + 5𝑠 3 + (𝑘 + 1)𝑠 2 + 2𝑘 + 8 𝑠 + 2𝑘 = 0

Find the value of k for which all the roots of Q(s) will be in the left of line at 𝝈 = −𝟏 on the s-plane.

Shifting the axis of 𝒋𝝎 to the left by 𝜶 = 𝟏 𝒑=𝒔+𝟏 𝒔=𝒑−𝟏

The characteristic equation 𝑄 𝑠 = 𝑠 4 + 5𝑠 3 + (𝑘 + 1)𝑠 2 + 2𝑘 + 8 𝑠 + 2𝑘 = 0
𝑝4 + 𝑝3 + 3 + 𝑘 𝑝2 − 5𝑝 + 5𝑘 + 12 = 0

Problem: The characteristics equation of a polynomial is

𝑄 𝑠 = 𝑠 3 + 10.1𝑠 2 + 21𝑠 + 2 = 0

Define the system is stable or not for the shifting 𝒋𝝎-axis to the left by 𝝈 = 𝟎. 𝟐 on the s-plane.