Lecture Notes on Control Systems/D.

Ghose/2012 87

11.3.1 Some special cases

Routh table is easy to form in most cases, but there could be some cases when we need
to do some extra work.
Case 1: The first element in any one row of the Routh array is zero, but the other
elements are not.
Why is this a problem? Because, subsequent rows have to be obtained by dividing by this
element and if it is zero, we end up with ∞.
An Example:

D(s) = s3 − 3s + 2
= s3 + 0s2 − 3s + 2
= a0 s3 + a1 s2 + a2 s + a3

Then the Routh array is given by,

s3 1 −3

s2 0 2

s1 ∞ 0

s0

We can perhaps guess that there is at least one unstable pole. But how many are there?
There are many ways to handle this situation. One is to multiply (s + a) to the original
polynomial, where a could be any number. But remember that if you take a to be
negative then you must account for it when you are counting the number of poles on
the RHS of the s-plane. So, to avoid any confusion select a positive a.
In the above example, let us select a = 3. Then, the new characteristic polynomial is,

D̂(s) = (s + 3)D(s) = (s + 3)(s3 − 3s + 2)

= s4 + 3s3 − 3s2 − 7s + 6
= a0 s4 + a1 s3 + a2 s2 + a1 s + a0

The Routh array is now,

Lecture Notes on Control Systems/D. Ghose/2012 88

s4 1 −3 6

s3 3 −7 0

2
s2 − 3
6 0

s1 20 0 0

s0 6 0 0

So, there are 2 changes in sign and so there are two poles. Note that since a is positive,
it does not affect the number of poles on the RHS of the s-plane and hence the number
of sign changes.
It is easy to verify that

s3 − 3s + 2 = (s − 1)2 (s + 2)

A word of caution here. The value of a has to be selected carefully or through trial and
error. Not all values of a will serve our purpose. For example, consider a = 1. Then,

(s3 − 3s + 2)(s + 1) = s4 + s3 − 3s2 − s + 2

The corresponding Rout table is

s4 1 −3 2

s3 1 −1 0

s2 −2 2 0

s1 0 0 0

s0

So we end up with a row of zeros.

There is another method to tackle the above situation. This is by putting a small number
 > 0 in place of the zero in the Routh array, and then continue. If we do this then, the
original Routh array becomes.
Lecture Notes on Control Systems/D. Ghose/2012 89

s3 1 −3

s2  2

−3−2
s1 
0

s0 2

Now, for  > 0, the term (−3 − 2)/ is a negative number that approaches −2/ as
 → 0. So, there are two sign changes.
A word of caution: The  method does not always work, especially when there are two
pure imaginary roots.
Case 2: When all the numbers in a row in the Routh array are zeros, then one of the
following may be true.

1. There is a pair of real roots with opposite signs (See Figure 11.5(a)).
2. There is a pair of pure imaginary roots (See Figure 11.5(b)).
3. There are pairs of complex conjugate roots placed symmetrically about the origin
of the s-plane (See Figure 11.5(c)).

Figure 11.5: Pole locations for Case 2

To handle this case, follow the procedure given below.

1. Form an auxiliary equation by taking its coefficients from the row just above the
row of zeros1 .
1
If the auxiliary equation has order two, then there are two equal and opposite roots. If it has order
four, then there must be two pairs of equal and opposite roots. If the auxiliary equation has odd order,
then one root is on the origin and the rest are distributed according to the rules governing even order
auxiliary equation. Moreover, all these roots can be obtained by simply solving for the roots of the
auxiliary equation.
Lecture Notes on Control Systems/D. Ghose/2012 90

2. Take its derivative with respect to s.

3. Replace the row of zeros with the coefficients of the derivative of the auxiliary
equation.

4. Continue with the Routh array. The number of sign changes indicates the number
of poles on the RHS of the s-plane2 .

An Example:

D(s) = s4 + s3 − 3s2 − s + 2

The Routh array is,

s4 1 −3 2

s3 1 −1 0

s2 −2 2 0 ← coefficients of auxiliary equation

s1 0 0 0

−4 0 0 ← new s1 row

s0 2 0

The s1 row is all zeros, so the auxiliary equation is formed from the s2 row as,

A(s) = −2s2 + 2

which on differentiating,

dA(s)
= −4s
s
This is used to obtain the new fourth row.
So, there are two sign changes and the equation has two poles on the RHS. In fact, it
can be verified that,

D(s) = s4 + s3 − 3s2 − s + 2 = (s − 1)2 (s + 2)(s + 1)

2
However, note that there may not be any sign changes at all. In which case, it implies that there
are no roots with positive real parts or no roots on the RHS of the s-plane. But the solution of the
auxiliary equation will reveal that there are roots on the imaginary axis. So, just examining the sign of
the first column is not enough but we need to also evaluate the roots of the auxiliary equation.
Lecture Notes on Control Systems/D. Ghose/2012 91

Another Example:

D(s) = s6 + s5 − 2s4 − 3s3 − 7s2 − 4s − 4

The Routh array is,
s6 1 −2 −7 −4

s5 1 −3 −4 0

s4 1 −3 −4 0 ← auxiliary equation

s3 0 0 0 0

4 −6 0 0 ← new row

s2 −6/4 −4 0 0

s1 −50/3 0 0 0

s0 −4 0 0 0

The auxiliary equation and its derivative are,

A(s) = s4 − 3s2 − 4 = 0
dA(s)
= 4s3 − 6s
s

From the Routh array, we can see that there is one sign change and so there is one root
on the RHS of the s-plane.
It can be verified that,

D(s) = s6 + s5 − 2s4 − 3s3 − 7s2 − 4s − 4

= (s + 2)(s − 2)(s2 + 1)(s2 + s + 1)

Note that there is a pair of roots on the imaginary axis. Examine the roots of the
auxiliary equation.

A(s) = s4 − 3s2 − 4 = (s2 − 4)(s2 + 1)

Hence, some of the roots of the original equation (especially those that were causing the
zero row!) can be identified by just solving the auxiliary equation.
Lecture Notes on Control Systems/D. Ghose/2012 92

A word of caution: Routh-Hurwitz criterion can be applied only when the coefficients
are real (not complex).
Another interesting result: If any element in any row of the Routh array is negative then
the system is unstable.

11.4 Application of Routh-Hurwitz Criterion

Consider the following feedback system.

Figure 11.6: A feedback system with proportional control

Let
s+1
G1 (s) =
(s + 2)(s − 3)
G2 (s) = k2
The closed-loop transfer function is
k1 G1 (s)
G(s) =
1 + k1 G1 (s)G2 (s)
k1 (s + 1)
=
(s + 2)(s − 3) + k1 k2 (s + 1)
k1 (s + 1)
= 2
s + (k1 k2 − 1)s + (k1 k2 − 6)
The denominator polynomial is,
D(s) = s2 + (k1 k2 − 1)s + (k1 k2 − 6)
The Routh array is,
s2 1 k1 k2 − 6

s1 k1 k2 − 1 0

s0 k1 k2 − 6 0
Lecture Notes on Control Systems/D. Ghose/2012 93

For the system to be stable

k1 k2 > 6 and k1 k2 > 1
which implies that any value of k1 and k2 such that k1 k2 > 6 should make the system
stable.
Another Example:

Figure 11.7: A more complicated feedback system

Let
s+1
G1 (s) =
s(s − T )
1
G2 (s) =
1 + sQ
The overall transfer function is,
(s + 1)(1 + sQ)
G(s) =
s(s − T )(1 + sQ) + (s + 1)
The characteristic polynomial,
D(s) = s(s − T )(1 + sQ) + (s + 1)
= Qs3 + (1 − T Q)s2 + (1 − T )s + 1
Just going by the necessary condition that says that all coefficients have to be positive
in order for the system to be stable we can write
Q > 0, T Q < 1, and T < 1
But this is not sufficient. So, let us form the Routh array,
s3 Q 1−T

s2 1 − TQ 1

Q
s1 (1 − T ) − 1−T Q
0

s0 1 0
Lecture Notes on Control Systems/D. Ghose/2012 94

So, suppose we assume Q > 0 then,

Q > 0
1 − TQ > 0 ⇒ TQ < 1
Q Q
(1 − T ) − > 0 ⇒ (1 − T ) >
1 − TQ 1 − TQ
Since T Q < 1, we can write the last inequality as,

(1 − T )(1 − T Q) > Q
⇒ (1 − T ) − (1 − T )T Q > Q
⇒ (1 − T ) > Q[(1 − T )T + 1]
1−T
⇒ Q<
1 + T − T2
Can T > 1? No, because then 1 − T will be negative and the previous inequality will be
violated as Q > 0.
√
5−1
Can T < 0? Yes, in fact so long as 0 < T < − 2
, we have a feasible solution for Q.
This can be easily shown.
Look at the following table where we get the range of Q for some given values of T that
will satisfy the inequality.

T (1 − T )(1 − T Q) > Q Q

0 Q<1 0<Q<1

0.1 Q < 0.82 0 < Q < 0.82

0.2 Q < 0.487 0 < Q < 0.487

0.4 Q < 0.483 0 < Q < 0.483

0.8 Q < 0.174 0 < Q < 0.174

−0.1 Q < 1.23 0 < Q < 1.23

Lecture Notes on Control Systems/D. Ghose/2012 95

Figure 11.8: The feasible region in the (T, Q)-space

PROBLEM SET 5

1. Use the Hurwitz determinants to find the stability of systems having the following
characteristic equations:
(a) s3 + 20s2 + 9s + 100 = 0
(b) 3s4 + 10s3 + 5s2 + s + 2 = 0
(c) s4 + 2s3 + 6s2 + 8s + 8 = 0

2. Use the Routh-Hurwitz criterion to find the stability of systems having the follow-
ing characteristic equations:
(a) s3 + 20s2 + 9s + 200 = 0
(c) s6 + 2s5 + 8s4 + 12s3 + 20s2 + 16s + 16 = 0

3. Consider the following characteristic equations. Determine the values of K that

corresponds to a stable system. Use the Routh-Hurwitz criterion.
(a) s4 + 22s3 + 10s2 + 2s + K = 0
(b) s4 + 20Ks3 + 5s2 + (10 + K)s + 18 = 0
(c) s3 + (K + 0.5)s2 + 4Ks + 50 = 0

4. Consider the following system with a unity feedback:

Lecture Notes on Control Systems/D. Ghose/2012 96

K
G(s) =
s(1 + T s)

(a) What would be the values of K and T to ensure stability?

(b) Suppose we want the poles to be on the left of the line s = −3, then what
would be the values of K and T to satisfy this requirement?
(c) Plot the feasible values of K and T on the (K, T ) plane.
(d) Repeat (a), (b), and (c) when

K(s + 1)
G(s) =
s(1 + T s)(1 + 2s)