1
Routh – Hurwitz (RH) Stability Test
A little history:
http://www.cai.cam.ac.uk/students/study/engineering/engineer03l/c
erouth.htm
General Statement
A BIBO system is stable if for all time ,t, an input r(t )  M , results
with an output c(t )  P for finite M and P. It is a necessary
condition for stability that all poles of the transfer function be
located in the --LHP--------. A system with poles on ---jw-----is
defined as marginally stable system.
Motivation (Rational)
Consider the following closed loop feedback control system.
Correction: H = 1/(s+2) not 1/(s+1)
Let us study the transient response for three different values of the
forward gain, K.
2
A-
K =1
Pole-Zero Map
4
3
System: Gclosed
Pole : -0.445 + 2.97i
Damping: 0.148
Overshoot (%): 62.4
Frequency (rad/sec): 3
Imaginary Axis
2
1
0
System: Gclosed
Pole : -1
Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 1
-1
-2
-3
-4
-2
-1
0
Real Axis
1
2
3
Step Response K =1
2.5
Amplitude
2
1.5
1
0.5
0
0
1
2
3
Time (sec)
4
5
6
4
K = 32/9
Pole - Zero Map K = 32/9
3
System: Gclosed
Pole : 9.16e-016 + 2.58i
Damping: -3.55e-016
Overshoot (%): 100
Frequency (rad/sec): 2.58
2
Imaginary Axis
B-
1
0
-1
System: Gclosed
Pole : -1.5 - 1.76i
Damping: 0.65
Overshoot (%): 6.83
Frequency (rad/sec): 2.31
-2
-3
-2
-1.5
-1
-0.5
Real Axis
0
0.5
5
Step Response K = 32/9
4.5
4
3.5
Amplitude
3
2.5
2
1.5
1
0.5
0
-0.5
0
5
10
15
Time (sec)
20
25
6
K=5
Pole-Zero Map, K =5
3
2
Imaginary Axis
C-
1
System: Gclosed
Pole : -1.8 + 2.03i
Damping: 0.663
Overshoot (%): 6.2
Frequency (rad/sec): 2.71
System: Gclosed
Pole : 0.299 + 2.59i
Damping: -0.115
Overshoot (%): 144
Frequency (rad/sec): 2.6
0
-1
-2
-3
-2
-1.5
-1
-0.5
Real Axis
0
0.5
7
20
Step Response K = 32/9
15
Amplitude
10
5
0
-5
-10
-15
-20
0
2
4
Time (sec)
6
8
8
Observation:
The systems poles moved from Stable to marginally stable to
unstable for slight changes in the forward gain, K.
Questions:
Is there an analytical way of determining the range of K for
which the system is stable?
9
Routh – Hurwitz Test
How is the RH test performed?
First:
Recall:
1-
For a closed loop system:
GCL ( s) 
2-
G( s)
1  G( s) H ( s)
The poles of the closed loop system are the roots of the
denominator of the transfer function, GCL(s):
Define:
D( s)  1  G( s) H ( s) : This is referred to as the
Characteristic Polynomial
3-
Let us standardize a form by expressing D(s) in
polynomial form:
Let
D( s )  an s n  an 1s n 1  an  2 s n 1  ....  a1s  a0
At this point, it is time to explore the RH test.
10
Second:
The RH Test Steps:
1- Build the starting RH array
Sn
an
an-2
an-4
.
.
.
an-1
an-3
an-5 .
.
.
b1
b2
b3
.
.
.
Sn-3
c1
c2
c3
.
.
.
Sn-4
.
.
.
S0
d1
d2
d3
.
.
.
.
.
.
.
.
.
Sn-1
Sn-2
.
.
.
Formed from the
coefficients of D(s)
These
coefficients are
calculated
11
2-
Calculate the remaining coefficients of the RH array
Note: The first column in the RH array is the “basis” vector.
a a
 an an 3
b1  n 1 n  2
an 1
an
an-2
an-4
an-1
an-3
an-5
a a
 an an 5
b2  n 1 n  4
an 1
b1
b2
c1 
b1 an 3  b2 an 1
b1
c2 
b1 an 5  b3 an 1
b1
d1 
c1 b2  c2 b1
c1
12
3-
Now the RH array is formed, what to do next?
All one has to do is look at the first column in the RH array. The number of sign changes will be
the number of poles in the right half s-plane. (Thus unstable system if the number is non-zero)
Before exploring some special cases of the RH array, let us revisit the original system of
today’s lecture.
13
Example – Determine the range of K to have a stable system.
The system: Correction: H= 1/(s+2) not 1/(s+1)
G(s) 
10 K
s s  s  10
GCL 
10 K ( s  2)
s ( s  s  10)( s  2)  10 K

2

2
D( s)  s 4  3s3  12s 2  20s  10 K
H (s) 
1
s2
14
1- The RH array
S4
1
12
10K
S3
3
20
0
S2
16/3
10K
0
S1
20-5.625K
0
10K
S
0
D( s)  s 4  3s3  12s 2  20s  10 K
Auxiliary
Equation
15
2-
Investigate the first column.
a. Whenever possible, it is desired to have no sign changes in the vector.
b. Coming down the basis vector, the signs are: +, +, + ,?,?
The last two terms have to be positive in order to have a stable system:
10K > 0, then
K> 0
(1)
20 – 5.625K > 0
K < 3.556
Therefore, it is concluded that the system is stable for
0 <K<3.556
Now it is obvious why when K =5, the system resulted in unstable response !!
16
For the next lecture:
12-
Investigate the auxiliary Equation (new idea)
Look at the special cases.
17
18
Recap with an in-class Example:
For the characteristic polynomial of the closed loop transfer
function:
D(s)=1+G(s)H(s) = s 4  s 3  3 s 2  2 s  4  K
(1) Determine the range of K necessary for stable response (if any)
(2) Determine the auxiliary equation
16/3 S2 +10K = 0, K= 3.556, therefore, S =  j 2.58 rad / sec .
(3) Determine the frequency of oscillation for a marginally stable
system.
19
Special Cases When the RH tabulation array terminates
prematurely:
1- The first element in any one row of the RH array is zero while
other elements in the same array are NOT
Example:
D(s) =
s 4  s 3  2s 2  2s  3  0
The RH tabulation is:
S4
S3
S2
:
:
:
1
1
0
2
2
3
3
0
Well, the remaining calculated elements for s1 and s0 will be
infinity  have a problem.
Solution:
When this happens, substitute the zero with a small positive
number,  , and continue building the table.
Which results in,
S4
S3
S2
S1
:
:
:
:
1
1
0
:
3
S

2  3


3

2
2
3
0
3
0
0
20
Two sign changes in the first column – Thus, there are two poles
in the RHP.
For fun:
Solve for the roots of D(s), results in ?
(Note: the epsilon method may not work appropriately if D(s) has
pure imaginary roots.)
2-
All the elements in one row are zeros
Example:
D(s) =
S5 + 4S4 + 8S3 + 8S2 + 7s + 4 = 0
The RH array is:
S5
S4
S3
S2
S1
:
:
:
:
:
1
4
6
4
0
8
8
6
4
0
7
4
0
0
The next terms will be?
Solution:
1- Form an auxiliary equation with S2
A(s) =
2-
4s2
+
4
=
0
Take the derivative of A(s) w.r.t. s
21
d (4s 2  4)
 8s  0
ds
Thus, the coefficients of dA ds are the coefficients of: 8s + 0 =0
which are 8 and zero – Use these to replace the elements in the S1
row.
Thus
S1
S0
:
:
8
4
0
No sign changes  Stable
Recap Example:
Example
Determine the range of the values of K (if exists) for which the
system is stable.
R(s)
1
In 1
Y(s)
K
1
Out 1
s2+4s+8
G(s)
H(s)
3
s2+2s+12
22
Solution
K
2
s
 4s  8
GCL 



K
3
1


2
2
 s  4 s  8  s  2 s  12 
GCL 
K ( s 2  2 s  12)
( s 2  2 s  12)( s 2  4 s  8)  3 K
4
3
2
D(s) = s  6 s  28 s  64 s  (96  3 K )
RH Tabulation:
S4
S3
S2
S1
S0
:
:
:
:
:
1
6
17.3
30.82-1.039K
96+3K
28
64
96+3K
0
96+3K
0
0
96+3K>0  K>-32
30.82-1.039K>0  K<29.7
Thus,
The range of K for a stable response is:
-32 < K < 29.7
23
Final Value Theorem:
Recall:
Voltage across a cap in a simple series RC circuit (E is the applied
DC voltage)


t
RC


Vc(t)= E 1  e




t
 

RC

  E
lim Vc (t )  lim  E  1  e

t 
t   

This should make sense to us:
Vc after a long time [ t>5 ], the voltage across the cap is equal
to the Thevinin’s voltage. In this case Vc = E.
Let us visit this from LaPlace point of view:
1
Cs
1
1
RC
Vc ( s )  R( s )
 R( s )
 R( s )
1
1  RC s
s 1
R
RC
Cs
Note:
R(s) = E/s E is the magnitude of the DC voltage applied. r(t) is
the input.
Therefore,
24
1
E
RC
Vc ( s) 
s s 1
RC
According to the final value theorem:
lim c(t )  lim s.C ( s) c(t) is the output {if the limit exists}
t 
s 0
E 1

RC
  E (agrees with previous
lim sVc ( s )  lim s 
1
s
s 0
s 0  s 

RC 
result.)
Using the final value theorem, one can predict, analytically, the
steady state value without resorting to the time domain function.