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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 Mtr )( , results
with an output Ptc )( 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.
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The Matlab code I used is: clear all K =1; num1 = [1]; den1=[1 0]; num2 =[10]; den2=[1 1 10]; num=conv(num1,num2); den=conv(den1,den2); G=K*tf(num,den); % This is the open loop transfer function numH=[1]; denH=[1 2]; H=tf(numH,denH); GCL=feedback(G,H); % This is the closed loop transfer function pole(GCL) disp('hit any key to continue ...') pause
step(GCL)
% note: May use pzmap command to plot the poles and zeros.
A- K =1
Pole-Zero Map
Real Axis
Ima
gin
ary
Ax
is
-2 -1 0 1 2-4
-3
-2
-1
0
1
2
3
4
System: Gclosed
Pole : -0.445 + 2.97i
Damping: 0.148
Overshoot (%): 62.4
Frequency (rad/sec): 3
System: Gclosed
Pole : -1
Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 1
Poles of the closed loop system, GCL(s) are:
-0.4452 + 2.9688i
-0.4452 - 2.9688i
-1.1096 + 0.0000i
-1.0000 + 0.0000i
And the the step response is given below
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0 1 2 3 4 5 60
0.5
1
1.5
2
2.5
Step Response K =1
Time (sec)
Am
plitu
de
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B- K = 32/9
Pole - Zero Map K = 32/9
Real Axis
Imag
ina
ry A
xis
-2 -1.5 -1 -0.5 0 0.5-3
-2
-1
0
1
2
3
System: Gclosed
Pole : 9.16e-016 + 2.58i
Damping: -3.55e-016
Overshoot (%): 100
Frequency (rad/sec): 2.58
System: Gclosed
Pole : -1.5 - 1.76i
Damping: 0.65
Overshoot (%): 6.83
Frequency (rad/sec): 2.31
Run the code again but with K = 32/9.
Poles are:
0.0000 + 2.5820i
0.0000 - 2.5820i
-1.5000 + 1.7559i
-1.5000 - 1.7559i
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0 5 10 15 20 25-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5Step Response K = 32/9
Time (sec)
Am
plitu
de
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C- K = 5
Pole-Zero Map, K =5
Real Axis
Imag
inary
Axis
-2 -1.5 -1 -0.5 0 0.5-3
-2
-1
0
1
2
3
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
The poles are:
-1.7993 + 2.0330i
-1.7993 - 2.0330i
0.2993 + 2.5873i
0.2993 - 2.5873i
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Typo: The title should say: Step response for K = 5.
0 2 4 6 8-20
-15
-10
-5
0
5
10
15
20Step Response K = 32/9
Time (sec)
Am
plitu
de
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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?
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Routh – Hurwitz Test
How is the RH test performed?
First:
Recall:
1- For a closed loop system:
)()(1
)()(
sHsG
sGsGCL
2- The poles of the closed loop system are the roots of the
denominator of the transfer function, GCL(s):
Define:
)()(1)( sHsGsD : This is referred to as the
Characteristic Polynomial
3- Let us standardize a form by expressing D(s) in
polynomial form:
Let
011
21
1 ....)( asasasasasD nn
nn
nn
At this point, it is time to explore the RH test.
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Second:
The RH Test Steps:
1- Build the starting RH array
Sn
Sn-1
Sn-2
Sn-3
Sn-4
.
.
.
S0
an an-2 an-4 . . .
an-1 an-3 an-5 . . .
b1 b2 b3 . . .
c1 c2 c3 . . .
d1 d2 d3 . . .
. . . .
. . . .
.
Formed from the
coefficients of D(s)
These
coefficients are
calculated
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2- Calculate the remaining coefficients of the RH array
Note: The first column in the RH array is the “basis” vector.
1
3211
n
nnnn
a
aaaab
1
5412
n
nnnn
a
aaaab
1
12311
b
ababc nn
1
13512
b
ababc nn
1
12211
c
bcbcd
an an-2 an-4
an-1 an-3 an-5
b1 b2
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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.
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Example – Determine the range of K to have a stable system.
The system: Correction: H= 1/(s+2) not 1/(s+1)
10
10)(
2
sss
KsG
2
1)(
ssH
Kssss
sKGCL
10)2)(10(
)2(102
KsssssD 1020123)( 234
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1- The RH array
KsssssD 1020123)( 234
S4
S3
S2
S1
S0
1 12 10K
3 20 0
16/3 10K 0
20-5.625K 0
10K
Auxiliary
Equation
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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 !!
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For the next lecture:
1- Investigate the auxiliary Equation (new idea)
2- Look at the special cases.
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Recap with an in-class Example:
For the characteristic polynomial of the closed loop transfer
function:
D(s)=1+G(s)H(s) = Kssss 423234
(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 = .sec/. radj 582
(3) Determine the frequency of oscillation for a marginally stable
system.
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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) = 0322234 ssss
The RH tabulation is:
S4 : 1 2 3
S3 : 1 2 0
S2 : 0 3
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 : 1 2 3
S3 : 1 2 0
S2 : 3 0
S1 :
332
0
S0 : 3
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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 : 1 8 7
S4 : 4 8 4
S3 : 6 6 0
S2 : 4 4 0
S1 : 0 0
The next terms will be?
Solution:
1- Form an auxiliary equation with S2
A(s) = 4s2 + 4 = 0
2- Take the derivative of A(s) w.r.t. s
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0844
2
sds
sd )(
Thus, the coefficients of ds
dA are the coefficients of: 8s + 0 =0
which are 8 and zero – Use these to replace the elements in the S1
row.
Thus
S1 : 8 0
S0 : 4
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)
Y(s)
Out 1
1
H(s)
3
s +2s+122
G(s)
K
s +4s+82In1
1
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Solution
122
3
841
84
22
2
ssss
K
ss
K
GCL
Kssss
ssKGCL
384122
122
22
2
))((
)(
D(s) = )( Kssss 39664286234
RH Tabulation:
S4 : 1 28 96+3K
S3 : 6 64 0
S2 : 17.3 96+3K 0
S1 : 30.82-1.039K 0
S0 : 96+3K
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
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Final Value Theorem:
Recall:
Voltage across a cap in a simple series RC circuit (E is the applied
DC voltage)
Vc(t)=
RC
teE 1
EeEtV RCt
tc
t
1lim)(lim
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:
RCs
RCsRsRC
sR
CsR
CssRsVc 1
1
1
1
1
1
)()()()(
Note:
R(s) = E/s E is the magnitude of the DC voltage applied. r(t) is
the input.
Therefore,
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RCs
RC
s
EsVc 1
1
)(
According to the final value theorem:
)(.lim)(lim sCstcst 0
c(t) is the output {if the limit exists}
E
RCs
RC
s
EsssV
sc
s
1
1
00lim)(lim (agrees with previous
result.)
Using the final value theorem, one can predict, analytically, the
steady state value without resorting to the time domain function.