EE481 Control Systems
Kunio Takaya
Electrical and Computer Engineering
University of Saskatchewan
October 15, 2009
Routh Hurwitz Stability Criterion
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University of Saskatchewan, Electrical Engineering
EE 481.3 Control Systems
April 2009, Kunio Takaya
Textbook: Norman S. Nise, “Control Systems Engineering” Fifth
Edition, John Wiley & Sons, Inc. 2008, ISBN-13 978-0471-79475-2.
Marks: Midterm Exam: 30%, Final Exam 55%, and Assignments
15%
1. Modeling in the frequency domain
• Laplace transform
• Transfer functions
2. Modeling in the time domain
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• Linear differential equations
• State-Space representation
3. Time response
• Second-Order Systems
• Poles and zeros
• Time domain solution of state equations
4. Reduction of multiple subsystems
• Block diagrams and Signal-Flow graphs
• Mason’s rule
• Similarity transformations
5. Stability
• Routh-Hurwitz criterion
• Stability in State-Space
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6. Steady-State errors
• Steady-State error for unity gain feedback
• Steady-State error for disturbance
7. Root Locus techniques
• Sketching the root locus
• Transient response design via gain adjustment
8. Design via root locus
• Cascade compensation
• Improving transient response and steady-state error
9. Frequency response techniques
• Bode plots
• Nyquist diagrams
• Systems with time delay
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10. Design via frequency response
• Lag compasation
• Lead compensation
Classes: MWF 8:30-9:30 a.m. 2B01 Engineering
My office: 3B31
Email: kunio.takaya@usask.ca
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Stability
The response of control systems consists of (1) natural response
and forced response, or (2) zero input response and zero state
response. For natural (zero input) response, a system is,
1. stable if the natural response approaches zero as time
approaches infinity.
2. unstable if the natural response approaches infinity as time
approaches infinity.
3. marginally stable if the natural response neither decays nor
grows but remians constant or oscillates.
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For the total response, a system is
1. stable if every bounded input yields a bounded output.
2. unstable if any bounded input yields an unbounded output.
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The Routh Hurwitz Stability Criterion
The Routh Hurwitz stability criterion is a tool to judge the
stability of a closed loop system without solving for the poles of the
closed loop system.
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Generating a Routh-Hurwitz table
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Example 6.1
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Any row of the Routh table may be multiplied by a positive
number.
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Interpreting the basic Routh table
The number of roots of the polynomial that are in the right half
plane is equal to the number of sign changes in the first column.
In Example 6.1,
s3
1
31
0
s2
1
103
0
s1
−72
0
0
s0
103
0
0
There are two sign changes in the first column.
1 =⇒ −72
and
− 72 =⇒ 103
Therefore, two of the four roots of s3 + 10s2 + 31s + 1030 = 0 are in
the right half plane of s-plane.
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Special case 1: Zero in the first column
T (s) =
10
s5 + 2s4 + 3s3 + 6s2 + 5s + 3
When zero is resulted in the first column, replace 0 with a small .
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Examine the first column by assuming is a either positive or
negative number.
Since there are two sign changes, the polynomial has two poles in
the right half plane.
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Special case 2: An entire row is zero.
T (s) =
10
s5 + 7s4 + 6s3 + 42s2 + 8s + 56
The row if s3 is entirely zero. Let the polynomial above (row of s4 )
be P (s). Then, differentiate P (s).
P (s)
=
s4 + 6s2 + 8
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dP (s)
ds
=
4s3 + 12
Then, replace the row of s3 with the coefficients of dP (s)/ds.
Then, proceed normally.
There are no sign changes, so the system T (s) is not unstable.
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More about the case of entire row is zero.
1. The row previous to the row of zeros contain the even
polynomial that is a factor of the original polynomial.
s4 + 6s2 + 8 = 0, in the example above.
2. The root to find is with respect to s2 , so s2 = a yields
√
s = ± a. Therefore, the roots are symmetric about the
imaginary axis. If s is complex, s = σ + jω, its complex
conjugate is also a root, meaning symmetric about the real axis
(because the original polynominal is of real coefficients.)
3. Everything from the row containing the even polynomial down
to the end of the Routh table is a test of only the even
polynomial. In the above example, s4 + 6s2 + 8 = 0 has no
roots in RHP. Then, all has to be on the imaginary axis.
√
4
2
2
2
s + 6s + 8 = (s + 2)(s + 4) = 0 ⇒ s = ±j 2, and ± j2.
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The above example has one root in LHP, and four roots on the
imaginary axis (marginally stable).
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Special case 2: One more example
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T (s) = 8
s + s7 + 12s6 + 22s5 + 39s4 + 59s3 + 48s2 + 38s + 20
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Stability Design for Feedback Systems
G(s)
K
T (s) =
= 3
1 + G(s)
s + 18s2 + 77s + K
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• If K > 1386, s1 term is negative. Hence, two poles are in the
right half plane (unstable).
• If K < 1386, there are no sign changes. Hence, the feedback
system is stable.
• If K = 1386, the row of s1 is zero. By replacing with the
coefficients of the dP (s)/ds, there are no sign changes. The
system T (s) is, however, marginally stable.
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Stability of a State Equation
Conversion from a state equation to a transfer function
The vector/matrix form of the state equation and output equation
for single input and single output is,

 ẋ = Ax + bu
 y = cx + du
Take the Laplace transform,

 sX(s) − x(0) = AX(s) + bU (s)
 Y (s) = cX(s) + dU (s)
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Transfer function assumes x(0) = 0.

 sX(s) − x(0) = AX(s) + bU (s)
 Y (s) = cX(s) + dU (s)
sX(s) − AX(s)
=
(sI − A)X(s) = bU (s)
X(s)
=
(sI − A)−1 bU (s)
Y (s)
=
cX(s) + dU (s)
=
c(sI − A)−1 bU (s) + dU (s)
=
[c(sI − A)−1 b + d] U (s)
Therefore,
Y (s)
U (s)
=
c(sI − A)−1 b + d
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=
=
c(sI − A)−1 b, if d = 0
c adj(sI − A) b
det(sI − A)
Since
(sI − A)−1 =
The poles of the system
adj(sI − A)
det(sI − A)
Y (s)
are, therefore,
U (s)
det(sI − A) = jsI − Aj = 0
which is the eigen values of A.
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Example: Apply Routh-Hurwitz stability criterion.


ẋ = 

0
2
−10
3

1

10






8
1 x +  0 
 u(t)
−5 −2
0
y = [1 0 0] x

1 0

sI−A = s 
 0 1
0 0
0
 
0
 

0 
− 2
1
−10
3
1


s
−3
 

8
1 
 =  −2 s − 8
−5 −2
10
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det(sI − A) = s3 − 6s2 − 7s − 52
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−1


−1 

s+2
The Routh table is,
s3
1
s2
−3
s1
−1
0
0
s0
−26
0
0
−7
0
−26 0
Since there is one sign change, one of three poles is in the right half
plane (unstable).
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