Routh Hurwitz Analysis

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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 SystemsApril 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 equations4. Reduction of multiple subsystems

Block diagrams and Signal-Flow graphs Masons rule

Similarity transformations5. 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 disturbance7. Root Locus techniques

Sketching the root locus

Transient response design via gain adjustment8. 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: [email protected]

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1 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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2 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 halfplane 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 0s0 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) = 10s5 + 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 evenpolynomial 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 theimaginary axis. If s is complex, s = +j, 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.

s4 + 6s2 + 8 = (s2 + 2)(s2 + 4) = 0 s = j2, 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

T(s) =20

s8 + s7 + 12s6 + 22s5 + 39s4 + 59s3 + 48s2 + 38s + 20

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Stability Design for Feedback Systems

T(s) =G(s)

1 + G(s) =K

s3 + 18s2 + 77s + K

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If K > 1386, s1 term is negative. Hence, two poles are in theright half plane (unstable).

If K < 1386, there are no sign changes. Hence, the feedbacksystem is stable.

If K = 1386, the row of s1 is zero. By replacing with thecoefficients 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,

x = 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)1bU(s)

Y(s) = cX(s) + dU(s)

= c(sI A)1bU(s) + dU(s)= [c(sI A)1b + d] U(s)

Therefore,

Y(s)

U(s)= c(sI A)1b + d

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= c(sI A)1b, if d = 0=

c adj(sI A) bdet(sI A)

Since

(sI A)1 = adj(sI A)det(sI A)

The poles of the systemY(s)

U(s)are, therefore,

det(sI

A) =jsI

Aj

= 0

which is the eigen values of A.

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Example: Apply Routh-Hurwitz stability criterion.

x =

0 3 1

2 8 1

10

5

2

x +

10

0

0

u(t)

y = [1 0 0] x

sIA = s

1 0 0

0 1 0

0 0 1

0 3 1

2 8 1

10 5 2

=

s 3 12 s 8 110 5 s + 2

det(sI A) = s3 6s2 7s 52

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The Routh table is,

s3 1 7 0

s2

3 26 0s1 1 0 0s0 26 0 0

Since there is one sign change, one of three poles is in the right half

plane (unstable).

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Routh Hurwitz Analysis