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Q1: Determine the stability of a closed loop control system whose characteristic equation is S6+s5+3s4+3s3+2s2+s+1=0. Investigate the stability using Routh Hurwitz criterion. Write a C/C++ program to determine the stability

Answers

Answered by 123utkarsh5678
0

Answer:ECE 680 Modern Automatic Control Routh’s Stability Criterion June 13, 2007 1

ROUTH’S STABILITY CRITERION

Consider a closed-loop transfer function

H(s) = b0s

m + b1s

m−1 + · · · + bm−1s + bm

a0s

n + a1s

n−1 + · · · + an−1s + an

=

B(s)

A(s)

(1)

where the ai

’s and bi

’s are real constants and m ≤ n. An alternative to factoring the

denominator polynomial, Routh’s stability criterion, determines the number of closedloop poles in the right-half s plane.

Algorithm for applying Routh’s stability criterion

The algorithm described below, like the stability criterion, requires the order of A(s) to

be finite.

1. Factor out any roots at the origin to obtain the polynomial, and multiply by −1 if

necessary, to obtain

a0s

n + a1s

n−1 + · · · + an−1s + an = 0 (2)

where a0 6= 0 and an > 0.

2. If the order of the resulting polynomial is at least two and any coefficient ai

is zero

or negative, the polynomial has at least one root with nonnegative real part. To

obtain the precise number of roots with nonnegative real part, proceed as follows.

Arrange the coefficients of the polynomial, and values subsequently calculated from

them as shown below:

s

n a0 a2 a4 a6 · · ·

s

n−1 a1 a3 a5 a7 · · ·

s

n−2

b1 b2 b3 b4 · · ·

s

n−3

c1 c2 c3 c4 · · ·

s

n−4 d1 d2 d3 d4 · · ·

.

.

.

.

.

.

.

.

.

s

2

e1 e2

s

1

f1

s

0

g0

(3)

where the coefficients bi are

b1 =

a1a2 − a0a3

a1

(4)

b2 =

a1a4 − a0a5

a1

(5)

b3 =

a1a6 − a0a7

a1

(6)

.

.

.

ECE 680 Modern Automatic Control Routh’s Stability Criterion June 13, 2007 2

generated until all subsequent coefficients are zero. Similarly, cross multiply the

coefficients of the two previous rows to obtain the ci

, di

, etc.

c1 =

b1a3 − a1b2

b1

(7)

c2 =

b1a5 − a1b3

b1

(8)

c3 =

b1a7 − a1b4

b1

(9)

.

.

.

d1 =

c1b2 − b1c2

c1

(10)

d2 =

c1b3 − b1c3

c1

(11)

.

Explanation:

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