Modern Control Engineering

526 Chapter 7 / Control Systems Analysis and Design by the Frequency-Response Method

Solution.The given sinusoidal transfer function G(jv)can be written as follows:

where

Then

Hence, we see that the plot of G(jv)is a circle centered at (0.5,0)with radius equal to 0.5. The

upper semicircle corresponds to 0vq, and the lower semicircle corresponds to

• qv0.

A–7–6. Prove the following mapping theorem: Let F(s)be a ratio of polynomials in s. Let Pbe the num-

ber of poles and Zbe the number of zeros of F(s)that lie inside a closed contour in the splane,

with multiplicity accounted for. Let the closed contour be such that it does not pass through any

poles or zeros of F(s). The closed contour in the splane then maps into the F(s)plane as a closed

curve. The number Nof clockwise encirclements of the origin of the F(s)plane, as a representa-

tive point straces out the entire contour in the splane in the clockwise direction, is equal to Z-P.

Solution.To prove this theorem, we use Cauchy’s theorem and the residue theorem. Cauchy’s

theorem states that the integral of F(s)around a closed contour in the splane is zero if F(s)is

analytic#within and on the closed contour, or

Suppose that F(s)is given by

whereX(s)is analytic in the closed contour in the splane and all the poles and zeros are located

in the contour. Then the ratio F¿(s)/F(s)can be written

(7–30)

This may be seen from the following consideration: If is given by

then has a zero of kth order at s=–z 1. Differentiating F(s)with respect to syields

Hence,

(7–31)

We see that by taking the ratio , the kth-order zero of becomes a simple pole of

Fˆ¿(s)Fˆ(s).

Fˆ¿(s)Fˆ(s) Fˆ(s)

Fˆ¿(s)

Fˆ(s)

=

k

s+z 1

+

X¿(s)

X(s)

Fˆ¿(s)=kAs+z 1 Bk-^1 X(s)+As+z 1 BkX¿(s)

Fˆ(s)

Fˆ(s)=As+z 1 BkX(s)

Fˆ(s)

F¿(s)

F(s)

= a

k 1

s+z 1

+

k 2

s+z 2

+pb- a

m 1

s+p 1

+

m 2

s+p 2

+pb+

X¿(s)

X(s)

F(s)=

As+z 1 Bk^1 As+z 2 Bk^2 p

As+p 1 Bm^1 As+p 2 Bm^2 p

X(s)

I

F(s)ds= 0

aX-

1

2

b

2

+Y^2 =

Av^2 T^2 - 1 B^2

4 A 1 +v^2 T^2 B^2

+

v^2 T^2

A 1 +v^2 T^2 B^2

=

1

4

X=

v^2 T^2

1 +v^2 T^2

, Y=

vT

1 +v^2 T^2

G(jv)=X+jY

#For the definition of an analytic function, see the footnote on page 447.

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