Modern Control Engineering

Example Problems and Solutions 527

If the last term on the right-hand side of Equation (7–31) does not contain any poles or zeros
in the closed contour in the splane,F¿(s)/F(s)is analytic in this contour except at point s=–z 1.
Then, referring to Equation (7–30) and using the residue theorem, which states that the integral
ofF¿(s)/F(s)taken in the clockwise direction around a closed contour in the splane is equal to
–2pjtimes the residues at the simple poles of F¿(s)/F(s),or

we have

where total number of zeros of F(s)enclosed in the closed
contour in the splane

total number of poles of F(s)enclosed in the closed

contour in the splane

[The kmultiple zeros (or poles) are considered kzeros (or poles) located at the same point.]
SinceF(s)is a complex quantity,F(s)can be written

and

Noting that F¿(s)/F(s)can be written

we obtain

If the closed contour in the splane is mapped into the closed contour in the F(s)plane, then

The integral is zero since the magnitude is the same at the initial point and the final
point of the contour Thus we obtain

The angular difference between the final and initial values of uis equal to the total change in
the phase angle of F¿(s)/F(s)as a representative point in the splane moves along the closed
contour. Noting that Nis the number of clockwise encirclements of the origin of the F(s)plane
andu 2 - u 1 is zero or a multiple of 2prad, we obtain

u 2 - u 1

2 p

=-N

u 2 - u 1

2 p

=P-Z

G.

DGdln^ ∑F∑ ln^ ∑F∑

I

F¿(s)

F(s)

ds=

I

G^ dln^ ∑F∑+j

I

Gdu=j

3

du= 2 pj(P-Z)

G

F¿(s)

F(s)

=

dln ∑F∑

ds

+j

du

ds

F¿(s)

F(s)

=

dlnF(s)

ds

lnF(s)=ln ∑F∑+ju

F(s)=∑F∑eju

P =m 1 +m 2 +p=

Z=k 1 +k 2 +p=

I

F¿(s)

F(s)

ds=- 2 pjCAk 1 +k 2 +pB-Am 1 +m 2 +pBD=- 2 pj(Z-P)

I

F¿(s)

F(s)

ds=- 2 pjaa residuesb