Fundamentals of System Analysis and Synthesis - Digital Signal Processing in Power System Protection and Control - page 39

Solution

X ð z Þ¼ X

1

1
1 z 1 ¼

z
z 1 ;

z n ¼

n ¼ 0

\1or jj [ 1 or should hold.

z 1

where for assuring series convergence

Example 4.6 Basing on the result of Example 4.5 and the Z transform features
(Table 4.3 )

determine Z equivalents of the signals: (a) y ð t Þ¼ t1 ð t Þ ;

(b) y ð t Þ¼ exp ð at Þ 1 ð t Þ ; (c) y ð t Þ¼ cos ð x 0 t Þ 1 ð t Þ:
Solutions

(a)

¼ zT S

d
dz

z
z 1

ð z 1 Þ z
ð z 1 Þ 2

zT S
ð z 1 Þ 2 :

Y ð z Þ¼ z f t1 ð t Þg ¼ zT S

¼

(b)

z
z 1

z
z exp ð aT S Þ :

Y ð z Þ¼ Z f exp ð at Þ 1 ð t Þg ¼

z ¼ z exp ð aT S Þ ¼

(c) Using an appropriate decomposition of the cosine function and previous

result (example b) one may derive:

Y ð z Þ¼ Z f cos ð x 0 t Þ 1 ð t Þg ¼ 0 : 5Z f 1 ð t Þ exp ð jx 0 t Þþ 1 ð t Þ exp ð jx 0 t Þg

¼

z
z exp ð jx 0 T S Þ

z
z exp ð jx 0 T S Þ

zz cos ð x 0 T S ½
z 2 2z cos ð x 0 T S Þþ 1 :

¼ 0 : 5

The opposite task to presented above is calculation of the signal samples for
given Z transform. There are various possibilities to realize the task. Three most
frequently used are the following: the equivalent of power series, the fractional
decomposition and the method of residua.

If Z transform can be represented as a power series given by the equation:

X ð z Þ¼ a 0 þ a 1 z 1 þ a 2 z 2 þþ: ð 4 : 13 Þ

then comparing it with the sum ( 4.8 ) results in equalities:

u ð 0 Þ¼ a 0 ; u ð T S Þ¼ a 1 ; u ð 2T S Þ¼ a 2 ; etc : ð 4 : 14 Þ

The sampled signal can thus be written in the form:

u ð t Þ¼ a 0 d ð t Þþ a 1 d ð t T S Þþ a 2 d ð t 2T S Þþþ;

ð 4 : 15 Þ

which is the solution of the problem. It can be added here that there are many
methods to find power series and one of them is division of numerator by
denominator.