Fundamentals of System Analysis and Synthesis - Digital Signal Processing in Power System Protection and Control - page 50

One can notice that the same result can be reached by substituting
z ¼ exp ð jxT S Þ in the transfer function ( 4.39 ). In the same way one can obtain
spectrum of a signal directly from its Z transform. It is illustrated further with
another examples.
Whatever methods are used to present the frequency response one can notice
that it can also be expressed in the form:

H ð jx Þ¼ H ð jx Þ

j exp f j arg ½ H ð jx Þg;

j ð 4 : 42 Þ

H ð jx Þ j is a magnitude of H ð jx Þ , and arg ½ H ð jx Þ is a argument of

where j

H ð jx Þ:
Substituting that in Eq. 4.40 one obtains:

y ð n Þ¼ H ð jx Þ j exp f j ½ nxT S þ arg ð H ð jx ÞÞg j ð 4 : 43 Þ

and for instance, for x ð n Þ¼ cos ð nxT S Þ one gets:
y ð n Þ¼ H ð jx j j cos f nxT S þ arg ½ H ð jx Þg:
Sometimes, especially when magnitudes of frequency response for some fre-
quencies differ substantially, the logarithmic frequency response defined as:

L ð x Þ¼ 20 log ð H ð jx Þ j jÞ ð 4 : 44 Þ

can be used. The units of ( 4.44 ) are decibels [dB].

Example 4.15 Determine frequency responses of the systems described by the
following difference equations:

(a) y ð n Þ¼ 0 : 5 ½ x ð n Þþ x ð n 1 Þ

(b) y ð n Þ¼ 0 : 5 ½ x ð n Þ x ð n 1 Þ

(c) y ð n Þ¼ x ð n Þþ 0 : 9y ð n 1 Þ

Solutions From ( 4.31 ) results that the corresponding systems' impulse responses

are:

(a) h ð n Þ¼ 0 : 5 ½ d ð n Þþ d ð n 1 Þ

(b) h ð n Þ¼ 0 : 5 ½ d ð n Þ d ð n 1 Þ

h ð n Þ¼ 0 :ð n (c)

Coming out either from the difference equations or from the impulse responses
one can determine the transfer functions for all the cases, in the form:

H ð z Þ¼ 0 : 5 ð 1 þ z 1 Þ

(a)

H ð z Þ¼ 0 : 5 ð 1 z 1 Þ

(b)

1 0 : 9z 1 ¼ z

1

(c) H ð z Þ¼

z 0 : 9
Now, substituting z ¼ exp ð jxT S Þ the frequency responses of the considered
system are reached: