Discrete and Digital Signals and Systems - Digital Signal Processing: An Introduction with MATLAB and Applications - page 104

2 f s

z −1
h o = 1/T s

x ( n )

h 1 = − 1/T s
y ( n )

=
ω T s

Ω

−

π π

0

π /2

P − Z
Diagram

Ω

X Re ( z )

0

− π

− π /2

π

Fig. 2.28 Digital differentiator-I with frequency response and p-z diagram

y ð n Þ¼ h ð n Þ x ð n Þ¼ X

N 1

h ð k Þ x ð n k Þ

k ¼ 0

¼ h 0 ð n Þþ h 1 x ð n 1 Þþþ h N 1 x ½ n ð N 1 Þ

It can be seen that ( 2.32 ) represents a FIR filter with two non-zero samples impulse
response {h(n) = (1/T s )d(n) - (1/T s )d(n - 1)}, i.e., h(0) =-h(1) = 1/T s . From
this one can find the system transfer function by taking the z-transform of both
sides of ( 2.32 ):

Y ð z Þ¼ 1
T s

½ X ð z Þ X ð z Þ z 1 ¼½ð 1 z 1 Þ= T s X ð z Þ

X ð z Þ ¼ ð 1 z 1 Þ

) H ð z Þ¼ Y ð z Þ

T s
Now consider the magnitude and phase response:

H ð e jX Þ¼ H ð z Þj z ¼ exp ð jX Þ ¼ 1
T s

½ 1 e jX

¼ 1

T s e 2 jX ½ e þ 2 jX e 2 jX
¼ 1

:

T s e 2 jX ½ 2j sin ð X = 2 Þ
¼ 2j
T s

e 2 jX sin

1
2 X

Hence, the magnitude response is given by:

j sin j¼ 2f s sin

j H ð e jX Þj¼ 2
T s

1
2 X

1
2 xT s

¼ 2f s j sin ð pf = f s Þj;

and the phase response is: