Digital Signal Processing Systems, Basic Filtering Types, and Digital Filter Realizations - Digital Signal Processing: Fundamentals and Applications - page 178

Poles: z ¼ 0:5, jzj ¼ 0:5 < 1; z ¼0:707 j0:707,
jzj¼

q

ð0:707Þ

2 2

þð0:707Þ ¼ 1.

The zero and poles are plotted in Figure 6.10 . Since the outmost poles are first order at z ¼0:707 j 0:707
and are on the unit circle, the system is marginally stable.
d. Zeros are z ¼0:5 j 0:5.
Poles: z ¼ 1, jzj ¼ 1; z ¼ 1, jzj ¼ 1; z ¼1, jzj ¼ 1;, z ¼ 0:6
jzj¼0:6 < 1.

The zeros and poles are plotted in Figure 6.10 . Since the outmost pole is a multiple order (second order) pole at
z ¼ 1 and is on the unit circle, the system is unstable.

6.4 DIGITAL FILTER FREQUENCY RESPONSE
From the Laplace transfer function, we can achieve the analog filter steady-state frequency response
HðjuÞ by substituting s ¼ ju into the transfer function HðsÞ . That is,

HðsÞj s¼ju ¼ HðjuÞ
Then we can study the magnitude frequency response jHðjuÞj and phase response : HðjuÞ . Similarly,

in a DSP system, using the mapping in Equation (6.16) , we substitute z ¼ e sT s¼ju ¼ e juT into the
z-transfer function HðzÞ to acquire the digital frequency response, which is converted into the

magnitude frequency response Hðe juT Þ and phase response : jHðe juT Þj . That is,
HðzÞj z¼e juT ¼ Hðe juT Þ¼

Hðe juT Þ : Hðe juT Þ

(6.18)

Let us introduce a normalized digital frequency in radians in the digital domain:

U ¼ uT (6.19)

Then the digital frequency response in Equation (6.18) becomes
Hðe j U Þ¼HðzÞ z¼e j U ¼ Hðe j U Þ : Hðe j U Þ

(6.20)

The formal derivation for Equation (6.20) can be found in Appendix D.
Now we verify the frequency response via the following simple digital filter. Consider a digital
filter with a sinusoidal input of amplitude K ( Figure 6.11 ) .
We can determine the system output yðnÞ , which consists of the transient response y tr ðnÞ and the
steady-state response y ss ðnÞ . We find the z-transform output as

FIGURE 6.11
System transient and steady-state frequency responses.