Finite Impulse Response Filter Design - Digital Signal Processing: Fundamentals and Applications - page 234

h w ð1Þ¼hð1Þw ham ð1Þ¼0:1871 0:08 ¼ 0:01497

h w ð1Þ¼hð1Þw ham ð1Þ¼0:1871 0:08 ¼ 0:01497

Thus delaying h w ðnÞ by M ¼ 1 sample gives

b 0 ¼ b 2 ¼ 0:01496 and b 1 ¼ 0:2

b. The transfer function is

HðzÞ¼0:01497 þ 0:2z 1 þ 0:01497z 2

Using the technique described in Chapter 6, we have
Y ð z Þ
X ðzÞ ¼ HðzÞ¼0:01497 þ 0:2z 1 þ 0:01497z 2

Multiplying X(z) leads to

Y ðzÞ¼0:01497X ðzÞþ0:2z 1 X ðzÞþ0:01497z 2 X ðzÞ

Applying the inverse z-transform on both sides, the difference equation is obtained as

yðnÞ¼0:01497xðnÞþ0:2xðn 1Þþ0:01497xðn 2Þ

c. The magnitude frequency response and phase response can be obtained using the technique introduced in
Chapter 6. Substituting z ¼ e jU into HðzÞ, it follows that

Hðe jU Þ¼0:01497 þ 0:2e jU þ 0:01497e j2U
¼ e jU ð0:01497e jU þ 0:2 þ 0:01497e jU Þ

Using Euler's formula leads to

Hðe jU Þ¼e jU ð0:2 þ 0:02994 cos UÞ

Then the magnitude frequency response and phase response are found to be
Hðe jU Þ

¼ j0:2 þ 0:2994 cos Uj

Table 7.4 Frequency Response Calculation in Example 7.5

Hðe jU Þ Hðe jU Þ dB

: Hðe jU Þ
degrees

f [ Uf s =ð 2 pÞ 0 : 2D0 : 02994 cos U dB

U radians Hz

0 0 0.2299 0.2299 12.77 0

p=4 1,000 0.1564 0.2212 13.11 45

p=2 2,000 0.2000 0.2000 13.98 90

3p=4 3,000 0.1788 0.1788 14.95 135

p 4,000 0.1701 0.1701 15.39 180