Finite Impulse Response Filter Design - Digital Signal Processing: Fundamentals and Applications - page 228

EXAMPLE 7.3
a. Calculate the filter coefficients for a 5-tap FIR bandpass filter with a lower cutoff frequency of 2,000 Hz and an
upper cutoff frequency of 2,400 Hz and a sampling rate of 8,000 Hz.
b. Determine the transfer function and plot the frequency responses with MATLAB.
Solution:
a. Calculating the normalized cutoff frequencies leads to

U L ¼ 2pf L =f s ¼ 2p 2; 000=8; 000 ¼ 0:5p radians

U H ¼ 2pf H =f s ¼ 2p 2; 400=8; 000 ¼ 0:6p radians
Since 2M þ 1 ¼ 5 in this case, using the equation in Table 7.1 yields

<
:

U H U L
p

n ¼ 0

hðnÞ¼ (7.14)

sinðU H n Þ
np

sinðU L n Þ
np

n s 0 2 n 2

Calculations for noncausal FIR coefficients are listed as

hð0Þ¼ U H U L
p

¼ 0:6p 0:5p
p

¼ 0:1

The other computed filter coefficients via Equation (7.14) are

hð1Þ¼ sin½0:6p 1
1 p

sin½0:5p 1
1 p

¼0:01558

hð2Þ¼ sin½0:6p 2
2 p

sin½0:5p 2
2 p

¼0:09355

Using symmetry leads to

hð1Þ¼hð1Þ¼0:01558

hð2Þ¼hð2Þ¼0:09355

Thus, delaying hðnÞ by M ¼ 2 samples gives

b 0 ¼ b 4 ¼0:09355

b 1 ¼ b 3 ¼0:01558; and b 2 ¼ 0:1

b. The transfer function is achieved as
HðzÞ¼0:09355 0:01558z 1 þ 0:1z 2 0:01558z 3 0:09355z 4
To complete Example 7.3, the magnitude frequency response plotted in terms of Hðe jU Þ dB ¼ 20 log 10 jHðe jU Þj
using MATLAB Program 7.1 is displayed in Figure 7.8 .