Example: Design of 9-coefficient LP FIR filter: Design a nine-coefficient (or 9-point or 9-tap) FIR digital filter to related an ideal low-pass filter with a cut-off frequency ωc= 0.2p. The value response, ∠Hd(ω),is provided below. Take Hd(ω) = 0.
Solution The impulse response of the provided filter is
Since hd(n) ≠ 0 for n < 0 that is a noncausal filter. The rest of the design is aimed at taking up with a noncausal relation of the above impulse response.
For a rectangular window of size 9, the related impulse response is calculated by evaluating hd(n) for -4 ≤ n ≤ 4 on a computer. In the MATLAB segment below, which creates the hd(n) coefficients for -50 ≤ n ≤ 50, division by zero for n = 0 provides "NaN" (Not a Number), while all the other coefficients are correct. In preparing the frequency response |H(ω)| we paste and copy all the coefficients except for hd(0) which is shown by hand.
Aside (MATLAB) The segment below creates and stem-plots the hd(n) coefficients.
%Calculate hdn = (sin (0.2*pi*n)) / (pi*n) and stem plot n = -50: 50, hdn = (sin(0.2*pi*n)) ./(pi*n), stem(n, hdn)
xlabel('n'), ylabel('hd(n)'); grid; title ('hd(n) = (sin (0.2*pi*n)) / (pi*n)')
n = -50 to 50
Warning: Divide by zero.
hdn = -0, -0.0038, -0.0063, -0.0064, -0.0041, 0, 0.0043 0.0070 0.0072,
0.0046 -0, -0.0048 -0.0080 -0.0082 -0.0052 0, 0.0055 0.0092, 0.0095
0.0060 -0, -0.0065 -0.0108 -0.0112 -0.0072 0, 0.0078, 0.0132 0.0138
0.0089 -0, -0.0098 -0.0168 -0.0178 -0.0117 0, 0.0134 0.0233 0.0252
0.0170 -0, -0.0208 -0.0378 -0.0432 -0.0312, 0, 0.0468 0.1009 0.1514
0.1871 NaN 0.1871 0.1514 0.1009, 0.0468 0.0000 -0.0312 -0.0432
-0.0378 -0.0208 -0.0000 0.0170 0.0252, 0.0233 0.0134 0.0000 -0.0117
-0.0178 -0.0168 -0.0098 -0.0000 0.0089, 0.0138 0.0132 0.0078 0.0000
-0.0072 -0.0112 -0.0108 -0.0065 -0.0000, 0.0060 0.0095 0.0092 0.0055
0.0000 -0.0052 -0.0082 -0.0080 -0.0048, -0.0000 0.0046 0.0072 0.0070
0.0043 0.0000 -0.0041 -0.0064 -0.0063, -0.0038 -0.0000
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