FIR filter design - Continuous and Discrete Time Signals and Systems - page 690

Example 15.6
Design a bandpass FIR filter with the following specifications:

(i) pass-band edge frequencies, Ω p1 = 0 . 375 π and Ω p2 = 0 . 5 π radians/s;

= 0 . 625 π radians/s;
(iii) stop-band attenuations, δ s1 > 50 dB and δ s2 > 50 dB.

(ii) stop-band edge frequencies, Ω s1 = 0 . 25 π and Ω s2

Plot the gain-frequency characteristics of the designed bandpass filter.

Solution
The cut-off frequencies of the bandpass filter are given by

= 0 . 5 (0 . 25 π + 0 . 375 π ) = 0 . 3125 π

Ω c1

and

= 0 . 5 (0 . 5 π + 0 . 625 π ) = 0 . 5625 π.

Ω c2

The normalized cut-off frequencies are given by Ω n1 =

Ω c1 /π

= 0 . 3125 and

=

Ω c2 /π

= 0 . 5625. The impulse response of an ideal bandpass filter is

Ω n2

given by

h ibp [ k ] = 0 . 5625 sinc[0 . 5625( k − m )] − 0 . 3125 sinc[0 . 3125( k − m )] .
(15.35)
Since only the stop-band attenuations are specified, and these are both equal to
50 dB, the minimum attenuation A = 50 dB.
The shape parameter β of the Kaiser window is computed to be

β = 0 . 1102(50 − 8 . 7) = 4 . 5513 .

The transition bands Ω c1 and Ω c2 for the bandpass FIR filter are given by

Ω c1 = 0 . 375 π − 0 . 25 π = 0 . 125 π

and

Ω c2 = 0 . 625 π − 0 . 5 π = 0 . 125 π,

which lead to the normalized transition BW of Ω n = 0 . 125 .

The length N of the Kaiser window is given by

50 − 7 . 95
2 . 285 π 0 . 125

N ≥ = 46 . 8619 .

Rounded to the closest higher odd number, N = 47, and the value of m in Eq.
(15.35) is 23. The expression for the Kaiser window is as follows:

1 − [( k − 23) / 23] 2

I 0

4 . 5513

0 ≤ k ≤ 46

w kaiser [ k ] =

(15.36)

I 0 [4 . 5513]

0 otherwise.