A secondorder bandpass filter is required to satisfy the fol

A second-order bandpass filter is required to satisfy the following specifications: f_s = 8,000 Hz f_0= 1000 Hz Bandwidth = 60 Hz Find the transfer function and difference equation by the pole- zero placement method.

Solution

H(z)=(1-a)(1-z^-2)/[1-b(1+a)z^-1-az^-2]

Given fo=1000 Hz, Fs=8000, BW= 60

Hence wo= 2*fo/fs= /4

Bw=2*60/8000= 3/200

b=cos(/4)

2a/(1+a^2) = cos(3/200)

After solving. a=0.868

H(z)= (1-0.868)(1-z^-2)/[1-1.321z^-1-0.868z^-2]

H(z)=0.074(1-z^-2)/(1.15-1.52z^-1-z^-2)

1.15y(n)-1.52y(n-1)-y(n-2)=0.074[x(n)-x(n-2)]