Question

What is the system function for a two pole band pass filter that has the centre of its pass band at ω=π/2, zero its frequency response characteristic at ω=0 and at ω=π, and its magnitude response is 1/√2 at ω=4π/9?

(a) \(0.15\frac{1-z^{-2}}{1+0.7z^{-2}}\)

(b) \(0.15\frac{1+z^{-2}}{1-0.7z^{-2}}\)

(c) \(0.15\frac{1-z^{-2}}{1-0.7z^{-2}}\)

(d) \(0.15\frac{1+z^{-2}}{1+0.7z^{-2}}\)

The question was asked during an online interview.

This is a very interesting question from LTI System as Frequency Selective Filters topic in section Frequency Analysis of Signals and Systems of Digital Signal Processing

Answer 1

Correct option is (a) \(0.15\frac{1-z^{-2}}{1+0.7z^{-2}}\)

To explain: Clearly, the filter must have poles at P1,2=re^±jπ/2 and zeros at z=1 and z=-1. Consequently the system function is

H(z)=\(G\frac{(z-1)(z+1)}{(z-jr)(z+jr)} = G \frac{(z^2-1)}{(z^2+r^2)}\)

The gain factor is determined by evaluating the frequency response H(ω) of the filer at ω=π/2. Thus we have,

H(π/2) = \(G \frac{2}{1-r^2} = 1=>G = \frac{1-r^2}{2}\)

The value of r is determined by evaluating the H(ω) at ω=4π/9. Thus we have

|H(4π/9)|^2=\(\frac{(1-r^2)^2}{4}\frac{2-2cos⁡(8π/9)}{1+r^4+2r^2 cos⁡(8π/9)}\)=1/2

On solving the above equation, we get r^2=0.7.Therefore the system function for the desired filter is

H(z)=\(0.15\frac{1-z^{-2}}{1+0.7z^{-2}}\)