Question

If H(s)=\(\frac{1}{s^2+s+1}\) represent the transfer function of a low pass filter (not Butterworth) with a pass band of 1 rad/sec, then what is the system function of a band pass filter with a pass band of 10 rad/sec and a center frequency of 100 rad/sec?

(a) \(\frac{s^2}{s^4+10s^3+20100s^2+10^5 s+1}\)

(b) \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+1}\)

(c) \(\frac{s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)

(d) \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)

I had been asked this question in unit test.

My question is from Design of Low Pass Butterworth Filters in chapter Digital Filters Design of Digital Signal Processing

Answer 1

Right answer is (d) \(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\)

The best I can explain: The low pass-to-band pass transformation is

\(s\rightarrow\frac{s^2+\Omega_u \Omega_l}{s(\Omega_u-\Omega_l)}\)

Thus the required band pass filter has a transform function as

Ha(s)=\(\frac{100s^2}{s^4+10s^3+20100s^2+10^5 s+10^8}\).