Question

What is the system function of the Butterworth filter with specifications as pass band gain KP=-1 dB at ΩP=4 rad/sec and stop band attenuation greater than or equal to 20dB at ΩS=8 rad/sec?

(a) \(\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)

(b) \(\frac{1}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+1}\)

(c) \(\frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)

(d) None of the mentioned

I got this question during an online exam.

The doubt is from Design of Low Pass Butterworth Filters in chapter Digital Filters Design of Digital Signal Processing

Answer 1

Right answer is (c) \(\frac{2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)

Easy explanation: From the given question,

KP=-1 dB, ΩP=4 rad/sec, KS=-20 dB and ΩS=8 rad/sec

We find out order as N=5 and ΩC=4.5787 rad/sec

We know that for a 5th order normalized low pass Butterworth filter, system equation is given as

H5(s)=\(\frac{1}{(s+1)(s^2+0.618s+1)(s^2+1.618s+1)}\)

The specified low pass filter is obtained by applying low pass-to-low pass transformation on the normalized low pass filter.

That is, Ha(s)=H5(s)|s→s/Ωc

=H5(s)|s→s/4.5787

upon calculating, we get

Ha(s)=\({2012.4}{s^5+14.82s^4+109.8s^3+502.6s^2+1422.3s+2012.4}\)