1.

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 mentionedI 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»

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



Discussion

No Comment Found

Related InterviewSolutions