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