The transfer function for a lowpass Butterworth filter is|H(j omega)|2 = 1/1+(om
ID: 1806252 • Letter: T
Question
The transfer function for a lowpass Butterworth filter is|H(j omega)|2 = 1/1+(omega / omega c)2N. We would like to design a filter with a passband frequency omega p and stopband frequency omega s such that |H(j omega)| 1- delta 1,|omega| omega p |H(j omega)| delta 2,|omega|> omega s Since|H(j omega)|is monotonically decreasing, the specifications will be met if we choose|H(j omega p)|= 1 - delta 1 |H(j omega s)|= delta 2 We would like the filter to have the following design specification: An attenuation of no more than 2 dB for|delta| 100 rad/s, and at least 30 dB for|delta| 600 rad/s. (Note: the formula for gain in dB is 20log10(|H(j omega)|). For example, a gain of 1/2 is a gain of about 20log10(l/2) -6 dB (an attenuation of about 6 dB).) Based upon the specifications for attenuation, what should delta 2 be?Explanation / Answer
2It is given that attenuation should be greater than or equal to 30 dB for |w|>= 600 rad/s
This means the Gain should be -30 dB or less in the region
=>
G(w) <= -30dB for |w|>= 600 rad/s
20log(|H(jw)| <= -30dB for |w|>= 600 rad/s
taking antilog
|H(jw)| <= 10^(-30/20) for |w|>= 600 rad/s
|H(jw)| <= 10^(-1.5) = 0.0316227766 for |w|>= 600 rad/s
=> 2 = 0.0316227766
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