hw8soln - ECE 202 - Linear Circuit Analysis II Purdue...

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ECE 202 - Linear Circuit Analysis II Purdue University, Spring 2009 Homework Set 3 Solutions Solution 29 (i) H ( s ) = K ( s 2 + 64) ( s + 0.2) 2 + 64 lim s →∞ H ( s ) = K = 1 (ii) MATLAB code » num=[1 0 64]; » den=[1 0.4 64.04]; » w=0:0.01:20; » freqs(num,den,w) 1
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The range of frequencies 7 ω < 9 is rejected. Solution 30 (i) H ( s ) = K s 2 [( s + 1) 2 + 64][( s + 2) 2 + 256] | H ( j 6) | = 1 K = 196.4002 (ii) (iii) The filter has a bandpass characteristic. If we look at the frequency response, the frequencies 6 ω 20 approximately have magnitude greater than 1. Hence these frequencies are passed without attenuation. Other frequencies outside this range are rejected. 2
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Solution 31 (a) H ( s ) = 0.5 s 3 + 2 s 2 + 2 s + 1 = 0.5 ( s + 1)( s 2 + s + 1) p 1 , p 2 , p 3 = 1, - 0.5 + 0.866 i , - 0.5 - 0.866 i | p 1 |=| p 2 |=| p 3 | = 1 (b) (c) K m = 10 C new = C old K m K f K f = C old K m C new = 6.2834 = 2 π L new = K m L K f 3
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= 3.1831 H H new ( s ) = 0.5
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hw8soln - ECE 202 - Linear Circuit Analysis II Purdue...

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