hw 2 solutoins

# Be sure to indicate important horizontal frequencies

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Draw the asymptotic magnitude Bode plots for the following transfer functions. Be sure to indicate important horizontal frequencies (in rad/s), vertical gain levels (in dB) and slopes (in dB/dec). Solution : First simplify the transfer functions to put them in a more appropriate form for drawing the asymptotic Bode plots. (a) H ( s ) = ( s + 1 , 000) ( s + 8 , 000) ( s + 4 , 000) ( s + 40 , 000) = 0 . 05 (1 + s/ 1 , 000) (1 + s/ 8 , 000) (1 + s/ 4 , 000) (1 + s/ 40 , 000) Solution : The low freqency gain is 0 . 05 → - 26dB. The vertical difference between 1000 and 4000 rad/s is computed by +20dB/dec × log(4000 / 1000)dec = 14dB (this is the slope times the number of decades). Similarly the vertial difference between 8000 and 40000 rad/s is given by 20 log(40000 / 8000) = 14dB. Using these differences and the starting point of - 26dB the horizontal levels of - 14dB and 0dB can be computed as shown. - 26dB - 14dB 0dB 12dB 14dB +20dB/dec +20dB/dec 1000 4000 8000 40000 rad/s (b) H ( s ) = ( s + 40 , 000) ( s + 1 , 000) 2 ( s + 400) ( s + 10 , 000) 2 = (1 + s/ 40 , 000) (1 + s/ 1 , 000) 2 (1 + s/ 400) (1 + s/ 10 , 000) 2 Solution : The low freqency gain is 1 0dB. The vertical difference between 1000 and 400 rad/s is computed by - 20dB/dec × log(1000 / 400)dec = - 8dB; therefore, the gain at 1000 rad/s is 8dB lower than 0dB or just - 8dB. The vertical difference between 10000 and 1000 rad/s is just 20dB since the frequency difference is 1 decade. The vertical difference between 10000 and 40000 rad/s is computed by - 20 log(40000 / 10000) = - 12dB; therefore, the gain at 40000 rad/s is 12dB lower than 12dB or just 0dB. - 8dB 8dB 0dB 0dB 20dB 12dB +20dB/dec - 20dB/dec - 20dB/dec 400 1000 10000 40000 rad/s Dr. Vahe Caliskan 7 of 8 Posted: February 8, 2013

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ECE 412 Introduction to Filter Synthesis Homework #2 Solutions University of Illinois at Chicago Spring 2013 (c) H ( s ) = 10 ( s + 5 , 000) 4 ( s + 20 , 000) 4 ( s + 500) 4 ( s + 200 , 000) 4 = 10 (1 + s/ 5 , 000) 4 (1 + s/ 20 , 000) 4 (1 + s/ 500) 4 (1 + s/ 200 , 000) 4 Solution : The low freqency gain is 10 20dB. The vertical difference between 500 and 5000 rad/s (also 20000 and 200000 rad/s) is one decade; therefore the vertical differences are 80dB since the magnitude of the slopes are 80dB/dec. - 60dB 20dB 20dB 80dB 80dB +80dB/dec - 80dB/dec 500 5000 20000 200000 rad/s (d) H ( s ) = 500 s 2 ( s + 100) 2 ( s + 10 , 000) 2 = 500 × 10 12 s 2 (1 + s/ 100) 2 (1 + s/ 10 , 000) 2 Solution : The low freqency behavior is proportional to 500 × 10 12 s 2 which is a line with slope of +40dB/dec with gain of ≈ - 186dB at 1rad/s. Going from 1 to 100 rad/s (2 decades), the gain increases by 80dB due to the 40dB/dec slope. As a result, the slope flattens out to - 106dB at 100rad/s. At 10000rad/s, the slope starts falling off at - 40dB/dec due to the double pole. - 106dB - 186dB 80dB +40dB/dec - 40dB/dec 1 100 10000 rad/s Dr. Vahe Caliskan 8 of 8 Posted: February 8, 2013

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