HW 08_examples - 1 ME360 - Solved Examples for Homework #8...

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1 ME360 - Solved Examples for Homework #8 Problem 1. 1) Derive 1st, 2nd, and 3rd order Butterworth denominator polynomials for ω = ω c =3 . 5 π rad/s. Find the poles of the polynomials for each case, draw s-plane map with pole location for each case, comment on the distribution of poles in the s-plane. Plot the magnitude spectral densities of the corresponding three low pass filters on the same graph and comment on their relative filtering properties. Plot the phase responses of these three filters on the same graph and comment on their relative phase distortion. First order: T ( s )= 1 as +1 = | T ( ) | 2 = | 1 jaω +1 | 2 = 1 1+( ) 2 = 1 1+ a 2 ω 2 | T ( ) | ω =3 . 5 π = 1 2 = 1 1+(3 . 5 πa ) 2 = 1 2 = 3 . 5 πa =1= a = 1 3 . 5 π So, T ( s )= 1 as +1 = 1 s/ 3 . 5 π +1 = 3 . 5 π s +3 . 5 π pole: s = 3 . 5 π .I t l i e so nac i r c l eo f R =3 . 5 π in the s-plane. -15 -10 -5 0 5 10 15 -10 -5 0 5 10 Second order: T ( s )= 1 as 2 + bs +1 = | T ( ) | 2 = | 1 2 + jbω +1 | 2 = 1 (1 2 ) 2 + b 2 ω 2 = 1 a 2 ω 4 +( b 2 2 a ) ω +1 thus, b 2 2 a =0 and | T ( ) | ω =3 . 5 π = 1 2 = 1 1+(3 . 5 π ) 4 a 2 = 1 2 = (3 . 5 π ) 4 a 2 =1= a = 1 (3 . 5 π ) 2 b 2 =2 a = b = 2 3 . 5 π So, T ( s )= 1 as 2 + bs +1 = 1 s 2 / (3 . 5 π ) 2 + s 2 / (3 . 5 π )+1 = (3 . 5 π ) 2 s 2 +3 . 5 π 2 s +(3 . 5 π ) 2 poles: s =3 . 5 π ( 2 2 ±
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2 -15 -10 -5 0 5 10 15 -10 -5 0 5 10 Third order: T ( s )= 1 as 3 + bs 2 + cs +1 = | T ( ) | 2 = | 1 jaω 3 jbω 2 + jcω +1 | 2 = 1 (1 2 ) 2 +( 3 ) 2 = 1 a 2 ω 6 +( b 2 2 ac ) ω 4 +( c 2 2 b ) ω 2 +1 thus, c 2 2 b =0 ,b 2 2 ac =0 . and | T ( ) | ω =3 . 5 π = 1 2 = 1 1+(3 . 5 π ) 6 a 2 = 1 2 = (3 . 5 π ) 6 a 2 =1= a = 1 (3 . 5 π ) 3 c 2 =2 b = c 4 =4 b 2 =8 ac = c 3 =8 a = c = 2 3 . 5 π b = c 2 2 = 2
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HW 08_examples - 1 ME360 - Solved Examples for Homework #8...

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