# hw5sol - Purdue University, School of Electrical and...

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #5 Solution (17–20) Problem 17 (a) Transfer function H ( s ) E ( s )= X re f F ( s ) Y ( s ) E ( s X re f F ( s ) G 1 ( s ) E ( s ) E ( s )[ 1 + F ( s ) G 1 ( s )] = X re f E ( s X re f 1 + F ( s ) G 1 ( s ) H ( s 1 1 + F ( s ) G 1 ( s ) = 1 1 + n ( s ) d ( s ) = d ( s ) d ( s )+ n ( s ) = d f ( s ) d g ( s ) d f ( s ) d g ( s n f ( s ) n g ( s ) (b) If E ( s ) has poles only in the open left half of complex plane, we can apply final-value theorem to find e ( t ) as t ) . lim t e ( t lim s 0 sE ( s ) = lim s 0 ± K 0 d f ( s ) d g ( s ) d f ( s ) d g ( s n f ( s ) n g ( s ) ² = 0 Case 1 : The numerator and denominator have no common factors d f ( s ) d g ( s sd 1 ( s s ( s z 1 )( s z 2 ) ... for some polynomial d 1 ( s ) . i.e., d f ( s ) d g ( s ) has a zero at the origin. Case 2 : The numerator and denominator have some common factors Same condition should be met after the cancellation. 1/ 8

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (c) As t , e ( t ) can only become a nonzero constant when d f ( 0 ) d g ( 0 ) and n f ( 0 ) n g ( 0 ) are constants. So the condi- tions are: 1. d f ( s ) d g ( s ) have no zero at the origin. i.e., d f ( s ) d g ( s ) 6 = s ( s z 1 )( s z 2 ) ... 2. All the roots of d f ( s ) d g ( s )+ n f ( s ) n g ( s ) in the open left half complex plane. (d) When E ( s ) is a strictly proper rational function of s , we can apply initial-value theorem. In our case, E ( s )= K 0 d ( s ) s ( d ( s n ( s ) ) is already a strictly proper rational function. So, we can find initial value by the below theorem: e ( 0 + lim s sE ( s ) (e) n f ( s ) to have a pair of zeros at the location where the poles are.
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hw5sol - Purdue University, School of Electrical and...

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