Hw07_Sol_Su09-1

# Hw07_Sol_Su09-1 - ECE220 Homework Assignment #7 Solution...

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ECE220 Homework Assignment #7 Solution Solution to Problem 8.1 V ( s ) = ± 0 4 e - 6 t cos(7 t ) u ( t ) e - st dt = 4 ± 0 e - ( s +6) t cos(7 t ) dt = 4 · 1 2 · ± 0 e - ( s +6) t ( e j 7 t + e - j 7 t ) dt = 2 ± 0 ( e - ( s +6 - 7 j ) t + e - ( s +6+7 j ) t ) dt = 2( e - ( s +6 - 7 j ) t - ( s + 6 - 7 j ) + e - ( s +6+7 j ) t - ( s + 6 + 7 j ) ) | 0 = 2( 1 s + 6 - 7 j + 1 s + 6 + 7 j ) = 2 s + 6 + 7 j + s + 6 - 7 j ( s + 6) 2 + 49 = 4( s + 6) ( s + 6) 2 + 49 Solution to Problem 8.3 The signal v ( t ) can be decomposed into two pulses of unit duration starting at time 0 and 3 respectively. Thus v ( t ) = ( u ( t ) - u ( t - 1)) + ( u ( t - 3) - u ( t - 4)) Note that v ( t ) is a linear combination of time shifted unit step functions. 1

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u ( t ) = 1 s u ( t - t 0 ) = e - st 0 1 s V ( s ) = 1 s - e - s s + e - 3 s s - e - 4 s s Solution to Problem 8.6 y ( t ) = e - at v ( t ) Y ( s ) = ± 0 e - at v ( t ) e - st dt = ± 0 e - ( s + a ) t v ( t ) dt = V ( s + a ) Solution to Problem 8.9 Rewrite V ( s ) = K 1 s + K 2 s + 1 + K 3 s + 3 Now K 1 = sV ( s ) | s =0 = 7 s + 3 ( s + 1)( s + 3) | s =0 = 3 1 · 3 = 1 K 2 = ( s + 1) V ( s ) | s = - 1 = 7 s + 3 s ( s + 3) | s = - 1 = - 4 - 1 · 2 = 2 2
3 = ( s + 3) V ( s ) | s = - 3 = 7 s + 3 s ( s + 1) | s = - 3 = - 18 - 3 · - 2 = - 3 Therefore, V ( s ) = 1 s + 2 s + 1 + - 3 s + 3 and v ( t ) = (1 + 2 e - t - 3 e - 3 t ) u ( t ) Solution to Problem 8.10 (c,d) (c) V ( s ) = s - 4 s 2 + 25 = s s 2 + 5 2 - 4 5 · 5 s 2 + 5 2 set V 1 ( s ) = s s 2 + 5 2 , V 2 ( s ) = - 4 5 · 5 s 2 + 5 2 from table 1 on p.282, we found v 1 ( t ) = cos(5 t ) u ( t

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## This note was uploaded on 08/24/2009 for the course ECE 220 taught by Professor Nilson during the Summer '08 term at N.C. State.

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Hw07_Sol_Su09-1 - ECE220 Homework Assignment #7 Solution...

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