hmw7_sol_sp08

# hmw7_sol_sp08 - i ECE220 Homework Assignment#7 Solutions...

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i ECE220 Homework Assignment #7 Solutions Problem 8.1 Find the Laplace transform of v ( t ) = 4 e - 6 t cos(7 t ) u ( t ) by direct integration. (Hint: using Euler’s identity may be helpful.) Solution We have V ( s ) = Z 0 4 e - 6 t cos(7 t ) u ( t ) e - st dt = 4 Z 0 e - ( s +6) t cos(7 t ) dt = 4 · 1 2 · Z 0 e - ( s +6) t ( e j 7 t + e - j 7 t ) dt = 2 Z 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 ————————————————————————————– Problem 8.3 Find the Laplace transform of the signal v ( t ) in Figure 0.1, using Equation 8.5 in Example 8.3, and the linearity and time-shifting properties. Solution 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)) (0.0) Note that v ( t ) is a linear combination of time shifted unit step functions.

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ii -1 0 1 2 3 4 5 6 7 8 9 10 -1 -0.5 0 0.5 1 1.5 2 The signal v(t) v(t) time t Figure 0.1. The signal v ( t ) in problem 8.3. L ( u ( t )) = 1 s (0.1) L ( u ( t - t 0 )) = e - st 0 1 s (0.2) V ( s ) = 1 s - e - s s + e - 3 s s - e - 4 s s (0.3) ————————————————————————————– Problem 8.6 Let a be a real number. Let y ( t ) = e - at v ( t ) . Show that Y ( s ) = V ( s + a ) Solution We have
iii y ( t ) = e - at v ( t ) Y ( s ) = Z 0 e - at v ( t ) e - st dt = Z 0 e - ( s + a ) t v ( t ) dt = V ( s + a ) ————————————————————————————– Problem 8.10 Use table 1 in page 404 and the linearity property to ﬁnd v ( t ) for the following Laplace transforms. (a)

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## This note was uploaded on 03/30/2009 for the course ECE 220 taught by Professor Nilson during the Spring '08 term at N.C. State.

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hmw7_sol_sp08 - i ECE220 Homework Assignment#7 Solutions...

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