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hmw9_sol_s08

# hmw9_sol_s08 - i ECE220 Homework Assignment#9 Solutions...

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i ECE220 Homework Assignment #9 Solutions Problem 10.1 Using direct integration, ﬁnd the Fourier transform of the signal s ( t ) = e - 25 t cos(2 π 100 t + π/ 4) u ( t ) You can check your result by using the Laplace transform table in Chapter 8. Solution s ( t ) = e - 25 t cos(2 π 100 t + π/ 4) u ( t ) (0.0) S ( ω ) = Z -∞ e - 25 t cos(2 π 100 t + π/ 4) u ( t ) e - jωt dt (0.1) = Z 0 e - 25 t cos(2 π 100 t + π/ 4) e - jωt dt (0.2) = Z 0 e ( - - 25) t cos(2 π 100 t + π/ 4) dt (0.3) We now proceed to develop a useful integral using integration by parts. Z e at cos( bt + c ) = e at 1 b sin( bt + c ) - Z ae at 1 b sin( bt + c ) (0.4) But Z e at sin( bt + c ) = e at - 1 b cos( bt + c ) + Z ae at 1 b cos( bt + c ) (0.5) Z e at cos( bt + c ) = 1 b e at sin( bt + c ) + a b 2 e at cos( bt + c ) - Z a 2 b 2 e at cos( bt + c ) (0.6) (1 + a 2 b 2 ) Z e at cos( bt + c ) = e at ( a cos( bt + c ) + b sin( bt + c )) b 2 (0.7) Z e at cos( bt + c ) = e at ( a cos( bt + c ) + b sin( bt + c )) a 2 + b 2 (0.8) Thus

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ii S ( ω ) = Z 0 e ( - - 25) t cos(2 π 100 t + π/ 4) dt (0.9) = e ( - - 25) t (( - - 25)cos(2 π 100 t + π/ 4) + 2 π 100sin(2 π 100 t + π/ 4)) ( - - 25) 2 + (2 π 100) 2 | 0 (0.10) = 0 - e 0 (( - - 25)cos( π/ 4) + 2 π 100sin( π/ 4)) ( - - 25) 2 + (2 π 100) 2 (0.11) = + 25 - 2 π 100 2(( + 25) 2 + (2 π 100) 2 ) (0.12) ————————————————————————————– Problem 10.4 Consider a signal s ( t ) with Fourier transform S ( ω ) . Let ω 0 be a real number. Deﬁne a new (complex-valued) signal y ( t ) = e 0 t s ( t ) . Show that
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hmw9_sol_s08 - i ECE220 Homework Assignment#9 Solutions...

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