102_1_hw05_sol - 1 EE102 Spring 2009-10 Systems and Signals...

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1 EE102 Spring 2009-10 Lee Systems and Signals Homework #5 Due: Tuesday, May 18, 2010 5PM 1. Given the signal f ( t ) = sinc ( t ) , evaluate the Fourier transforms of the following signals. Provide a labeled sketch for each function and its Fourier transform. (a) f ( t ) Solution: From class, sinc ( t ) rect ( ω / 2 π ) which is plotted below. (b) f ( t - 1) Solution: Using the shift theorem, and the previous result F [ sinc ( t - 1)] = e - j ω rect ( ω / 2 π ) which is plotted below. (c) 1 2 ( f ( t - 1) - f ( t + 1)) Solution: Again by the shift theorem, and linearity F 1 2 ( sinc ( t - 1) - sinc ( t + 1)) = 1 2 e - j ω rect ( ω / 2 π ) + e - j ω rect ( ω / 2 π ) = - j sin( ω ) rect ( ω / 2 π ) which is plotted below. (d) t f ( t ) Solution: There are two solutions. The first is to use the frequency domain derivative theorem. The second is to note that t sinc ( t ) = t sin π t π t = 1 π sin( π t ) and then use the generalized Fourier transform of sin to find F [ t sinc ( t )] = F 1 π sin( π t ) = 1 π ( j π ( δ ( ω + π ) - δ ( ω - π ))) = j δ ( ω + π ) - j δ ( ω - π )
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2 This is plotted below. The second approach uses the frequency domain differentiation theorm, tf ( t ) j d d ω F ( j ω ) The Fourier transform F ( j ω ) is rect ( ω / 2 π ) which has discontinuties at ± π . Differen- tiating produces a positive impulse at ω = - π and a negative impulse at ω = π . The result is the same. (e) f ( t ) cos(10 π t ) Solution: Using the modulation theorem: F [ t sinc ( t ) cos(10 π t )] = 1 2 ( rect (( ω - 10 π ) / 2 π ) + rect (( ω + 10 π ) / 2 π )) This is plotted below. a) ! 4 ! 3 ! 2 ! 1 0 1 2 3 4 ! 1 ! 0.5 0 0.5 1 1.5 time (s) f(t) ! 8 ! 6 ! 4 ! 2 0 2 4 6 8 ! 1 ! 0.5 0 0.5 1 1.5 ! F(j ! ) b) ! 4 ! 3 ! 2 ! 1 0 1 2 3 4 ! 1 ! 0.5 0 0.5 1 1.5 time (s) f(t ! 1) ! 8 ! 6 ! 4 ! 2 0 2 4 6 8 ! 4 ! 2 0 2 4 ! F(j ! ) Magnitude Phase c) ! 4 ! 3 ! 2 ! 1 0 1 2 3 4 ! 0.8 ! 0.6 ! 0.4 ! 0.2 0 0.2 0.4 0.6 time (s) 0.5(f(t !
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