hw 5 sol

hw 5 sol - 1 EE102 Spring 2009-10 Systems and Signals Lee...

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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 - 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)) ² = 1 2 ³ e - rect ( ω/ 2 π ) + e - 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 πt πt = 1 π sin( πt ) and then use the generalized Fourier transform of sin to find F [ t sinc ( t )] = F ± 1 π sin( πt ) ² = µ 1 π ( ( δ ( ω + π ) - δ ( ω - π ))) = ( ω + π ) - ( ω - π )
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2 This is plotted below. The second approach uses the frequency domain differentiation theorm, tf ( t ) j d F ( ) The Fourier transform F ( ) 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 πt ) Solution: Using the modulation theorem: F [ t sinc ( t ) cos(10 πt )] = 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 ω F(j ω ) 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 ω Magnitude
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hw 5 sol - 1 EE102 Spring 2009-10 Systems and Signals Lee...

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