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hw3_sol - ECE438 Homework 3 Solution February 7 2013...

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ECE438 Homework 3 Solution February 7, 2013
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Problem 1 a) x ( t ) can be written as a windowed cosine function. x ( t ) = cos( π 2 t )rect( t 2 ) (1) Using the product theorem of CTFT and transform pairs, we have X ( f ) = F{ cos( π 2 t ) } * F{ rect( t 2 ) } (2) = 1 2 ( δ ( f - 1 4 ) + δ ( f + 1 4 )) * 2sinc(2 f ) = sinc(2( f - 1 4 )) + sinc(2( f + 1 4 )) Figure 1: Sketch of problem 1 part a 1
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b) y ( t ) = rep 4 { x ( t ) } = rep 4 { cos( π 2 t )rect( t 2 ) } (3) Using transform relations we have Y ( f ) = 1 4 comb 1 4 { X ( f ) } (4) = 1 4 comb 1 4 { sinc(2( f - 1 4 )) + sinc(2( f + 1 4 )) } Figure 2: Sketch of problem 1 part b c) z ( t ) = rect( t 20 ) y ( t ) = rect( t 20 )rep 4 { cos( π 2 t )rect( t 2 ) } (5) Using transform relations we have Z ( f ) = 20sinc(20 f ) * 1 4 comb 1 4 { X ( f ) } (6) Figure 3: Sketch of problem 1 part c. The zero crossings occur at interval 0.05. The lobes are drawn wider than they should be. 2
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Problem 2 a) x [ n ] = x ( nT ) = sin( π 2 n )sinc( n 2 ) (7) Figure 4: Sketch of problem 2 part a. 3
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b) X ( f ) = F{ x ( t ) } = sin(2 π (500 t )) } * F{ sinc(1000 t ) } (8) = [ 1 2 j ( δ ( f - 500) - δ ( f + 500))] * [ 1 1000 rect( f 1000 )] = 1 2000 j [rect( f - 500 1000 ) - rect( f + 500 1000 )] Figure 5: Sketch of problem 2 part b. c) X s ( f ) = x s ( t ) } = comb T ( x ( t )) } (9) = 1 T rep 1 T { X ( f ) } = rep 2000 [ - j [rect( f - 500 1000 ) - rect( f + 500 1000 )] = X k = -∞ [ - j [rect( f - 500 - 2000 k 1000 ) - rect( f + 500 - 2000
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hw3_sol - ECE438 Homework 3 Solution February 7 2013...

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