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

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE438 Homework 3 Solution February 7, 2013
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Problem 2 a) x [ n ] = x ( nT ) = sin( π 2 n )sinc( n 2 ) (7) Figure 4: Sketch of problem 2 part a. 3
Image of page 4
b) X ( f ) = F{ x ( t ) } = F{ 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.
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern