HW_8_sol - This problem using the properties of the Fourier...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
DKL:11/10/06 HW 8 1/2 HW 8 ECE 2704 Due 11-10-06 7.4.6 Find the inverse Fourier transform of each of the Fourier transforms shown in Figures P7.4.6ac. |H( ! )| ! 1 " #" $ H( ! ) ! -t d " #" Figure P7.4.6c (a) (5) Find a function describing H ( ! ) = H ( ! ) e " H ( ! ) . (b) (10) Set up the integral for finding the inverse Fourier transform using the definition. (c) (10) Evaluate the integral to find the inverse Fourier transform. Solution The inverse Fourier transform is found using Definition 7.4.1: h ( t ) = 1 2 ! H ( " ) e j " t d " #$ $ % In Section 14.6 we will use these functions to define ideal filters. (a) H LP ( ! ) = 1 e " jt d ! # ! 2 $ % & ( ) * (b) h LP ( t ) = 1 2 ! H ( " ) e j " t d " #$ $ % = 1 2 ! 1 e # jt d " ( ) e j " t d " #& & % (c) h LP ( t ) = 1 2 ! 1 e " jt d # ( ) e j # t d # "$ $ % = 1 2 ! 1 e j ( t " t d ) # d # "$ $ % = 1 2 ! 1 j ( t " t d ) e j ( t " t d ) # & ( ) * + "$ $ = 1 ! 1 ( t " t d ) e j ( t " t d ) $ " e " j ( t " t d ) $ ( ) 2 j = $ ! sin ( t " t d ) $ ( ) ( t " t d ) $ = $ ! Sa ( t " t d ) $ ( )
Image of page 1

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

View Full Document Right Arrow Icon
DKL:11/10/06 HW 8 2/2 7.5.1 (10) Find the Fourier transform of each of the following signals using the Table and properties of the Fourier transform. x ( t ) = ! ( t + t 0 ) + ! ( t " t 0 ) Solution This problem using the properties of the Fourier transform in Table 7.6.2 and the Fourier
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: This problem using the properties of the Fourier transform in Table 7.6.2 and the Fourier transform pairs in Table 7.6.3. (x) Using the Time Shift Property X ( ! ) = 1 e j ! t + 1 e " j ! t = 2 e j ! t + e " j ! t 2 # $ % & ’ ( = 2cos( ! t ) 7.5.3 (10) Find the inverse Fourier transform of the following functions. X ( ! ) = Sa(4 !" 5) + Sa(4 ! + 5) Solution This problem using the properties of the Fourier transform in Table 7.6.2 and the Fourier transform pairs in Table 7.6.3. (i) X ( ! ) = 2 1 2 Sa 4 ! " 5 4 # $ % & ’ ( ) * + ,-. + Sa 4 ! + 5 4 # $ % & ’ ( ) * + ,-. / 1 2 3 4 Sa 4 ! [ ] = 1 8 " t 8 # $ % & ’ ( x ( t ) = 1 4 ! t 8 " # $ % & ’ cos 5 4 t " # $ % & ’...
View Full 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