102_1_Lecture13 - UCLA Winter 2009-2010 Systems and Signals...

Info icon This preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
UCLA Winter 2009-2010 Systems and Signals Lecture 13: Impulse trains, Periodic Signals, and Sampling May 12, 2010 EE102: Systems and Signals; Spr 09-10, Lee 1
Image of page 1

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

View Full Document Right Arrow Icon
Fourier Transforms of Periodic Signals So far we have used Fourier series to handle periodic signals, they do not have a Fourier transform in the usual sense (not finite energy). We can generalize Fourier transform to such signals. Given a periodic signal f ( t ) with period T 0 , f ( t ) has a Fourier Series. f ( t ) = n = -∞ D n e jn ω 0 t where D n = 1 T T 0 f ( t ) e - jn ω 0 t dt and ω 0 = 2 π /T . EE102: Systems and Signals; Spr 09-10, Lee 2
Image of page 2
Fourier series resembles an inverse Fourier transform of f ( t ) , but it is a and not an . We can make the connection much clearer using the Fourier transform for complex exponentials, and extended linearity: f ( t ) = n = -∞ D n e jn ω 0 t F ( j ω ) = n = -∞ D n 2 πδ ( ω - n ω 0 ) ! ! ! 0 2 ! 0 - 2 ! 0 - ! 0 0 ! 0 2 ! 0 - 2 ! 0 - ! 0 0 D n Fourier Series Coefficients Fourier Transform F ( j ! ) The Fourier series coe ffi cients and Fourier transform are the same! (with a scale factor of 2 π ). EE102: Systems and Signals; Spr 09-10, Lee 3
Image of page 3

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

View Full Document Right Arrow Icon
Example: Square Wave Consider the square wave f ( t ) = n = -∞ rect( t - 2 n ) This is the square pulse of width T = 1 defined on the interval of width τ = 2 and then replicated infinitely often. 0 1 2 3 - 1 - 2 - 3 t f ( t ) EE102: Systems and Signals; Spr 09-10, Lee 4
Image of page 4
The Fourier series from before (Lecture 7, page 38) is f ( t ) = n = -∞ D n e j 2 π nt/ τ = n = -∞ D n e j π nt with Fourier coe ffi cients D n = T τ sinc n T τ = 1 2 sinc( n/ 2) so that f ( t ) = n = -∞ 1 2 sinc( n/ 2) e j π nt . The Fourier transform is then F ( j ω ) = n = -∞ 1 2 sinc( n/ 2)(2 πδ ( ω - n π )) EE102: Systems and Signals; Spr 09-10, Lee 5
Image of page 5

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

View Full Document Right Arrow Icon
= π n = -∞ sinc( n/ 2) δ ( ω - n π ) Note that this can also be written: F ( j ω ) = π n = -∞ sinc( ω / 2 π ) δ ( ω - n π ) . This is the Fourier transform of the rect, multiplied by an array of evenly spaced δ ’s. EE102: Systems and Signals; Spr 09-10, Lee 6
Image of page 6
1 / 2 ! ! ! - 8 ! 8 ! 4 ! - 4 ! - 12 ! 12 ! 0 - 8 ! 8 ! 4 ! - 4 ! - 12 ! 12 ! 0 1 2 sinc ( ! / 2 " ) ! " # n = - " sinc ( n / 2 ) $ ( % - n ! ) EE102: Systems and Signals; Spr 09-10, Lee 7
Image of page 7

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

View Full Document Right Arrow Icon
Impulse Trains – Sampling Functions Define δ T ( t ) to be a sequence of unit δ functions spaced by T , δ T ( t ) = n = -∞ δ ( t - nT ) which looks like 2T -T -3T -2T T 0 3T t ! T ( t ) 1 What do we get if we expand this function as a Fourier series over - T/ 2 to T/ 2 ? EE102: Systems and Signals; Spr 09-10, Lee 8
Image of page 8
The Fourier coe ffi cients are D n = 1 T T/ 2 - T/ 2 f ( t ) e - j 2 π nt/T dt = 1 T T/ 2 - T/ 2 δ ( t ) e - j 2 π nt/T dt = 1 T . All of the coe ffi cients are the same! The Fourier series is then f ( t ) = n = -∞ 1 T e j 2 π nt/T = 1 T n = -∞ e jn ω 0 t . EE102: Systems and Signals; Spr 09-10, Lee 9
Image of page 9

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

View Full Document Right Arrow Icon
The Fourier transform of δ T ( t ) is then F [ δ T ( t )] = 1 T n = -∞ 2 πδ ( ω - n ω 0 ) = ω 0 n = -∞ δ ( ω - n ω 0 ) = ω 0 δ ω 0 ( ω ) since ω 0 = 2 π /T .
Image of page 10
Image of page 11
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