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Lec 13

# Lec 13 - UCLA Winter 2009-2010 Systems and Signals Lecture...

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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

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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 ) = X n = -∞ D n e jnω 0 t where D n = 1 T Z T 0 f ( t ) e - jnω 0 t dt and ω 0 = 2 π/T . EE102: Systems and Signals; Spr 09-10, Lee 2
Fourier series resembles an inverse Fourier transform of f ( t ) , but it is a and not an R . We can make the connection much clearer using the Fourier transform for complex exponentials, and extended linearity: f ( t ) = X n = -∞ D n e jnω 0 t F ( ) = X n = -∞ D n 2 πδ ( ω - 0 ) ω ω ω 0 2 ω 0 - 2 ω 0 - ω 0 0 ω 0 2 ω 0 - 2 ω 0 - ω 0 0 D n Fourier Series Coefficients Fourier Transform F ( j ω ) The Fourier series coefficients and Fourier transform are the same! (with a scale factor of 2 π ). EE102: Systems and Signals; Spr 09-10, Lee 3

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Example: Square Wave Consider the square wave f ( t ) = X 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
The Fourier series from before (Lecture 7, page 38) is f ( t ) = X n = -∞ D n e j 2 πnt/τ = X n = -∞ D n e jπnt with Fourier coefficients D n = T τ sinc n T τ = 1 2 sinc( n/ 2) so that f ( t ) = X n = -∞ 1 2 sinc( n/ 2) e jπnt . The Fourier transform is then F ( ) = X n = -∞ 1 2 sinc( n/ 2)(2 πδ ( ω - )) EE102: Systems and Signals; Spr 09-10, Lee 5

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= π X n = -∞ sinc( n/ 2) δ ( ω - ) Note that this can also be written: F ( ) = π X n = -∞ sinc( ω/ 2 π ) δ ( ω - ) . 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
1 / 2 ω ω π - 8 π 8 π 4 π - 4 π - 12 π 12 π 0 - 8 π 8 π 4 π - 4 π - 12 π 12 π 0 1 2 sinc ( ω / 2 π ) π n = - sinc ( n / 2 ) δ ( ω - n π ) EE102: Systems and Signals; Spr 09-10, Lee 7

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Impulse Trains – Sampling Functions Define δ T ( t ) to be a sequence of unit δ functions spaced by T , δ T ( t ) = X n = -∞ δ ( t - nT ) which looks like 2T -T -3T -2T T 0 3T t δ 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
The Fourier coefficients are D n = 1 T Z T/ 2 - T/ 2 f ( t ) e - j 2 πnt/T dt = 1 T Z T/ 2 - T/ 2 δ ( t ) e - j 2 πnt/T dt = 1 T . All of the coefficients are the same! The Fourier series is then f ( t ) = X n = -∞ 1 T e j 2 πnt/T = 1 T X n = -∞ e jnω 0 t . EE102: Systems and Signals; Spr 09-10, Lee 9

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The Fourier transform of δ T ( t ) is then F [ δ T ( t )] = 1 T X n = -∞ 2 πδ ( ω - 0 ) = ω 0 X n = -∞ δ ( ω - 0 ) = ω 0 δ ω 0 ( ω ) since ω 0 = 2 π/T .
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Lec 13 - UCLA Winter 2009-2010 Systems and Signals Lecture...

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