102_1_lecture18

# 102_1_lecture18 - UCLA Fall 2011 Systems and Signals...

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UCLA Fall 2011 Systems and Signals Lecture 18: Sampling Theorem II November 30, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1

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Administration Project due Wednesday Nov. 30 2011 at 10 am (grace period till 5 pm). Submission Guidelines: A short writeup of what was done for each task Include all labeled ﬁgures Attach a printout of all code Final Exam: Dec. 08, Thursday 11:30am-2:30pm 6 problems, 3 hours 3 hand-written letter sheets permitted closed book, closed notes, no calculator (no cell phone) Course Survey 1 % extra credit for those of you who complete the survey EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2
Agenda Today’s topics Sampling of continuous-time signals Course Review EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3

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Sampling Review Sampling of continuous-time signals can be represented by multiplication of impulse train with the signal of interest: Deﬁne δ T ( t ) to be a sequence of unit δ functions spaced by T , δ T ( t ) = X k = -∞ δ ( t - kT ) If f ( t ) is a signal, then f ( t ) δ T ( t ) = f ( t ) X k = -∞ δ ( t - ) = X k = -∞ f ( ) δ ( t - ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4
f ( t ) 2T -T -3T -2T T 0 3T t δ T ( t ) 1 2T -T -3T -2T T 0 3T t 2T -T -3T -2T T 0 3T t × = Original Signal Sampling Function Sampled Signal f ( t ) δ T ( t ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5

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Sampling Theorem Frequency content of the resulting signal ¯ f ( t ) = f ( t ) δ T ( t ) can be understood as: ¯ F ( ) = F [ f ( t ) δ T ( t )] = 1 2 π F [ f ( t )] * F [ δ T ( t )] = 1 T F ( ) * X k = -∞ δ ( ω - 0 ) ¯ F ( ) = 1 T X k = -∞ F ( j ( ω - 0 )) The spectrum of the sampled signal consists of shifted replicas of the original spectrum, scaled by 1 /T . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6
What this looks like exactly depends on the sampling frequency ω 0 . We’ll deﬁne the bandwidth of f ( t ) to be ± B Hz. The following plot shows the case where 2 πB ² ω 0 / 2 This is the case where the signal bandwidth is much less than the sampling rate. - ω 0 0 - ω 0 0 ω * = 1 T δ ω 0 ( ω ) ω - ω 0 0 ω ¯ F ( ω ) lowpass ﬁlter 2 π B - 2 π B 2 π B - 2 π B F ( j ω ) 1 T F ( j ω ) 1 T F ( j ( ω + ω 0 )) 1 T F ( j ( ω - ω 0 )) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7

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If we lowpass ﬁlter ¯ F ( ) , then we can perfectly recover F ( ) , and f ( t ) ! As the sampling frequency ω 0 decreases (sampling period T increases) the spectral replicas get closer: ω 0 - ω 0 0 ω 2 π B - 2 π B ω 0 - ω 0 0 ω 0 ω 2 ω 0 - 2 ω 0 2 ω 0 - ω 0 ω 0 - 2 ω 0 - 3 ω 0 3 ω 0 Eventually the replicas overlap, and F ( ) cannot be recovered. EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8
The overlap is called aliasing because the low frequencies of one band appear (alias) as high frequencies of the next band. High frequencies from one band also alias as low frequencies of the next band. - 2 π B 2 π B - ω 0 ω 0 0 ω ω 0 / 2 - ω 0 / 2 No aliasing occurs only if 2 πB < ω 0 / 2 , or 2 B < ω 0 / 2 π = 2 π T 1 2 π = 1 T Hz The signal can be recovered exactly only if the signal bandwidth 2 B Hz is less than or equal to the sampling rate 1 /T Hz.

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## This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.

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102_1_lecture18 - UCLA Fall 2011 Systems and Signals...

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