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Unformatted text preview: ECE600 Introduction to Digital Signal Processing Winter 2011 Homework #4 Feb. 11, 2011 HOMEWORK SOLUTIONS #4 1. For rate1 /T 1 sampling, the relationship between the CTFT of the continuoustime signal x ( t ) and the DTFT of the discrete time signal x [ n ] is X ( e j ) = 1 T 1 summationdisplay k = X c parenleftbigg j parenleftbigg T 1 + 2 k T 1 parenrightbiggparenrightbigg . (1) However, when there is no aliasing, the previous expression simplifies to X ( e j ) = 1 T 1 X c parenleftbigg j T 1 parenrightbigg for ( , ) . (2) For rate1 /T 2 sinc reconstruction, the relationship between the DTFT of the sampled signal x [ n ] and the CTFT of the reconstructed signal y ( t ) is Y c ( j ) = braceleftBigg T 2 X ( e j T 2 )   < T 2 otherwise . (3) Notice that the CTFT frequencies   < T 2 correspond to the DTFT frequencies = T 2 ( , ). For that range of DTFT frequencies, the simplified equation (2) connects us back to the original CTFT X c ( j ). In summary, this yields Y c ( j ) = T 2 T 1 X c parenleftbigg j T 2 T 1 parenrightbigg   < T 2 otherwise . (4) 2. (a) After rate 1 T 1 sampling, a frequency component at rad/sec in x ( t ) manifests as a frequency component at = T 1 rad/sample in x [ n ]. But, because the DTFT is 2 periodic, this component actually appears at all frequencies + 2 k for k Z . Lets refer to the copy that lives in the interval ( , ) as . After 1 T 2rate reconstruction, we know that the frequency component at manifests at = /T 2 rad/sec....
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 Spring '08
 Clymer,B
 Digital Signal Processing, Signal Processing

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