solns4_600

solns4_600 - ECE-600 Introduction to Digital Signal...

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Unformatted text preview: ECE-600 Introduction to Digital Signal Processing Winter 2011 Homework #4 Feb. 11, 2011 HOMEWORK SOLUTIONS #4 1. For rate-1 /T 1 sampling, the relationship between the CTFT of the continuous-time signal x ( t ) and the DTFT of the discrete time signal x [ n ] is X ( e jω ) = 1 T 1 ∞ summationdisplay k = −∞ X c parenleftbigg j parenleftbigg ω T 1 + 2 πk T 1 parenrightbiggparenrightbigg . (1) However, when there is no aliasing, the previous expression simplifies to X ( e jω ) = 1 T 1 X c parenleftbigg j ω T 1 parenrightbigg for ω ∈ ( − π,π ) . (2) For rate-1 /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 Ω) = braceleftBigg T 2 X ( e j Ω T 2 ) | Ω | < π T 2 otherwise . (3) Notice that the CTFT frequencies | Ω | < π T 2 correspond to the DTFT frequencies ω = Ω T 2 ∈ ( − π,π ). For that range of DTFT frequencies, the simplified equation (2) connects us back to the original CTFT X c ( j Ω). In summary, this yields Y c ( j Ω) = T 2 T 1 X c parenleftbigg j T 2 T 1 Ω parenrightbigg | Ω | < π T 2 otherwise . (4) 2. (a) After rate- 1 T 1 sampling, a frequency component at Ω rad/sec in x ( t ) manifests as a frequency component at ω = Ω T 1 rad/sample in x [ n ]. But, because the DTFT is 2 π-periodic, this component actually appears at all frequencies ω + 2 πk for k ∈ Z . Let’s refer to the copy that lives in the interval ω ∈ ( − π,π ) as ω ′ . After 1 T 2-rate reconstruction, we know that the frequency component at ω ′ manifests at Ω ′ = ω ′ /T 2 rad/sec....
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solns4_600 - ECE-600 Introduction to Digital Signal...

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