solns4_600

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

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Unformatted text preview: ECE-600 Introduction to Digital Signal Processing Autumn 2011 Homework #4 Oct. 21, 2011 HOMEWORK SOLUTIONS #4 1. The figure below shows sketches of the DTFT for various sampling frequencies. Note that, since x ( t ) is conjugate-symmetric, its CTFT is real-valued. Hence, when DTFT copies overlap due to aliasing, the overlapped spectra add together. 1 2 3 6 12 ω ω ω ω ω X ( e jω ) X ( e jω ) X ( e jω ) X ( e jω ) X ( e jω ) π 3 π 3 π 3 π 3 π 3 2 π 3 2 π 3 2 π 3 2 π 3 2 π 3 π π π π π 4 π 3 4 π 3 4 π 3 4 π 3 4 π 3 5 π 3 5 π 3 5 π 3 5 π 3 5 π 3 2 π 2 π 2 π 2 π 2 π − π 3 − π 3 − π 3 − π 3 − π 3 − 2 π 3 − 2 π 3 − 2 π 3 − 2 π 3 − 2 π 3 − π − π − π − π − π − 4 π 3 − 4 π 3 − 4 π 3 − 4 π 3 − 4 π 3 − 5 π 3 − 5 π 3 − 5 π 3 − 5 π 3 − 5 π 3 − 2 π − 2 π − 2 π − 2 π − 2 π 1 T = 60 MHz: 1 T = 30 MHz: 1 T = 15 MHz: 1 T = 10 MHz: 1 T = 5 MHz: P. Schniter, 2011 1 2. 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...
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## This note was uploaded on 10/29/2011 for the course ECE 600 taught by Professor Clymer,b during the Fall '08 term at Ohio State.

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

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