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solns4_600(2)

# solns4_600(2) - ECE-600 Homework#4 Introduction to Digital...

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ECE-600 Introduction to Digital Signal Processing Autumn 2010 Homework #4 Oct. 22, 2010 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 ) X ( e ) X ( e ) X ( e ) X ( e ) 0 0 0 0 0 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 = 12 MHz: 1 T = 6 MHz: 1 T = 3 MHz: 1 T = 2 MHz: 1 T = 1 MHz: P. Schniter, 2010 1

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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 ) = 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 ) = 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 0 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
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