solns5_600(2)

solns5_600(2) - ECE-600 Introduction to Digital Signal...

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Unformatted text preview: ECE-600 Introduction to Digital Signal Processing Autumn 2010 Homework #5 Nov. 5, 2010 HOMEWORK SOLUTIONS #5 1. (a) To resample a discrete-time signal from a sampling rate of 990 MHz to an effective rate of 999 MHz, we change the sampling rate by the ratio 999000 990000 = 999 990 = 111 110 , noticing that { 111 , 110 } are co-prime (i.e., we cannot simplify their ratio any further). Re- sampling then consists of the following three steps: i. Upsample by 111. ii. Lowpass filter the result with cutoff frequency min { π 111 , π 110 } = π 111 (radians/sample) and DC gain 111. iii. Downsample by 110. 2. (a) We start with the definition of the DTFT: ˜ X ( e jω ) = ∞ summationdisplay n = −∞ ˜ x [ n ] e − jωn (1) = ∞ summationdisplay l = −∞ N − 1 summationdisplay m =0 ˜ x [ m + lN ] e − jω ( m + lN ) via n = m + lN (2) = ∞ summationdisplay l = −∞ e − jωlN N − 1 summationdisplay m =0 x [ m ] e − jωm (3) where in the last equation we used the facts that x [ m ] = ˜ x [ m ] = ˜ x [ m + lN ] for any m = , 1 ,...,N − 1 and any l ∈ Z . Next, we recall from Homework 2 that ∞ summationdisplay l = −∞ e − jωlN = 2 π ∞ summationdisplay k = −∞ δ ( ωN − k 2 π ) (4) = 2 π N ∞ summationdisplay k = −∞ δ ( ω − k 2 π N ) (5) where the last equation used the Dirac scaling rule. Pluggin (5) into (3), we get ˜ X ( e jω ) = 2 π N ∞ summationdisplay k = −∞ δ ( ω − k 2 π N ) N − 1 summationdisplay m =0 x [ m ] e − jωm (6) which shows that the only values of ω yielding a non-zero DTFT are ω = k 2 π N for integers k . We can use this fact to rewrite the previous expression as ˜ X ( e jω ) = 2 π N ∞ summationdisplay k = −∞ δ ( ω − k 2 π N ) N − 1 summationdisplay m =0 x [ m ] e − j 2 π N km . (7) P. Schniter, 2010P....
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This note was uploaded on 02/14/2011 for the course ECE 600 taught by Professor Clymer,b during the Fall '08 term at Ohio State.

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solns5_600(2) - ECE-600 Introduction to Digital Signal...

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