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HW9sol_FA08 - UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN...

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UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Homework 9 Solutions Wednesday, October 22, 2008 Prof. Bresler, Prof. Jones Due by October 29, 2008 Problem 1 (30 points) (a) Given an ideal D/A with period T = 2 ms, and input sequence y [ n ] = cos( nπ/ 4) for all n , (i) Determine a closed-form expression for the output y ( t ). (ii) Sketch (by hand) the output for 0 t 20ms. (iii) Determine a closed-form expression for the Fourier Transform of the output. (iv) Sketch (by hand) the magnitude of the Fourier Transform of the output for 0 Ω 4000 π (v) Identify on your sketch from (iv) the spurious component with the largest magnitude, and determine a closed-form expression for its magnitude and frequency. (b) Repeat (ii), (iv), and (v) in (a) above, when the ideal D/A is replaced by a zero-order hold. (c) Repeat (i)-(v) in (a) above when the ideal D/A is replaced by a zero-order hold followed by an analog compensator with frequency response F a (Ω) = 2 · 10 3 e - j 6 · 10 - 3 Ω sinc(10 - 3 Ω) , | Ω | ≤ 500 π 0 , | Ω | > 500 π Compare your results to (a) and (b) above and explain. (d) Extra credit: Repeat (ii), (iv), and (v) in (a) above, when the ideal D/A is replaced by a zero-order hold followed by an analog compensator built of an RC circuit, with time constant R · C = T/π = (2 ) · 10 3 . Solution: (a) (i) For an ideal D/A, g a ( t ) = sinc ( π T t ) G a (Ω) = T rect ( Ω T 2 π ) . Also for y [ n ] = cos( nπ/ 4), Y d ( ω ) = n = −∞ π δ ( ω π 4 2 ) + δ ( ω + π 4 2 ) 1
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Therefore y a ( t ) = + n = −∞ y [ n ] g a ( t nT ) = + n = −∞ cos π 4 n sinc π T ( t nT ) Y a (Ω) = G (Ω) Y d T ) = πT δ T π 4 ) + δ T + π 4 ) , | Ω | ≤ π T 0 , otherwise = π δ π 4 T ) + δ (Ω + π 4 T ) , | Ω | ≤ π T 0 , otherwise = π [ δ 125 π ) + δ (Ω + 125 π )] y a ( t ) = cos(125 πt ) (ii) See Figure 1 Figure 1: y a ( t ) for 0 t 20ms for Problem 1(a) (iii) See solution to part (i). (iv) See Figure 2. Figure 2: | Y a (Ω) | for Problem 1(a) (v) No spurious component in this case. (b) In this case, y a ( t ) = + n = −∞ cos π 4 n p 0 ( t nT ) where p 0 ( t ) = 1 , 0 t T 0 , else P 0 (Ω) = T e j Ω T 2 sinc Ω T 2 2
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Therefore, Y a (Ω) = T e j Ω T 2 sinc Ω T 2 Y d T ) = T e j Ω T 2 sinc Ω T 2 n = −∞ π T δ π 4 T 2 T ) + δ (Ω + π 4 T 2 T ) = π e j Ω T 2 sinc Ω T 2 n = −∞ δ π 4 T 2 T ) + δ (Ω + π 4 T 2 T ) The spurious component with the largest magnitude are a set of Dirac delta at frequencies ± 1 T (2 π π 4 ) = ± 875 π , and amplitude | π sinc( ± 7 π 8 ) | = 0 . 44. See Figures 3 and 4 for sketches of y a ( t ) and Y a (Ω) in this case.
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