{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}