midexam_soln

midexam_soln - 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.341 Discrete-Time Signal Processing Fall 2005 MIDTERM EXAM Tuesday, November 8, 2005 SOLUTIONS Disclaimer: These are not meant to be full solutions, but rather an indication of how to approach each problem. Problem Grade Points Grader 1 /15 2 (a) /12 2 (b) /6 2 (c) /12 3 (a) /10 3 (b) /10 4 /15 5 (a) /10 5 (b) /10 Total /100
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± ² ³ ´ 2 Problem 1 (15%) A stable system with system function H ( z ) has the pole-zero diagram shown in Figure 1-1. It can be represented as the cascade of a stable minimum-phase system H min ( z ) and a stable all-pass system H ap ( z ). Re( z ) Im( z ) 1 2 3 4 Figure 1-1: Pole-zero diagram for H ( z ). Determine a choice for H min ( z ) and H ap ( z ) (up to a scale factor) and draw their corresponding pole-zero plots. Indicate whether your decomposition is unique up to a scale factor. We find a minimum-phase system H min ( z ) that has the same frequency response magnitude as H ( z ) up to a scale factor. Poles and zeros that were outside the unit circle are moved to their conjugate reciprocal locations (3 to 1 4 , to 0). 3 ,4to 1 (1 1 z 1 ) 4 H min ( z )= K 1 (1 1 z 1 )(1 1 z 1 ) 2 3 There is no need to include an explicit z term to account for the zero at the origin. now include all-pass terms in H ap ( z ) to move poles and zeros back to their original locations 1 z 3 in H ( z ). The term 1 3 z 1 moves the pole at 1 back to 3, the term z 1 moves the zero from 0 3 to , and so on: H ap ( z z 1 z 1 3 z 1 1 4 1 3 z 1 1 1 z 1 4 The decomposition is unique up to a scale factor. cannot include additional all-pass terms in H ap ( z ), since it is not possible for H min ( z ) to cancel the resulting poles and zeros outside the unit circle.
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3 Name: Re( z ) Im( z ) 1 2 1 3 1 4 Figure 1-2: Pole-zero diagram for H min ( z ). Re( z ) Im( z ) 1 3 1 4 3 4 zero at Figure 1-3: Pole-zero diagram for H ap ( z ).
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± ² 4 Problem 2 (30%) x c ( t ) x L [ n ] C/D x [ n ] 2 y H ( e ) y M [ n [ n ] 2 y c ( t ) ] D/C T T Figure 2-1: For parts (a) and (b) only , X c ( j Ω) = 0 for | | > 2 π × 10 3 and H ( e ) is as shown in Figure 2-2 (and of course periodically repeats). ) H ( e A ω π π 4 4 Figure 2-2: (12%) (a) Determine the most general condition on T , if any, so that the overall continuous-time system from x c ( t )to y c ( t )isLTI .
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This note was uploaded on 10/09/2009 for the course EE 6.341 taught by Professor Xyz during the Fall '05 term at Massasoit.

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midexam_soln - 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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