ps1soln

# ps1soln - Massachusetts Institute of Technology Department...

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Unformatted text preview: Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.341: Discrete-Time Signal Processing Fall 2005 Problem Set 1 Solutions Issued: Thursday, September 8, 2005 Problem 1.1 (OSB 2.1) Answers are in the back of the book but have a few typos: (d) The system is causal when n ≥ 0, not when n ≤ 0. (h) (not assigned but for your benefit) The system is also causal. Problem 1.2 (OSB 2.6) (a) 1 y [ n ] − 2 y [ n − 1] = x [ n ] + 2 x [ n − 1] + x [ n − 2] Y ( e jω ) 1 − 1 e − jω = X ( e jω ) 1 + 2 e − jω + e − j 2 ω 2 Y ( e jω ) − jω + e − j 2 ω ( ) 1 + 2 e H e jω = = X ( e jω ) − jω 1 − 1 e 2 (b) H ( e jω ) = − jω + e − j 3 ω Y ( e jω ) 1 − 1 e 2 = − jω + 3 − j 2 ω X ( e jω ) 1 + 1 e e 2 4 1 Y ( e jω ) 1 + 1 e − jω + 3 e − j 2 ω = X ( e jω ) 1 − e − jω + e − j 3 ω 2 4 2 1 3 1 y [ n ] + 2 y [ n − 1] + 4 y [ n − 2] = x [ n ] − x [ n − 1] + x [ n − 3] 2 Problem 1.3 (OSB 2.11) We can write x [ n ] as a sum of exponentials and compute the response of the system to each exponential: ( ( 4 4 πn ) For x [ n ] = sin 4 , 1 πn j e π n − e π − j [ n ] = sin n = x 4 2 j 4 π Response to e j n : 4 π − j 2 √ 1 − e π 4 j π 4 n = j π 4 π 4 n = 2 π π j j j 2 e j H ( e ) e n = 2(1 + j ) e n 4 4 e e 4 π − j 4 1 + 1 e 2 4 π − j Response to e n : 4 π 1 − e j 2 √ π 4 ) e − j π 4 n = π 4 π 4 π π − j − j e − j − j − j H ( e n = 2(1 − j ) e n = 2 2 e n 4 4 e 4 π e j 4 1 + 1 2 √ 1 1 j ( − j ( ) − e ) π π π π j π 4 ) e j π 4 n − H ( e − j π 4 ) e − j π 4 n = n + n + [ n ] = H ( e 2 2 4 4 4 4 y e 2 j 2 j √ π ( n + 1) y [ n ] = 2 2 sin 4 Note: Answer in the back of the book has a typo. For x [ n ] = sin 7 πn ) , we have to map the frequency 7 π into the − π to π range where H ( e jω ) 4 4 is defined. πn 7 πn πn x [ n ] = sin = sin − = − sin 4 4 4 Then from above we have: √ π ( n + 1) y [ n ] = − 2 2 sin 4 Problem 1.4 (OSB 2.55) Yes. Suppose x 1 [ n ] = cos( ωn ) and x 2 [ n ] = cos(( ω + 2 π ) n ). Then x 1 [ n ] = x 2 [ n ] and the inputs are identical. All three systems behave deterministically, so the intermediate signals and the respective outputs A 1 and A 2 will be identical. Thus A will be periodic in ω (with period 2 π ). Recall that all distinct frequencies in discrete time fall...
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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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ps1soln - Massachusetts Institute of Technology Department...

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