HW_Set_10_Solutions - Introduction to Signal Processing...

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Introduction to Signal Processing Hayder Radha - 1 - ECE 366 Homework Set #10 Wednesday, April 15, 2009 (In class) ECE 366 – Introduction to Signal Processing Spring 2009 Michigan State University Department of Electrical and Computer Engineering Please remember to follow the rules and policies outlined in the ECE 366 Syllabus [1] Problem 3.3-1 (a) For : [ ] () 1 n xn = − Using: [] 2 1 lim 21 N x N nN Px n N + →∞ =− = ++ 2 11 lim 1 lim 2 1 1 N n x NN PN + →∞ →∞ =+ = . (d) For : [ ] [ ] 1 n x nu n 2 0 1 lim 1 lim 2 N n x n + →∞ →∞ = = = . (e) For: cos 36 n π This is a periodic discrete-time signal with period 0 N : 0 2 6 3 N == ⎛⎞ ⎜⎟ ⎝⎠ . In other words, when evaluating the power of the signal, we need to consider a set of discrete-time samples that cover a complete period (i.e., six samples in this case), and take the average over such set of samples:
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Introduction to Signal Processing Hayder Radha - 2 - ECE 366 2 5 0 2 2 2222 1 cos 63 6 1 cos cos cos 2 cos 3 cos 4 cos 5 66 3 6 3 6 3 6 3 6 3 6 x n Pn ππ π = ⎛⎞ ⎡⎤ =+ ⎜⎟ ⎢⎥ ⎣⎦ ⎝⎠ = + ++ + () 22 2 2 13 3 3 3 1 00 3 0 . 5 62 2 2 2 6 x P = + +− +− + + = = . [2] Problem 3.3-4 (a) This signal can be represented as a linear combination of time-shifted ramp functions. There are a couple of ways of performing this representation. Below is an example of such expression. First, we have one increasing ramp that begins as a zero value at 3 n =− and it grows linearly until it reaches a value of 3 at 0 n = . Such function can be expressed as: [ ] ( ) [ ] [ ] 1 33 1 xn n un + . (Test the signal at different values of n from 3 n = − to 0 n = .) [ ] ( ) [ ] [ ] ( ) 1 1 n = n 3 2 1
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Introduction to Signal Processing Hayder Radha - 3 - ECE 366 Second, we have a decaying ramp that with a value of +2 at 1 n = and it goes to zero at 3 n = . [ ] () [ ] [ ] 2 31 3 xn n un =−+ − − . Therefore: [ ] [ ] [ ] 12 x nx n =+ (c) [ ] [ ] [ ] 34 nun [ ] x n n 3 2 1 3 2 1 n 3 2 1 3 2 1 [ ] ( ) [ ] [ ] 2 3 n
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Introduction to Signal Processing Hayder Radha - 4 - ECE 366 [3] Problem 3.4-8 In this problem we are given the following information: The response of the system is: [ ] [ ] [ ] [ ] 1 12 2 yn n n n δδ δ =− + − + , when the input is: [ ] [ ] [ ] 1 1 xn n n =+ Since the system is a linear time-invariant system, we can identify the response of the system [ ] 2 to any other input [ ] 2 x n if we can express that input ( [ ] 2 x n ) as a linear combination of time-shifted versions of the input [ ] 1 x n .
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This note was uploaded on 06/08/2009 for the course ECE 366 taught by Professor Staff during the Spring '08 term at Michigan State University.

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HW_Set_10_Solutions - Introduction to Signal Processing...

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