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HW_Set_10_Solutions

# 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 x n = − Using: [ ] 2 1 lim 2 1 N x N n N P x n N + →∞ =− = + + ( ) ( ) 2 1 1 lim 1 lim 2 1 1 2 1 2 1 N n x N N n N P N N N + →∞ →∞ =− = = + = + + + + . (d) For : [ ] ( ) [ ] 1 n x n u n = − ( ) ( ) 2 0 1 1 1 lim 1 lim 2 1 2 1 2 N n x N N n P N N N + →∞ →∞ = = = = + + + + . (e) For: [ ] cos 3 6 x n n π π = + This is a periodic discrete-time signal with period 0 N : 0 2 6 3 N π π = = . 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 2 2 2 2 1 cos 6 3 6 1 cos cos cos 2 cos 3 cos 4 cos 5 6 6 3 6 3 6 3 6 3 6 3 6 x n P n π π π π π π π π π π π π π = = + = + + + + + + + + + + ( ) 2 2 2 2 2 2 1 3 3 3 3 1 0 0 3 0.5 6 2 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 3 3 1 x n n u n u n = + + . (Test the signal at different values of n from 3 n = − to 0 n = .) [ ] ( ) [ ] [ ] ( ) 1 3 3 1 x n n u n u n = + + n 3 2 1