Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider a function

$y[n]= \cos[w n ]$, where $n$ is an integer.

I have to prove that this signal will have highest rate of oscillation at $w = \pi$. I was thinking I can take the derivatives (of rate of oscillation) and set condition of maxima, but the variable is a discrete variable and not a continuous one.

How do I actually prove this?

Update

This is what I observe:

at $w = 0$ all samples will have value 1 and at $w = \pi$ samples will have positive and negative values in succession- highest rate of oscillation

Please [see this, page number 16], but they have not proved it.

Here is the corresponding page from the book (Digital Signal Processing: Principles, Algorithms and Applications (3rd Edition) ):

enter image description here enter image description here

As can be seen they have just verified, and not proved. I want to prove this.

share|improve this question
    
Why did you change to $\cos?$ It doesn't change anything much except to make it $\frac \pi 2$ instead of $0$ in my last sentence. –  Ross Millikan Feb 5 '13 at 18:05
    
In my original question , it is actually cos –  gpuguy Feb 5 '13 at 18:07

1 Answer 1

What do you mean by "maximum oscillation"? If the range of $y$, it will be $(-1,1)$ for almost any choice of $w$. If by a local maximum, it isn't a function of $w$, but of $n$. If $\theta=0, y[n]$ will always be $0$ for $w=\pi$, certainly not a maximum oscillation.

share|improve this answer
    
Maximum oscillation means highest rate of oscillation –  gpuguy Feb 5 '13 at 18:10
    
@gpuguy: $3\pi$ will work just as well, as will $5 \pi \ldots$ –  Ross Millikan Feb 5 '13 at 18:29
    
please see page 16 here –  gpuguy Feb 5 '13 at 18:40
    
@gpuguy: It doesn't let me Look Inside –  Ross Millikan Feb 5 '13 at 19:02
    
the book is: Digital Signal Processing, by Proakis et al. –  gpuguy Feb 5 '13 at 19:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.