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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?


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.

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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

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.

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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

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