Determine if signal is periodic.

Discussion in 'Homework Help' started by shteii01, Sep 15, 2010.

1. shteii01 Thread Starter AAC Fanatic!

Feb 19, 2010
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x(n)=cos(n/3)cos(pi*n/3)

The cos times cos is throwing me off. I know how to determine if signal is periodic if I have one cos. How do I do it when I have product of two cos? Do I determine for each cos and if one is not periodic, then the whole thing is not periodic? Is that it?

2. Georacer Moderator

Nov 25, 2009
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The sinousoid signals (sines and cosines) are always periodic.
In general, in continuous time, the sum or product of two periodic signals is periodic, so your signal is periodic too.

It will look like a cosine whose envelope will be another cosine. That is, a cosine whose amplitude will change sinusoidaly.

3. shteii01 Thread Starter AAC Fanatic!

Feb 19, 2010
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In this book x(t) is continuous time, x(n) is discrete time. This is discrete time signal and I do not know how to deal with the product of cosines.

4. Georacer Moderator

Nov 25, 2009
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In the discrete time, things are a little different. In order for the signal $A \cdot \cos (\Omega_{\small0} \cdot t)$ to be periodical, the following should be valid:
$\frac{\Omega_{\small0}}{2 \pi}=\frac{m}{N}$, where m, an arbitrary integer, and N an integer representing the number of samples a full cycle will contain. In other words $\frac{\Omega_\small0}{2 \pi}$ should be rational, which isn't the case with $\cos (\frac{n}{3})$. I guess that makes your whole signal non-periodical.

However I cannot be 100% sure of that last sentence, as I don't know my Signals that well.

5. shteii01 Thread Starter AAC Fanatic!

Feb 19, 2010
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Ok, I asked the professor. For this one you use trig identity to rewrite the product as a sum. So then:
cosAcosB=0.5[cos(A+B)+cos(A-B)]

6. Georacer Moderator

Nov 25, 2009
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Ok, but what about its periodicity? Is it non-periodical as I said?

7. shteii01 Thread Starter AAC Fanatic!

Feb 19, 2010
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Both cosine are non-periodic so the whole signal non-periodic.

And now am curious. If one cosine was periodic and another was not, would that classify the whole signal as non-periodic?

8. JoeJester AAC Fanatic!

Apr 26, 2005
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They say a picture is worth a thousand words ....

File size:
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9. shteii01 Thread Starter AAC Fanatic!

Feb 19, 2010
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Can you explain what I am looking at?

All the explanations about periodic signal so far have been in regard to frequency. Frequency is rational, signal is periodic. Frequency is not rational, signal aperiodic. So just math. Nobody seem to want to put it in visual context.

I am beginning to dislike my textbook anyway, the thing is big enough you can kill with it and yet I still look for answers.

10. Georacer Moderator

Nov 25, 2009
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Why do you say that both signals are aperiodic? $\cos (\frac{\pi n}{3})$ is periodic as its frequency is $\frac{\pi}{3 \cdot 2 \cdot \pi}=\frac{1}{6}$, which is rational. Consequently, only one signal is aperiodical.

As for the image, well, as much as I like to use visual solutions in general, I don't think that in this particular problem it can help verify the solution. The shift of the cosine peak value, could be so small that we would mistake it for periodical. Math is our way out of here in my opinion.

11. JoeJester AAC Fanatic!

Apr 26, 2005
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n from 1 to 500 of the formulae you posted in the first posting.

Feb 19, 2010
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Here you go:

File size:
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13. Georacer Moderator

Nov 25, 2009
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I think you "cheated" a little. What I said is that from the two signals, whose product gave the final signal, one was periodical, and the other wasn't.
Instead, you converted them into a sum of two other signals, which, yes, no arguement there, are both aperiodical.

In the end, however, whichever method you choose, the resulting signal is aperiodical