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Here is my signal

Cos(n/2)*cos(pi*n/4)

cos(n/2) has period 4pi and cos(pi*n/4) has period 8

Now, the question is will the signal be periodic for fundamental period 32pi ?

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  • $\begingroup$ If $n$ is a discrete time variable (i.e. it assumes only integer values), then $\cos(n/2)$ is not a periodic signal, simply because $4\pi$ isn't rational. $\endgroup$ – Matt L. Jul 29 '14 at 10:17
  • $\begingroup$ @Matt-I yeah, that is where I was confused. So, the conclusion should be that it is periodic if it is analog and aperiodic if it is digital right ? $\endgroup$ – Sujit Maharjan Jul 29 '14 at 10:19
  • $\begingroup$ In discrete time it is indeed not periodic. $\endgroup$ – Matt L. Jul 29 '14 at 10:26
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You can try using this:

(1) cos(a+b) = cos(a)*cos(b)+sin(a)*sin(b)
(2) cos(a-b) = cos(a)*cos(b)-sin(a)*sin(b)

Then you can do (1)+(2) and get that

cos(a)*cos(b)=1/2(cos(a+b)+cos(a-b))

Then you can identify the period of cos(a+b)-cos(a-b) by using this for example http://fourier.eng.hmc.edu/e101/lectures/Fundamental_Frequency.pdf

Hope this helps

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  • $\begingroup$ $\cos(a)\cos(b)=\frac12[\cos(a+b)+\cos(a-b)]$. Please try for yourself to find the period, and then you might want to reconsider your answer. $\endgroup$ – Matt L. Jul 29 '14 at 10:24
  • $\begingroup$ summation or multiplication doesn't make difference in calculation of period. $\endgroup$ – Sujit Maharjan Jul 29 '14 at 10:30
  • $\begingroup$ Sorry, I mixed them up... Should have looked them up before writing. I also think I missed the point of the question, the answer is more a general approach to finding periods of multiplied sinusoid signals... $\endgroup$ – schvaba986 Jul 29 '14 at 11:05

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