What is the minimum period P (in samples) of the signal $e^{j(\frac{M}{N} )*2πn}$ for the following values of M and N?




I have got the answer for the first pair of values, which is 3. But I could not get the other two correct. Please help.

  • 2
    $\begingroup$ Hint: $\exp\left(j\frac MN2\pi n\right) = \exp(0) = 1$ when $n=0$, and this value will repeat when $n$ is the smallest positive integer such that $\frac MN n$ equals an integer. So, try the third case for $n=1,2,3,4,\ldots $ and report back to us what the smallest value of $n$ is such that $\frac MN n$ equals an integer. $\endgroup$ Oct 11 '13 at 22:00
  • $\begingroup$ @DilipSarwate I have got the correct answer, which is 3, 7 and 3. Thanks for your help! $\endgroup$
    – Yicheng Ye
    Oct 12 '13 at 3:48
  • $\begingroup$ could you please elaborate the explanation a bit more thanks $\endgroup$
    – user5667
    Oct 14 '13 at 18:40
  • $\begingroup$ @hamina Read this fourier.eng.hmc.edu/e101/lectures/Fundamental_Frequency/… $\endgroup$
    – Yicheng Ye
    Oct 15 '13 at 5:40

It's already answered in the comments so just to close it out: the answers are 3, 7 and 3. The periodicity is basically given by the denominator of the fraction in front of 2*pi*n after the fraction has been simplified (no common divisors between nominator and denominator). The simplified fractions for the cases are 1/3, 5/7 and 7/3 so the periodicity is 3, 7 and 3 respectively.


I cannot ask a question in the comment section therefore I decided to add my question here.

The website provided by Yicheng Ye somehow gives something and I believe people who know this stuff understand all of it but I still find it confusing. I don't understand how to work out the fundamental frequency or minimum period.

I tried my own workaround but I am not sure if I am right. Please correct me if I am wrong. Thanks. Fundamental frequency:

$\omega_0n=2\pi\frac{M}{N}n$, ..... dividing by n gives

To calculate minimum period, I have rearranged the formula $\frac{\omega_0}{2\pi{}M}=\frac{1}{N}$ and then got the answer:


Which means that $N$ is a minimum period. I don't feel if this is correct. Perhaps I don't because cannot really understand what $M$ represents here or what it does correspond to. I will be grateful for explanation. Cheers!


If we have : \begin{equation} x[n] = x[n + N]\\ e^{j(\omega n+\phi)} = e^{j(\omega(n+N)+\phi)}\\ e^{j\omega n}e^{j\phi} = e^{j\omega n}e^{j\omega N}e^{j\phi} \end{equation}

if we now divide both sides by : $e^{j\omega n}e^{j\phi}$, we get : \begin{equation} e^{j\omega N} = 1\\ e^{j\omega N} = e^{j2\pi M}\\ \omega N = 2\pi M\\ \omega = \frac{2\pi M}{N} \end{equation}


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