I want to compute the fundamental period of the following discrete time signal:

$ x[n]= \exp^{(j\frac{2\pi}{3})n} + \exp^{(j\frac{3\pi}{4})n}$

I know I can do this by taking the fundamental periods of the first term $\exp^{(j\frac{2\pi}{3})n}$, $N_1=3$ and of the second term $\exp^{(j\frac{3\pi}{4})n}$ $N_2=8$ and taking the least commmon multiple which gives $N=24$.

However I tried to express $x[n]$ as a single term as:

$ x[n]=\exp^{(j\frac{8.5\pi}{12})n} (\exp^{(j\frac{0.5\pi}{12})n}-\exp^{(j\frac{-0.5\pi}{12})n}) $

$ x[n]=2\exp^{(j\frac{8.5\pi}{12})n}.\cos(\frac{0.5 \pi n}{12}) $

Since the modulus of the complex exponential represents the magnitude of the signal, the fundamental period then equates to $N=48$ using only the cosine term. I tried this for varying frequencies and using this method I always got a the value for the period that was twice the correct value.

I know there is a flaw in my reasoning but could someone point that out ? And is there a reason why the fundamental period computed in this way always comes out to be twice that of the correct value ?

  • $\begingroup$ @cevans: I don't know the answer but reading your question led to the same question that I had when reading something in a text recently. You say that the term on the outside is a complex exponential. ( you left out the i's in your expressions but I get it ). My question is why the complex exponential doesn't get multiplied out because then a cosine term would multiply the cosine term that you have already and then there would be 2 cosine terms multiplied which I imagine would change the fundamental frequency. thanks. $\endgroup$
    – mark leeds
    Commented May 7, 2019 at 23:56
  • $\begingroup$ @markleeds Thanks for pointing out my silly mistake. Fixed it now. If we were to multiply the complex exponential then it would have a real cosine component and an imaginary sine compononent. Using the trignometric product to sum formulae we would end up again with the sum of two complex exponentials, that is we would end up at the first equation I typed out. $\endgroup$
    – acevans
    Commented May 8, 2019 at 0:17
  • $\begingroup$ Thanks for your explanation. I understand what you mean in that you'd end up back where you started from. But , in a slightly different context, the intro books often use what you did to show that a LTI system has the property than an impulse response can only effect an inputs amplitude and phase but not not its frequency. They end up with something similar to your last expression.. But, my confusion is always How do we know that the frequency won't change if everything got multiplied out into sines and cosines. I can find it in a textbook if I'm not explaining it clearly. Thanks. $\endgroup$
    – mark leeds
    Commented May 8, 2019 at 14:02
  • $\begingroup$ No problem. I'm not exactly sure what you mean. A LTI system does change the frequency of an input signal in most cases. Consider for example low pass filters: they remove higher frequency components present in the signal. An impulse response is used to determine the transfer function of the system. Maybe if you could point out what you are confused about in a textbook, as you said, that would be better. $\endgroup$
    – acevans
    Commented May 9, 2019 at 2:45
  • $\begingroup$ thanks. I'll try to find a textbook example of what I'm talking about and make a new question. Not sure when exactly but your efforts to help me are appreciated. $\endgroup$
    – mark leeds
    Commented May 9, 2019 at 14:18

1 Answer 1


The flaw in your reasoning is that you only consider the cosine term. The cosine term has a period of $48$. But there's also the complex exponential, which is not just a magnitude but a function of $n$. You can show that that exponential term is periodic with a period of $48$. However, you cannot conclude that the multiplication of the two terms must also have a period of $48$.

Multiplying the two terms results in a period of $24$ because



$$\cos\left(\frac{0.5\pi n}{12}\right)=-\cos\left(\frac{0.5\pi (n+24)}{12}\right)$$

so the sign changes cancel in the product.

  • $\begingroup$ very nice. thanks. $\endgroup$
    – mark leeds
    Commented May 9, 2019 at 19:14
  • $\begingroup$ @markleeds: welcome :) $\endgroup$
    – Matt L.
    Commented May 9, 2019 at 20:08
  • $\begingroup$ @MattL. Thanks a lot for that. Just an observation here: it is so stragne that x[n] and mod(x[n]) have the same period. (stack exchange discourages saying thank yous so I wasn't sure if I should or shouldn't even though I wanted to) $\endgroup$
    – acevans
    Commented May 10, 2019 at 1:15

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