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Given the discrete-time function $f[n] = 2\cos(\frac{\pi}{4}n) + \sin(\frac{\pi}{8}n) - 2\cos(\frac{\pi}{2}n + \frac{\pi}{6})$

How can I show that the function is periodic? I know that a discrete time function is periodic if $x[n] = x[n+N]$ where N and n are considered to be positive integers.

But even if I were to just plug [n+N] in I can't see how to determine equality from there.

I considered letting each trig function be the Re/Im part of their respective complex exponentials so that I could combine them and (potentially)more easily group the functions so I could show the equality for a given N, but I'm not sure if that's a valid way of doing it.

I'd also considered that the sum of periodic functions can be periodic, but not always, and I'm not completely clear on what conditions guarantee that a sum of periodic functions is also periodic.

Looking around at other similar questions and other sites, I can see that if they do not have a common multiple, then the sum is not periodic, but these do, which to me says that this may be a periodic sum, but also may not:

Is it always the case that if individual periodic functions have a common multiple, the sum is also periodic?

I can easily just let N be 16, based on the above, and see that it is the period of $f$, but I'm looking for more general information.

BigM
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Daniel B.
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1 Answers1

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As in this case you're dealing with discrete periodic functions, the sum of periodic functions is also periodic. To show that you only have to consider that the function $f[n]=\sum_{i=1}^Mf_i[n]$ will have period $N=\text{lcm}(N_1,N_2,...,N_M)$, where $N_i$ is the period of $f_i$ since $f_i[n+N]=f_i[n+kN_i]=f_i[N]$, for some natural $k$. In your exemple, we have $N_1=8, N_2=16$ and $N_3=4$. So we got $N=\text{lcm}(8,16,4)=16$. In the general case, this is not always true, since the periods may not even be intergers. For exemple, consider the functions $f(x)=\sin x$ and $g(x)=\sin \pi x$. It is easy to see, that $f$ and $g$ are periodic, with period $2\pi$ and $2$, respectively. But the sum $f(x)+g(x)$ is not periodic. Assume, that it has period $T$. then, we would have $f(x+T)+g(x+T)=f(x+2\pi k)+g(x+2j)$, for some $k,j\in \mathbb{Z}$. But this would lead to $$2\pi k =2j \Leftrightarrow \pi=\frac{j}{k},$$ which is a contradiction, since $\pi$ is irrational. By the above argument, we can see that in the general case, the sum will be periodic iff the ratio of the periods is a rational number.

Mateus Sampaio
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