# Fourier Series of Aperiodic convolution of periodic functions

we were given the following classic exercise:

Given two periodic signals $x(t), y(t)$ with fundamental period $T$ with Fourier series coefficients $c_m^x, c_m^y$ respectively, find the Fourier coefficients of the signal $z(t) = x(t) * y(t)$ with relation to $T, c_m^x, c_m^y$.

Now, this can easily be solved when the aforementioned convolution is the circular convolution (integral over a period only).

However, in class our professor noted that it can be solved even when we have an aperiodic convolution (that is, convolution as an integral from $-\infty$ to $+\infty$). We argued that, in this case, that infinite integral doesn't converge, and he responded that, even though the convolution integral doesn't converge (i.e. might be infinite), the Fourier Series coefficients are still finite and can be calculated!!

Is this true? If yes, then is the relation the usual one: $c_m^z=Tc_m^xc_m^y$ or another and how do you prove that? If not, why? Austere mathematical proofs would be appreciated.

• if $x(t)$ and $y(t)$ are both infinite energy signals and you're convolving them, you gotta problem. even a bigger problem if they're exactly the same period. – robert bristow-johnson Jan 14 '18 at 2:20
• your professor is wrong and i'll take him on any day. – robert bristow-johnson Jan 14 '18 at 2:23
• maybe with finite bandwidth? someone can crank it out with the Fourier series. – robert bristow-johnson Jan 14 '18 at 2:27
• @robertbristow-johnson Ok, I agree, but how do we prove that? Because, I guess that he kind of has a point, i.e. some combination of finite Fourier Series coefficients might eventually lead to infinite energy in one period. – user33199 Jan 14 '18 at 2:34
• no, not infinite energy in one period, but integrating a finite power over an infinite time. you can convolve a finite energy impulse response, $y(t)$, with a finite power (but infinite energy) input, $x(t)$, and get a finite power output $z(t)$. but if they're both power signals, you're screwded. – robert bristow-johnson Jan 14 '18 at 2:52

## 2 Answers

okay, now that the popcorn is eaten and my fingers are clean...

$$x(t) = \sum\limits_{m=-\infty}^{+\infty} c_m^x e^{j 2 \pi (m/T) t}$$

$$y(t) = \sum\limits_{m=-\infty}^{+\infty} c_m^y e^{j 2 \pi (m/T) t}$$

then

\begin{align} z(t) &= \sum\limits_{m=-\infty}^{+\infty} c_m^z e^{j 2 \pi (m/T) t} \\ \\ &= x(t) \circledast y(t) \\ \\ &= \int\limits_{-\infty}^{+\infty} x(\tau) y(t-\tau) \, d\tau \\ \\ &= \int\limits_{-\infty}^{+\infty} \sum\limits_{m=-\infty}^{+\infty} c_m^x e^{j 2 \pi (m/T) \tau} \sum\limits_{n=-\infty}^{+\infty} c_n^y e^{j 2 \pi (n/T) (t-\tau)} \, d\tau \\ \\ &= \sum\limits_{n=-\infty}^{+\infty} \sum\limits_{m=-\infty}^{+\infty} c_m^x c_n^y \int\limits_{-\infty}^{+\infty} e^{j 2 \pi (m/T) \tau} e^{j 2 \pi (n/T) (t-\tau)} \, d\tau \\ \\ &= \sum\limits_{n=-\infty}^{+\infty} \sum\limits_{m=-\infty}^{+\infty} c_m^x c_n^y \int\limits_{-\infty}^{+\infty} e^{j 2 \pi ((m-n)/T) \tau} \, d\tau \ e^{j 2 \pi (n/T) t} \\ \end{align}

this integral: $\int\limits_{-\infty}^{+\infty} e^{j 2 \pi ((m-n)/T) \tau} \, d\tau$ ain't converging when $n \ne m$ and certainly not when $n=m$.

you will not be getting finite values for $c_m^z$.

• Well, you cannot exchange summation with integration when integration doesn't absolutely converge, so that proof has a problem.... – user33199 Jan 14 '18 at 2:53
• @Jason it won't even converge in the case of finite summations, when you can swap the order. it won't converge. you cannot convolve two power signals against each other and get a finite power signal back. – robert bristow-johnson Jan 14 '18 at 3:07

You can in fact define a convolution product between periodic signals and power signals in general by using regularisation.

Because of its the Fourier invariance and closed point wise product, the Gauss distribution is a good choice for a regularisation window $W$.

$$A \star B := \lim_{b\to\infty} b^{-1} \int_\mathbb{R} W(b^{-1}(t-t'))\, A(t-t') \cdot W(b^{-1}(t))\, B(t) \, dt$$

For power signals the limit exists and for periodic signals with shared period the result coincides with the circular convolution of the periods.

• Would the downvoter please have the decency to leave a comment and point to the problem with the answer? Robert? – Jazzmaniac Jan 14 '18 at 17:47