Fourier Series of Aperiodic convolution of periodic functions

Question

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.

Answer 1

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 $c_m^z$.

Answer 2

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.