# What is the Fourier convolution theorem range of application (example of Dirac comb times rectangular window)?

$$\DeclareMathOperator{\sinc}{sinc}$$

I have questions regarding the Fourier transform of the product of functions or distributions and the range of application of the convolution theorem.

# Context

When measuring a signal $$x$$, measurements are made at a certain frequency $$f_s=\frac{1}{T_s}$$ over a certain period $$L$$. The measured signal then equals the following distribution: $$W_{[\frac{-L}{2};\frac{L}{2}]}.\operatorname{\text{Ш}}_{T_s}.x$$, with $$W_{[\frac{-L}{2};\frac{L}{2}]}$$ being the rectangular window. Let consider a simple case that illustrates my understanding and my pending questions (Q1 and Q2): $$x=1$$. Thus, I want to know the Fourier transform of the product between a Dirac comb and a rectangle window: $$W_{[\frac{-L}{2};\frac{L}{2}]}.\operatorname{\text{Ш}}_{T_s}$$.

# Convolution theorem

Unless I am wrong, the convolution theorem has two variants, for functions and for tempered distributions.

• The convolution theorem for functions can be applied for two integrable functions (which have a Fourier transform): $$\hat{fg}=\hat{f}\ast\hat{g}$$.
• The convolution theorem for tempered for distributions can be applied to a product of one tempered distribution and one function of the Schwartz space [https://www-fourier.ujf-grenoble.fr/~edumas/td3-1617.pdf].

Q1: First of all, could you please confirm or correct me please?

If I am correct, it is then impossible to apply any of the variants. The Dirac comb is not a function, thus the first variant is inapplicable. The rectangle window does not belong to the Schwartz space, the second variant is then also inapplicable. If so, I would be surprised to often see the argument of the convolution theorem when computing the Fourier transform of the product of two signals, in particular in Wikipedia for the product of the Dirac comb with any signal (without restricting them to the Schwartz space) [https://en.wikipedia.org/wiki/Dirac_comb#Fourier_transform].

# Fourier transforms

The Fourier transform of the product, which is a finite number of Dirac, is then the sum of the Fourier transform of these Dirac thanks to the linearty property of the Fourier transform. Noting $$k$$, the highest integer such that $$k<\frac{L}{2}$$: $$\hat{W_{[\frac{-L}{2};\frac{L}{2}]}.\operatorname{\text{Ш}}_{T_s}}(\omega) = \sum_{n=-k}^{k} e^{-j\omega nT_s}$$

We know that the Fourier transforms of the Dirac comb and the rectangular window are the following: \begin{align} \hat{\operatorname{\text{Ш}}_{T_s}} &= \frac{2\pi}{T_s} \operatorname{\text{Ш}}_{\frac{2\pi}{T_s}} \\ \hat{W_{[\frac{-L}{2};\frac{L}{2}]}}(\omega) &= L\sinc(\frac{L\omega}{2}) \end{align}

Even if the convolution cannot be applied (according to my understanding), let's try to compute the convolution of the two Fourier transforms, which is $$\frac{2\pi}{T_s}$$ periodic: \begin{align} \hat{\operatorname{\text{Ш}}_{T_s}} \ast \hat{W_{[\frac{-L}{2};\frac{L}{2}]}}(\omega) &= \frac{2\pi}{T_s} \sum_{n=-\infty}^{\infty} L\sinc(\frac{L}{2}(\omega+\frac{2\pi n}{T_s})) \end{align}

Since it is periodic, the Fourier series can then be calculated: $$c_n = \frac{T_s}{2\pi} \int_{0}^{\frac{2\pi}{T_s}} \frac{2\pi}{T_s} \sum_{k=-\infty}^{\infty} L\sinc(\frac{L}{2}(\omega+\frac{2\pi n}{T_s})) e^{-j 2\pi \frac{T_s}{2\pi} n \omega} \, d\omega$$

If we could exchange the sum and the integral, we would get (like I saw on this post for instance https://math.stackexchange.com/questions/1242280/what-is-the-sum-over-a-shifted-sinc-function): \begin{align} c_n &= \sum_{k=-\infty}^{\infty} \int_{0}^{\frac{2\pi}{T_s}} L\sinc(\frac{L}{2}(\omega+\frac{2\pi n}{T_s})) e^{-j 2\pi \frac{T_s}{2\pi} n \omega} \, d\omega \\ &= \sum_{k=-\infty}^{\infty} \int_{k \frac{2\pi}{T_s}}^{(k+1) \frac{2\pi}{T_s}} L\sinc(\frac{Lu}{2}) e^{-j 2\pi \frac{T_s}{2\pi} n (u - k \frac{2\pi}{T_s})} \, du \\ &= \sum_{k=-\infty}^{\infty} \int_{k \frac{2\pi}{T_s}}^{(k+1) \frac{2\pi}{T_s}} L\sinc(\frac{Lu}{2}) e^{-j 2\pi \frac{T_s}{2\pi} n u} \, du \\ &= \int_{-\infty}^{\infty} L\sinc(\frac{Lu}{2}) e^{-j T_s n u} \, du \\ &= W_{[\frac{-L}{2};\frac{L}{2}]}(-T_s n) \\ \end{align}

Hence we can observe that: $$\operatorname{\text{Ш}}_{\frac{2\pi}{T_s}} \ast L\sinc(\frac{L\omega}{2}) \stackrel{?}{=} \sum_{n=-\infty}^{\infty} c_n e^{-j \frac{2\pi}{\frac{2\pi}{T_s}} n \omega} = \sum_{n=-k}^{k} e^{-j T_s n \omega} = \hat{W_{[\frac{-L}{2};\frac{L}{2}]}.\operatorname{\text{Ш}}_{T_s}}(\omega)$$

However this result relies on the assumption we can exchange the two symbols. I know the dominated convergence theorem can justify such an exchange. However, it is applicable if the absolute function is integrable. But $$\sinc$$ is not $$L_1$$ integrable.

Q2: Am I right? Am I missing something? Does another theorem can be applied to justify the symbol exchange?

Once again, if I am right, I cannot assert that The Fourier transform of the product equals the convolution of the Fourier transforms since I cannot prove the first part of the previous equality: $$\operatorname{\text{Ш}}_{\frac{2\pi}{T_s}} \ast L\sinc(\frac{L\omega}{2}) \stackrel{?}{=} \sum_{n=-k}^{k} e^{-j T_s n \omega}$$ which relies of the symbol exchange.