# Proof of fourier transformation of multiplication of two signals

I've been trying to find a proof of the following, but still I m unable to proof it, can someone help me? $$ℱ[x(t)g(t)] = \frac{1}{2\pi} [X(\omega)*G(\omega)]$$ • You can’t have looked very hard for the proof. Almost any text on signal processing will do. – Dilip Sarwate Jun 7 at 22:18
• @DilipSarwate O dear, Can you suggest me text that has proof of this? – Gurpreet Singh Jun 8 at 6:41
• Try replacing only $x(t)$ by its representation as the inverse Fourier transform of $X(\omega)$ so that the integral formula for $X_q(\omega)$ involves only two integrals instead of the three that you have. Then interchange the order of integration. When you finish solving the question, write up a nicely LaTexed answer and post it. Yes, it is permissible (even encouraged) to post an answer to one's own question. – Dilip Sarwate Jun 8 at 13:52

I am very thankful to Dilip-sarwate and Gilles, who took their precious time to understand my problem and guide me.
So, Now I'm going to write the correct solution to my question. Which is as follows: $$ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}[x(t)g(t)]e^{-j\omega t}dt$$
As we know : $$x(t) = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha t}d\alpha$$
Therefore, $$ℱ[x(t)g(t)] = \int_{t=-\infty}^{\infty}\bigg[\frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)e^{j\alpha t}d\alpha\bigg]g(t) e^{-j\omega t}dt$$
After rearranging, we have $$ℱ[x(t)g(t)] = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)\bigg[\int_{t=-\infty}^{\infty}g(t)e^{-j(\omega-\alpha)t}dt\bigg]d\alpha$$
As we also know that : $$G(\omega) = \int_{t=-\infty}^{\infty}g(t)e^{-j\omega t}dt$$ and replacing $$\omega$$ with $$\omega-\alpha$$, we get: $$G(\omega-\alpha) = \int_{t=-\infty}^{\infty}g(t)e^{-j(\omega-\alpha)t}dt$$
Therefore, $$ℱ[x(t)g(t)] = \frac 1{2\pi}\int_{\alpha=-\infty}^{\infty}X(\alpha)G(\omega-\alpha)d\alpha$$
Finally, we get $$ℱ[x(t)g(t)] = \frac 1{2\pi}[X(\omega)*G(\omega)]$$
[$$\because$$ $$x_1(t)*x_2(t) = \int_{\tau=-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau$$]

• Good job. After a couple of days, you can even accept your own answer as the most satisfactory answer. – Dilip Sarwate Jun 9 at 0:49
• You're welcome, well done! – Gilles Jun 9 at 7:11

HINT:

Write it as: $$\mathcal{F}\big\{x_1(t)x_2(t)\big\} = \frac 1{2\pi}\int_{-\infty}^{\infty}\bigg[\mathcal{F}^{-1}\big\{X_1(\alpha)\big\}x_2(t)\bigg]e^{-j\Omega t}dt$$

You correctly used the inverse Fourier transform of $$X_1(\alpha)$$ for a new expression of $$x_1(t)$$ as in $$(1)$$

$$x_1(t) = \frac 1{2\pi}\int_{-\infty}^{\infty}X_1(\alpha)e^{j\alpha t}d\alpha\tag{1}$$ Don't express $$x_2(t)$$ in the form of the inverse Fourier transform, leave it as it is and you'll end up with an exponential to the power $$-j(\Omega - \alpha)t$$. The expression can be further simplified and give equation $$(2)$$ $$\frac 1{2\pi}\int_{-\infty}^{\infty}X_1(\alpha)X_2(\Omega - \alpha)d\alpha\tag{2}$$