Fourier transform: duality property and convolution

[fatpotato]

Homework Statement

Using duality, prove that $F(t)G(t) \overset{\mathscr{F}}{\longleftrightarrow} \frac{1}{2\pi} f(-\omega) \ast g(-\omega)$

Relevant Equations

Definition of Fourier transform : $\mathscr{F} \{x(t)\} = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt$

Definition of inverse Fourier transform : $\mathscr{F}^{-1} \{X(\omega)\} = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega$

Duality property : If $x(t) \overset{\mathscr{F}}{\longleftrightarrow}X(\omega)$ then $X(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi x(-\omega)$

Fourier transform of a convolution : $f(t) \ast g(t) \overset{\mathscr{F}}{\longleftrightarrow} F(\omega) G(\omega)$

Hello,

First of all, I checked several other threads mentioning duality, but could not find a satisfying answer, and I don't want to revive years old posts on the subject; if this is bad practice, please notify me (my apologies if that is the case).

For the past few days, I have had a lot of troubles with the duality property of the Fourier transform (to put it lightly, I am desperate). I want to prove that a product in the time domain corresponds to a convolution in the frequency domain, but I cannot get it to work.

What I understand (I hope):

Duality states that if $f(t) \overset{\mathscr{F}}{\longleftrightarrow} F(\omega)$ then $F(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi f(-\omega)$, meaning that we can easily find the Fourier transform of a function whose morphology is known from a table of transforms, for example.

Thus, knowing that $\delta(t) \overset{\mathscr{F}}{\longleftrightarrow} 1$, by duality we have $1 \overset{\mathscr{F}}{\longleftrightarrow} 2\pi \delta(-\omega) = 2\pi \delta(\omega)$, by parity of the Dirac's delta function.

What I don't understand :

We know that $f(t) \ast g(t) \overset{\mathscr{F}}{\longleftrightarrow} F(\omega) G(\omega)$, so I would want to compute the Fourier transform of $F(t) G(t)$ using duality like stated before:
$$ F(t)G(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi f(-\omega) \ast g(-\omega)$$
But this is wrong, and my lessons mention that instead I should have:
$$ F(t)G(t) \overset{\mathscr{F}}{\longleftrightarrow} \frac{1}{2\pi} f(-\omega) \ast g(-\omega)$$
This is the part that is making me lose sleep. Why is my first try erroneous? Further, if I try to define $h(t) = f(t) \ast g(t)$, then its Fourier transform should be $H(\omega) = F(\omega)G(\omega)$, and now going purely by duality I should find:
$$ F(t)G(t) = H(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi h(-\omega) = 2\pi \big( f(-\omega) \ast g(-\omega) \big)$$
This is contradictory, because we should have $H(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi h(-\omega)$, which is basically how the property is stated in my lesson (except for using another letter which does not matter). However, the result is wrong.

What I have tried :

I suspected a misinterpretation of the duality (although my symbolic manipulation looks correct to me) and that I should rather think of "putting the Fourier transform of each function on the right side", yielding:
$$ F(t)G(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi \big( 2\pi f(-\omega) \ast 2\pi g(-\omega) \big)$$
Where $2\pi f(-\omega)$ if the Fourier transform of $F(t)$, likewise for $G(t)$. Still, in this scenario, I have $2\pi$ to the cube, whereas I should have $\frac{1}{2\pi}$...

Any generous mind to put an end to my torment?

Thank you

 

[fatpotato]

Ok, I think I managed to get some modicum of understanding, but any comment is more than welcome, I am not sure of my steps.

Looking again at the following:
$$ F(t)G(t) = H(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi h(-\omega) = 2\pi \big( f(-\omega) \ast g(-\omega) \big)$$
I believe that there are missing steps. First, let us define the Fourier transform of $F(t)$ and $G(t)$ using the overscript hat symbol to denote the Fourier transform:
$$F(t) \overset{\mathscr{F}}{\longleftrightarrow} \hat{F}(\omega) = 2\pi f(-\omega) \,\,\,\,\, , \,\,\,\,\, G(t) \overset{\mathscr{F}}{\longleftrightarrow} \hat{G}(\omega) = 2\pi g(-\omega)$$
Thus, I can replace $f(-\omega)$ by $\frac{\hat{F}(\omega)}{2\pi}$ and $g(-\omega)$ by $\frac{\hat{G}(\omega)}{2\pi}$ to get:
$$ 2\pi \big( f(-\omega) \ast g(-\omega) \big) = 2\pi \big( \frac{\hat{F}(\omega)}{2\pi} \ast \frac{\hat{G}(\omega)}{2\pi} \big) = \frac{1}{2\pi}\big( \hat{F}(\omega) \ast \hat{G}(\omega) \big)$$
Now, reverting back to the first symbolic notation, where time-domain functions are written with lowercase characters, and their Fourier transforms are written in the frequency-domain using uppercase characters, we are allowed to write:
$$ f(t)g(t) \overset{\mathscr{F}}{\longleftrightarrow}\frac{1}{2\pi}\big( F(\omega) \ast G(\omega) \big)$$
Thus proving the initial frequency-domain convolution property.

If this proof is correct, I believe that I now understand how to prove properties using duality, but applying duality to actual transform pairs will surely confuse my poor small brain. Tips and advice are welcome.

Edit: reword