- #1

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.

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.

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.

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

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