Duality says that
if $$ x(t) \iff X(\omega) $$ then $$ X(t) \iff 2\pi \cdot x(-\omega) $$

Given that $$ f(t)\circledast g(t) \iff F(\omega)G(\omega) $$ By duality, shouldn't that imply that $$ f(t)g(t) \iff 2\pi \cdot F(\omega) \circledast G(\omega) $$ since $$ f(-t) \circledast g(-t) \cdot 2 \pi = 2 \pi \cdot f(t) \circledast g(t) $$

When I look at the theorem in my textbook it states that $$ f(t)g(t) \iff \tfrac{1}{2\pi} \cdot F(\omega) \circledast G(\omega) $$

I'm confused as to why the constant is divided by $2\pi$ instead of multiplied by $2\pi$. Does that mean that duality is violated?


Be careful on the signals which are forward Fourier transforms vs inverse Fourier transforms. So you want to show that $$x(t)y(t) \longleftrightarrow \frac{1}{2\pi} X(\omega) \star Y(\omega) \tag{1}$$

where $\star$ denotes convolution, and the signals to the right ($X(\omega)$ and $Y(\omega)$) are the forward Fourier transforms of the signals to the left ($x(t)$ and $y(t)$); a fact that you seem to ignore.

Looking at the inverse Fourier transform relationship $$ x(t) = \frac{1}{2\pi} \int X(\omega) e^{j \omega t} d\omega \tag{2} $$

we can manipulate it to see that $$ 2\pi ~ x(-\omega) = \int X(t) e^{-j \omega t} dt \tag{3} $$

Eq.3 states that forward Fourier transform of $X(t)$ is $2\pi ~ x(-\omega)$; this is indeed the duality realtionship.

Turning back into your proof it should be completed like this:

Let $$x(t) = f(t) \star g(t) \longleftrightarrow X(\omega) = F(\omega)G(\omega) \tag{4}$$

then by duality we should have $$X(t) = F(t)G(t) \longleftrightarrow 2\pi ~x(-\omega) = 2\pi ~ f(-\omega) \star g(-\omega) \tag{5}$$

Now in Eq.5, the signals to the right are NOT the forward Fourier transforms of the signals to the left. To correct this, and to make Eq.5 in line with Eq.1, you shall insert the forward Fourier transforms of $F(t)$ and $G(t)$ into the right of the equation. Again note that Forward Fourier transforms of $F(t)$ and $G(t)$ are $2\pi ~f(-\omega) $ and $2\pi ~g(-\omega) $ respectively.

$$F(t)G(t) \longleftrightarrow \frac{1}{2\pi} ( 2\pi ~f(-\omega) ) \star (2\pi ~g(-\omega) ) \tag{6}$$

or better stated $$F(t)G(t) \longleftrightarrow \frac{1}{2\pi} \mathcal{FT} \{ F(t) \} \star \mathcal{FT} \{ G(t) \} \tag{7}$$

you can replace the dummy function names $F,G,f,g$ with $x,y,X,Y$ in Eq.7 to get Eq.1

$$x(t)y(t) \longleftrightarrow \frac{1}{2\pi} \mathcal{FT}\{x(t) \} \star \mathcal{FT} \{ y(t) \} \tag{8}$$

Or stated finally $$x(t)y(t) \longleftrightarrow \frac{1}{2\pi} ~X(\omega) \star Y(\omega) \tag{9}$$

Now Eq.9 is the Eq.1.

  • $\begingroup$ Nice and to the point answer. $\endgroup$ Jul 17 '19 at 18:47
  • $\begingroup$ @DilipSarwate thanks prof. Sarwate... $\endgroup$
    – Fat32
    Jul 17 '19 at 21:36

Try using this definition of the continuous Fourier Transform and its inverse:

$$ \mathscr{F} \Big\{ x(t) \Big\} \triangleq X(f) \triangleq \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2 \pi f t} \ \mathrm{d}t $$

$$ \mathscr{F}^{-1} \Big\{ X(f) \Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty} X(f) \, e^{j2 \pi f t} \ \mathrm{d}f $$

see how well that works for your duality.

Okay, I'll be more explicit. Like your "$\omega$" definition, we have linearity:

$$ \mathscr{F} \Big\{ x(t) + y(t) \Big\} = \mathscr{F} \Big\{ x(t) \Big\} + \mathscr{F} \Big\{ y(t) \Big\} = X(f) + Y(f) $$

and derivative:

$$ \mathscr{F} \Big\{ \tfrac{\mathrm{d}}{\mathrm{d}t} x(t) \Big\} = j2\pi f \cdot X(f) $$

and delay:

$$ \mathscr{F} \Big\{ x(t-\tau) \Big\} = e^{-j2\pi f \tau} \cdot X(f) $$

But we get duality as:

$$\mathscr{F} \Big\{ X(t) \Big\} = x(-f)$$ $$\mathscr{F}^{-1} \Big\{ x(f) \Big\} = X(-t)$$

We get for convolution:

$$ \mathscr{F} \Big\{ h(t) \circledast x(t) \Big\} = H(f) \cdot X(f) $$

and it's the same in reverse:

$$ \mathscr{F}^{-1} \Big\{ W(f) \circledast X(f) \Big\} = w(t) \cdot x(t) $$

The other cool thing about this unitary definition of the Fourier Transform is how Parseval's theorem come out:

$$ \int\limits_{-\infty}^{\infty} \big| x(t) \big|^2 \ \mathrm{d}t = \int\limits_{-\infty}^{\infty} \big| X(f) \big|^2 \ \mathrm{d}f $$

And the more general Plancherel's theorem:

$$ \int\limits_{-\infty}^{\infty} x(t) \big(y(t)\big)^* \ \mathrm{d}t = \int\limits_{-\infty}^{\infty} X(f) \big(Y(f)\big)^* \ \mathrm{d}f $$

where $(\cdot)^*$ denotes the complex conjugate.

No funky scaling factors.

This is, in my opinion, clearly the most natural definition of the Fourier Transform if you are never relating it directly to the Laplace Transform. If you are bopping between the F.T. and the L.T. maybe your definition is better, but then I would say "$X(j \omega)$" so that the F.T. is simply identical to the L.T. but with the substitution of $s=j\omega$.


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