# Proving Fourier transform pair with derivatives using duality

I want to use duality to prove the Fourier transform pair $$t^nx(t) \overset{\mathscr{F}}{\longleftrightarrow} j^n\frac{d^nX(\omega)}{d\omega^n}$$ but I am struggling.

I learned that if $$x(t) \overset{\mathscr{F}}{\leftrightarrow} X(\omega)$$ then $$X(t) \overset{\mathscr{F}}{\leftrightarrow} 2\pi x(-\omega)$$, however I am not sure if I can apply it here. My take, going from the derivative first:

$$g(t) = \frac{d^n}{dt^n}x(t)\overset{\mathscr{F}}{\longleftrightarrow} (j\omega)^nX(\omega) = G(\omega)$$

My knowledge is still shaky, so I try to apply duality with pure symbolic manipulation, so I should find that $$G(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi g(-\omega)$$, which should get us back to the first expression I want to prove:

$$G(t) = (jt)^n X(t) \overset{\mathscr{F}}{\longleftrightarrow} 2\pi \frac{d^n}{dt^n}x(-\omega) = 2\pi g(-\omega)$$ Obviously, there are a few problems.

1. I want $$t^nx(t)$$ but I have $$(jt)^n X(t)$$. What should I do with this $$j$$?
2. The $$2\pi$$ should not appear, but the duality formula requires so. Why?
3. Why the negative sign on the $$\omega$$?

I want to really understand the duality property but for some reason, it just won't click...

Let's start with $$n=1$$. We know that

$$\frac{dx(t)}{dt}\overset{\mathscr{F}}{\longleftrightarrow} j\omega X(\omega)\tag{1}$$

where $$X(\omega)$$ is the Fourier transform of $$x(t)$$.

From the duality property we obtain

$$jtX(t)\overset{\mathscr{F}}{\longleftrightarrow}2\pi\frac{dx(-\omega)}{d(-\omega)}\tag{2}$$

Now the problem is that $$x(\omega)$$ is not the Fourier transform of $$X(t)$$. If we use $$\hat{X}(\omega)$$ to denote the Fourier transform of $$X(t)$$, we have, again from duality,

$$\hat{X}(\omega)=2\pi x(-\omega)\tag{3}$$

Plugging $$(3)$$ into $$(2)$$ gives

$$jtX(t)\overset{\mathscr{F}}{\longleftrightarrow}\frac{d\hat{X}(\omega)}{d(-\omega)}\tag{4}$$

Multiplying both sides of $$(4)$$ by $$-j$$ results in

$$tX(t)\overset{\mathscr{F}}{\longleftrightarrow}j\frac{d\hat{X}(\omega)}{d\omega}\tag{5}$$

Finally, applying $$(5)$$ $$n$$ times gives the desired relationship

$$t^nX(t)\overset{\mathscr{F}}{\longleftrightarrow}j^n\frac{d^n\hat{X}(\omega)}{d\omega^n}\tag{6}$$

• Thank you very much. One last point: could you please develop the argument that "$x(\omega)$ is not the Fourier transform of $X(t)$? At point (3), isn't that substitution exactly proving that it is? Apr 19, 2022 at 7:20
• @chuchvara: $X(t)$ is the Fourier transform of $x(\omega)$, so $x(\omega)$ is the inverse Fourier transform of $X(t)$. Not much of a difference, just a factor of $2\pi$ and a minus sign, as shown in Eq. (3). Apr 19, 2022 at 9:49
• Ok thank you for the precision. I often get confused with the notation and will need to work on this. It doesn't help that I had a first course in signal processing where we solely expressed the FT using $f$ instead of $\omega$, increasing the confusion... Apr 19, 2022 at 9:52