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...

Thank you for your time.


1 Answer 1


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


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


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


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


  • $\begingroup$ 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? $\endgroup$
    – chuchvara
    Apr 19, 2022 at 7:20
  • $\begingroup$ @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). $\endgroup$
    – Matt L.
    Apr 19, 2022 at 9:49
  • $\begingroup$ 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... $\endgroup$
    – chuchvara
    Apr 19, 2022 at 9:52

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