I know that Fourier transform of $t^n f(t)= i^n \frac{d}{d\omega^n} F(\omega)$.

But does this work when $n<0$?

Is there any direct relation to compute the Fourier transform of $\frac{f(t)}{t}$?

  • $\begingroup$ Try sketching $f(t)/t$ in the vicinity of 0 to see if you can discern any reason to suspect that the desired Fourier transform might not even exist. Nothing obvious springs to mind? Try reading your textbook especially the gobbledygook just before and after the definition of Fourier transform. $\endgroup$ Aug 15 '20 at 18:41
  • 1
    $\begingroup$ @DilipSarwate I got what you're trying to say. The function might blow up at t=0. But there can be nicer functions like sin(t)/t which does not blow up and hence Fourier transform do exist. So, I was trying to see if there is a general way to transform for the nicer functions. $\endgroup$
    – user824530
    Aug 15 '20 at 18:45

A unit step in the frequency domain $u(f)$ has the following inverse Fourier transform:

$$\mathcal{F}^{-1}\{u(f)\}=\frac12\delta(t)-\frac{1}{2\pi jt}\tag{1}$$

This follows from the well-known Fourier transform of the time-domain unit step $u(t)$, and from the basic Fourier transform property


Given $x(t)$ and its Fourier transform $X(f)$, we have

$$(X\star u)(f)=\int_{-\infty}^fX(\nu)d\nu\tag{3}$$

where $\star$ denotes convolution.

Using $(1)$, the inverse Fourier transform of $(3)$ is

$$x(t)\cdot \mathcal{F}^{-1}\{u(f)\}=\frac{x(0)}{2}\delta(t)-\frac{x(t)}{2\pi jt}\tag{4}$$


$$\mathcal{F}\left\{\frac{x(t)}{t}\right\}=2\pi j\left[\frac{x(0)}{2}-\int_{-\infty}^fX(\nu)d\nu\right]\tag{5}$$

As an exercise, you can try to verify $(5)$ with $x(t)=\sin(2\pi f_0t)/\pi$ (resulting in an ideal low-pass with cut-off frequency $f_0$), and with $x(t)=1/\pi$ (resulting in an ideal Hilbert transformer).

  • $\begingroup$ @Matt_L. Bracewell uses $\star$ for correlation (no time reversal) and $\ast$ for convolution. In Eqn. (3) what operation does $\star$ represent? $\endgroup$
    – Andy Walls
    Sep 15 '20 at 12:11
  • $\begingroup$ @AndyWalls: It's convolution; I've added a clarification to my answer. $\endgroup$
    – Matt L.
    Sep 15 '20 at 12:27

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