I am looking for the proof that the Hilbert transform of $\displaystyle\frac{\sin(at)}{at}$ is given by


How do we prove this? This is a $\operatorname{sinc}(at)$ function. I could not find a proof of this in any textbook so far. Preferably I want to arrive at this result using the analytic signal whose imaginary part would be the Hilbert transform:

$$f_a(t) = 2 \int^\infty_0 FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu$$

I tried this in a few different ways, but could not get to the answer.

  • $\begingroup$ Good question indeed, in fact I do not remember having used this formula for the analytic signal, so +1 $\endgroup$ May 4, 2016 at 8:09

2 Answers 2


I agree that one of the easiest ways to compute the Hilbert transform in this case is to use the analytic signal. This is most easily obtained via the Fourier transform. Note that the Fourier transform of




i.e., a rectangular function with value $\pi/a$ in the interval $[-a,a]$, and zero otherwise. The spectrum of the corresponding analytic signal equals the part of $F(j\omega)$ at positive frequencies, scaled by a factor of $2$:


The inverse Fourier transform of $(3)$ gives the analytic signal:

$$f_a(t)=\frac{1}{2\pi}\frac{2\pi}{a}\int_0^{a}e^{j\omega t}d\omega=\frac{1}{jat}\left(e^{jat}-1\right)=e^{jat/2}\frac{\sin(at/2)}{at/2}\tag{4}$$

The Hilbert transform of $f(t)$ is the imaginary part of $(4)$:


as stated in your question.


The integrand in the Hilbert transform formula is $h(t,u) = \frac{f(t)}{u-t}$. With a (non-dilated) cardinal sine, you get $$\frac{\sin(t)}{t(u-t)} = \frac{1}{u}\left( \frac{\sin(t)}{u-t}+\frac{\sin(t)}{t}\right)\,,$$ by splitting the rational fraction. The first term inside the parentheses, when integrated (with the principal value), is the Hilbert transform of a sine, $-\cos(u)$, finally divided by $u$. The second one is the integral of a cardinal sine, of value $1$.

So, the Hilbert transform is $$\frac{1-\cos(u)}{u}\,,$$ hence $$\frac{\sin^2(u/2)}{2u}$$ since $\sin^2(u/2) = \frac{1-\cos(u)}{2}$. Replace $u=at$, and you are done. No need to know the Fourier transform of a $\operatorname{sinc}$, no need to integrate complex functions, too complicated for me;)

Since I like to live in danger, let me try another one, even shorter, without resorting to the analytic signal (again).

Let $f$ and $g$ be functions whose Fourier transforms have disjoint supports. A nice property of the Hilbert transform is the Bedrosian theorem, often used in its simpler form called "Low-pass High-pass Products" (there exists $W$ such that $F(w) = 0$ if $w< W$ and $G(w) = 0$ if $w > W$). The result is: $$\mathcal{H}(f(t)g(t)) = f(t)\mathcal{H}(g(t))\,.$$ Write $\operatorname{sinc}(t) = 2\operatorname{sinc}(t/2)\cos(t/2)$ and apply the above result with $g(t) = \cos(t/2)$: $$\mathcal{H}(\operatorname{sinc}(t)) = 2\operatorname{sinc}(t/2)\sin(t/2)\,.$$

For a sounder use of the Bedrosian theorem, let me suggest A necessary and sufficient condition for a Bedrosian identity, 2010, using a generalized cardinal sine.

  • 1
    $\begingroup$ Is it possible to arrive at this result by first finding the analytic signal $f_a (t) = 2 \int^\infty_0 F(\nu) \exp(j 2 \pi \nu t) \ d\nu$? We can then find the Hilbert transform $ \hat{f} (t)$ from $f_a (t) = f(t) +j\hat{f} (t).$ I tried writing the integral as (1) $$f_a(t) = 2 \int^\infty_{-\infty} u(t) \ FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu$$ And (2) $$f_a(t) = 2 \int^\infty_0 FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu = 2 \int^\infty_0 \frac{1}{|a|} \Pi \left( \frac{\nu}{a} \right) \exp(j2 \pi \nu t) \ d\nu.$$ I didn't succeed using either approach $\endgroup$
    – Merin
    May 3, 2016 at 9:13
  • $\begingroup$ @Merin Very likely, though I won't do that this afternoon. I think that truncating the integral on the support of the "sinc" window will do the job. But the definition of the analytic signal (in the frequency) does not require you to compute it effectively there. You can check in Gabor 1946 paper that "It can be easily verified that the signal [associated with the real part] is given by the integral", follows the Hilbert transform $\endgroup$ May 3, 2016 at 12:21
  • $\begingroup$ Noted. Thank you so much for this explanation. The Hilbert transform formula is beyond the scope of the course I am taking. I am just trying to find Hilbert transform of some functions by computing the analytic signal first. $\endgroup$
    – Merin
    May 4, 2016 at 7:44
  • 1
    $\begingroup$ +1 for reminding me of Bedrosian's theorem. You're indeed pretty fearless in applying it, because the frequency of the cosine is exactly at the discontinuity of the transform of the sinc ... $\endgroup$
    – Matt L.
    May 4, 2016 at 7:53
  • $\begingroup$ @Matt L. Thank you. I was aware of the danger, and I should add that analycity of the functions is required in the version I know. $\endgroup$ May 4, 2016 at 8:06

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