Hilbert transform pair proof

Question

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

$$\frac{\sin^2(at/2)}{at/2}.$$

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.

Answer 1

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

$$f(t)=\frac{\sin(at)}{at}\tag{1}$$

is

$$F(j\omega)=\frac{\pi}{a}\text{rect}\left(\frac{\omega}{2a}\right)\tag{2}$$

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 $2$:

$$F_a(j\omega)=\frac{2\pi}{a}\text{rect}\left(\frac{\omega}{a}-\frac12\right)\tag{3}$$

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 $(4)$:

$$\hat{f}(t)=\text{Im}\{f_a(t)\}=\frac{\sin^2(at/2)}{at/2}\tag{5}$$

as stated in your question.

Answer 2

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.