Hilbert transform of Sinc

[roam]

Homework Statement

Show that the Hilbert transform of $\frac{\sin(at)}{at}$ is given by

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

Homework Equations

The analytic signal of a function is given by $f_a(t) = 2 \int^\infty_0 F(\nu) \exp(j2 \pi \nu t) \ d\nu,$ where $F(\nu)$ is the Fourier transform of the function. We have $f_a (t) = f(t) +j\hat{f}(t),$ where $\hat{f}(t)$ is the Hilbert transform.

$FT(sinc(at)) = \frac{1}{a} \Pi (\frac{\nu}{a})$, where $\Pi$ denotes the rectangular function.

The Attempt at a Solution

This is 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 to rewrite this as:

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

So, it looks like an inverse Fourier transform, so this becomes $\left( \frac{1}{j2 \pi \nu} + \frac{\delta(\nu)}{2} \right) \frac{\sin(at)}{at}.$ So this is not the correct solution. So here is another approach:

$$f_a(t) = 2 \int^\infty_0 FT \Big[ \frac{\sin(at)}{at} \Big] \exp(j2 \pi \nu t) \ d\nu = 2 \int^\infty_0 \frac{1}{|a|} \Pi \left( \frac{\nu}{a} \right) \exp(j2 \pi \nu t) \ d\nu$$

So, how do I continue from here? What method should I use?

 

[mighty2000]

One method you could use is to rewrite the rectangular function in terms of sines and cosines using the Fourier series expansion. This would give you:

$$f_a(t) = \frac{2}{|a|} \sum_{n=0}^\infty \frac{\sin(n\pi t/a)}{n\pi t/a} \cos(2\pi nt)$$

Then, you can use the fact that the Hilbert transform of a cosine is equal to the negative of the sine, and vice versa, to get the imaginary part of the analytic signal. This would give you:

$$\hat{f}(t) = -\frac{2}{|a|} \sum_{n=0}^\infty \frac{\cos(n\pi t/a)}{n\pi t/a} \sin(2\pi nt)$$

From here, you can simplify using trigonometric identities and eventually arrive at the desired solution:

$$\hat{f}(t) = -\frac{2}{a} \sum_{n=0}^\infty \frac{\sin^2(n\pi t/2a)}{n\pi t/2a}$$

which is equivalent to $$\frac{\sin^2(at/2)}{at/2}$$ after some algebraic manipulation.

Another method you could use is the residue theorem from complex analysis. You can show that the Fourier transform of $\frac{\sin(at)}{at}$ is given by $$F(\nu) = \frac{1}{2}\left( \delta(\nu-a) + \delta(\nu+a) \right).$$

Then, you can use the fact that the Hilbert transform of a function is equal to the convolution of the function with $\frac{1}{\pi t}$. This would give you:

$$\hat{f}(t) = \frac{1}{\pi t} * F(\nu) = \frac{1}{2\pi t}\left( \delta(\nu-a) + \delta(\nu+a) \right)$$

Using the fact that the integral of a delta function centered at $\nu_0$ is equal to 1, you can simplify this to:

$$\hat{f}(t) = \frac{1}{2\pi t}\left( e^{j2\pi