half cosine pulse

[logistic_guy]

Find the Fourier transform of the half-cosine pulse shown.

 

[logistic_guy]

This graph can be described by:

\(\displaystyle g(t) = \cos \frac{\pi t}{T} \ \text{rect}\frac{t}{T}\)

where \(\displaystyle \text{rect}\frac{t}{T}\) is the rectangular function.

 

[logistic_guy]

Let \(\displaystyle g_1(t) = A\cos \frac{\pi t}{T}\) and \(\displaystyle g_2(t) = \text{rect}\frac{t}{T}\)


\(\displaystyle G_1(f) = \mathcal{F}\{g_1(t)\} = \mathcal{F}\left\{A\cos \frac{\pi t}{T}\right\} = \frac{A}{2}\left[\delta\left(f - \frac{1}{2T}\right) + \delta\left(f + \frac{1}{2T}\right)\right]\)

 

[logistic_guy]

\(\displaystyle g_2(t) = \text{rect}\frac{t}{T}\)

\(\displaystyle G_2(f) = \mathcal{F}\{g_2(t)\} = \mathcal{F}\left\{\text{rect}\frac{t}{T}\right\} = T\text{sinc} fT\)

\(\displaystyle \textcolor{red}{\text{Definition.}}\)

\(\displaystyle \text{sinc} \lambda = \frac{\sin \pi \lambda}{\pi \lambda}\)

Then,

\(\displaystyle G_2(f) = T\text{sinc} fT = T\frac{\sin \pi fT}{\pi fT}\)

 

[logistic_guy]

Now we are ready to do the convolution which will give us the Fourier transform of \(\displaystyle g(t)\).


\(\displaystyle \mathcal{F}\{g(t)\} = G(f) = G_1(f)*G_2(f) = \frac{AT}{2}\int_{-\infty}^{\infty} \frac{\sin \pi \mu T}{\pi \mu T} \left[\delta\left(\mu - \left[f - \frac{1}{2T}\right]\right) + \delta\left(\mu - \left[f + \frac{1}{2T}\right]\right)\right] \ d\mu\)

Properties of \(\displaystyle \delta\) function gives:

\(\displaystyle G(f) = \frac{AT}{2}\left(\frac{\sin \pi \left[f - \frac{1}{2T}\right] T}{\pi \left[f - \frac{1}{2T}\right] T} + \frac{\sin \pi \left[f + \frac{1}{2T}\right] T}{\pi \left[f + \frac{1}{2T}\right] T}\right)\)

With a little simplification, we get:

\(\displaystyle G(f) = \frac{A}{2\pi}\left(\frac{\cos \pi fT}{f + \frac{1}{2T}} - \frac{\cos \pi fT}{f - \frac{1}{2T}}\right)\)