I want to find the Fourier transform of the raised cosine with $\alpha=1$, i.e.
$g(t) = \text{sinc}(t/T)\frac{\cos(\pi t/T)}{1-4t^2/T^2}$
We can with substitution $u=t/T$, a trig identity and partial fractions expansion rewrite it as
$\frac{1}{4\pi u}\Big(\frac{\sin(2\pi u)}{2u+1} – \frac{\sin(2\pi u)}{2u-1}\Big)$. I also tried various variants of $\cos(u) = \sin(\pi/2-u)$, obtaining
$\frac{\pi}{2}\frac{\text{sinc}(u)}{2}\Big(\text{sinc}(2u+1)+\text{sinc}(2u-1)\Big)$. But from here I'm not sure how to proceed.
Edit: actually got as far as
$= \frac{\sin(2\pi u)}{4\pi}\Big(\frac{1}{u}+\frac{1}{1-2u}-\frac{1}{1+2u}\Big)$ now, if I haven't made any mistakes…
Best Answer
The last line is wrong, it's supposed to be: $$\frac{\sin(2\pi u)}{2\pi}\Big(\frac{1}{u}-\frac{1}{1-2u}-\frac{1}{1+2u}\Big)$$ I'll work out the first term, the rest are similar: $$F(\omega)=\mathcal{F}_u\bigg[\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)\bigg]=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du$$
Keep in mind that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Hence: $$F(\omega)=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du=\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega-2\pi) u}du-\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega+2\pi) u}du$$
Using two facts $$\mathcal{F}_t[e^{i\omega_0t}\cdot f(t)]=F(\omega-\omega_0)$$ where $G(\omega)=\mathcal{F}_t[f(t)]$ and $$\mathcal{F}_t\bigg[\frac 1t\bigg]=-i\pi\text{sgn} (\omega)$$
You get $$F(\omega)=-\frac 1{4} \text{sgn}(\omega-2\pi)+\frac 1{4} \text{sgn}(2 \pi+\omega)$$
For the other parts, using the following property (very easy to prove): $$\mathcal{F}_t(f(t-t_0))=e^{-it_0\omega}F(\omega)$$ $$F_1(\omega)=\frac 12 e^{-i\omega/2}F(\omega)$$ $$F_2(\omega)=\frac 12 e^{i\omega/2}F(\omega)$$
The transform $G_u(\omega)$ of the function $f(u)$ is: \begin{align*} G_u(\omega)&=F(\omega)+F_1(\omega)+F_2(\omega)\\ &=(1+\cos(w/2)F(\omega)\\ &=2\cos^2(\omega/4)F(\omega) \end{align*}
So that $$\mathcal{F}_t(g(t))=G(\omega)=\frac T2 \cos^2(\frac{T\omega} 4) \left(\text{sgn}(2 \pi - T\omega) + \text{sgn}(2 \pi + T\omega)\right)$$