Fourier transform of the raised cosine pulse.

Question

How do I solve for the Fourier transform of the raised cosine function given below:

$$p(t) = \frac{\operatorname{sinc}(Rt)\cos(πaRt)}{1-4a^2R^2t^2},$$ where $0<a<1$.

P.S.- The final result is known to me but I don't know how to solve for it.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \on{p}\pars{t} & \equiv \bbox[5px,#ffd]{\on{sinc}\pars{Rt} \cos\pars{\pi aRt} \over 1 - 4a^{2}R^{2}t^{2}} \end{align} Lets $\ds{\pars{~\tau \equiv \pi\verts{aR}t \implies t = {\tau \over \pi\verts{aR}}~}}$ and $\ds{\beta \equiv {1 \over \pi\verts{a}}}$ such that \begin{align} \on{p}\pars{t} & \equiv {\on{sinc}\pars{\beta t} \cos\pars{\tau} \over 1 - 4\tau^{2}/\pi^{2}} = -\,{\pi^{2} \over 4}\, {\on{sinc}\pars{\beta\tau} \cos\pars{\tau} \over \tau^{2} - \pars{\pi/2}^{2}} \end{align} Then, \begin{align} &\bbox[#ffd,5px]{\int_{-\infty}^{\infty} \on{p}\pars{t}\expo{-\ic\omega t}\,\dd t} \\ = &\ -\,{\pi^{2} \over 4\pi\verts{aR}}\ \overbrace{\int_{-\infty}^{\infty} {\on{sinc}\pars{\beta\tau} \cos\pars{\tau} \over \tau^{2} - \pars{\pi/2}^{2}} \,\expo{-\ic\nu\tau}\,\dd\tau} ^{\ds{\equiv {\cal J}}} \\[2mm] &\ \mbox{where}\quad \nu \equiv {\omega \over \pi\verts{aR}} \end{align}

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\begin{align} {\cal J} & \equiv \int_{-\infty}^{\infty} {\on{sinc}\pars{\beta\tau} \cos\pars{\tau} \over \tau^{2} - \pars{\pi/2}^{2}} \,\expo{-\ic\nu\tau}\,\dd\tau \\[5mm] & = {1 \over \pi}\int_{-\infty}^{\infty} {\on{sinc}\pars{\beta\tau} \cos\pars{\tau} \over \tau - \pi/2} \,\expo{-\ic\nu\tau}\,\dd\tau \\[2mm] & - {1 \over \pi}\int_{-\infty}^{\infty} {\on{sinc}\pars{\beta\tau} \cos\pars{\tau} \over \tau + \pi/2} \,\expo{-\ic\nu\tau}\,\dd\tau \\[5mm] & = -{\expo{-\ic\nu\pi/2} \over \pi}\int_{-\infty}^{\infty} {\on{sinc}\pars{\beta\tau + \beta\pi/2} \sin\pars{\tau} \over \tau} \,\expo{-\ic\nu\tau}\,\dd\tau \\[2mm] & -{\expo{\ic\nu\pi/2} \over \pi}\int_{-\infty}^{\infty} {\on{sinc}\pars{\beta\tau - \beta\pi/2} \sin\pars{\tau} \over \tau} \,\expo{-\ic\nu\tau}\,\dd\tau \\[5mm] & = -{\expo{-\ic\nu\pi/2} \over \pi}\int_{-\infty}^{\infty} \on{sinc}\pars{\beta\tau + \beta\pi/2} \on{sinc}\pars{\tau} \,\expo{-\ic\nu\tau}\,\dd\tau \\[2mm] & -{\expo{\ic\nu\pi/2} \over \pi}\int_{-\infty}^{\infty} \on{sinc}\pars{\beta\tau + \beta\pi/2} \on{sinc}\pars{\tau} \,\expo{\ic\nu\tau}\,\dd\tau \\[5mm] & = -{2 \over \pi}\Re\bracks{\expo{-\ic\nu\pi/2} \int_{-\infty}^{\infty} \on{sinc}\pars{\beta\tau + \beta\pi/2} \on{sinc}\pars{\tau} \,\expo{-\ic\nu\tau}\,\dd\tau} \end{align}

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\begin{align} &\int_{-\infty}^{\infty} \on{sinc}\pars{\beta\tau + \beta\pi/2} \on{sinc}\pars{\tau} \,\expo{-\ic\nu\tau}\,\dd\tau \\[5mm] = &\ \int_{-\infty}^{\infty} \bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic k\pars{\beta\tau + \beta\pi/2}}\,\,\dd k} \bracks{{1 \over 2}\int_{-1}^{1}\expo{-\ic q\tau} \,\dd q} \expo{-\ic\nu\tau}\,\dd\tau \\[5mm] = &\ {\pi \over 2}\int_{-1}^{1}\expo{\ic k\beta\pi/2} \int_{-1}^{1}\ \overbrace{\int_{-\infty}^{\infty} \expo{\ic\pars{k\beta - q - \nu}\tau}\,\, {\dd\tau \over 2\pi}} ^{\ds{\delta\pars{k\beta - q - \nu}}}\ \dd q\,\dd k \\[5mm] = &\ {\pi \over 2}\int_{-1}^{1}\expo{\ic k\beta\pi/2}\ \bracks{-1 < k\beta - \nu < 1}\,\dd k \\[5mm] = &\ {\pi \over 2}\int_{-1}^{1}\expo{\ic k\beta\pi/2}\ \bracks{{\nu - 1 \over \beta} < k < {1 + \nu \over \beta}}\,\dd k \end{align} Now, you can finishes the job.