i want to prove that the forier transform of sinc(100t) is a pulse by integration ..

Question

I am trying to prove that the foreir transform of sinc(100t) is a pulse . I know that the forier transform of sinc is a normal pulse but am trying prove that by integration.. any help .....?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, $\ds{\,\mrm{sgn}}$ and $\ds{\,\mrm{H}}$ are the sign and the Heaviside Step Functions, respectively.

\begin{align} \int_{-\infty}^{\infty}{\sin\pars{100 t} \over 100 t}\,\expo{\ic\omega t}\,\dd t & = {1 \over 100} \int_{-\infty}^{\infty}{\sin\pars{100 t}\cos\pars{\omega t} \over t}\,\dd t \\[5mm] & = {1 \over 200} \int_{-\infty}^{\infty}{\sin\pars{\bracks{100 + \omega}t} \over t}\,\dd t + {1 \over 200} \int_{-\infty}^{\infty}{\sin\pars{\bracks{100 - \omega}t} \over t}\,\dd t \\[5mm] & = {\mrm{sgn}\pars{\omega + 100} \over 200} \int_{-\infty}^{\infty}{\sin\pars{t} \over t}\,\dd t - {\mrm{sgn}\pars{\omega - 100} \over 200} \int_{-\infty}^{\infty}{\sin\pars{t} \over t}\,\dd t \\[5mm] & = {\pi \over 200}\braces{\bracks{2\,\mrm{H}\pars{\omega + 100} - 1} - \bracks{2\,\mrm{H}\pars{\omega - 100} - 1}} \\[5mm] & = {\pi \over 100}\bracks{\mrm{H}\pars{\omega + 100} - \mrm{H}\pars{\omega - 100}} = \bbx{{\pi \over 100}\,\mrm{H}\pars{100 - \verts{\omega}}} \end{align}