Search Search Any Topic from Any Website Search
Fourier Transform of \(x(t)=\dfrac{t}{(1+t^2)^2}\) We want to determine the Fourier transform of $$ x(t)=\frac{t}{(1+t^2)^2} $$ using the Fourier transform definition $$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt $$ Step 1 : Let $$ x_1(t)=\frac{1}{1+t^2} $$ Then $$ X_1(\omega)=\pi e^{-|\omega|}. $$ Proof: $$ \int_{-\infty}^{\infty}\frac{1}{1+t^2}e^{-j\omega t}\,dt $$ Use the Residue Theorem. $$ \int_{-\infty}^{\infty}\frac{1}{1+z^2}e^{-j\omega z}\,dz $$ Or, $$ \int_{-\infty}^{\infty}\frac{1}{(z+j)(z-j)}e^{-j\omega z}\,dz $$ Poles are at \(z=-j\) and \(z=j\). For the pole at \(z=-j\): $$ \operatorname*{Res}_{z=-j}\left(\frac{e^{-j\omega z}}{(z+j)(z-j)}\right) = \lim_{z\to -j}(z+j)\frac{e^{-j\omega z}}{(z+j)(z-j)} = \frac{e^{-\omega}}{-2j}. $$ Therefore, $$ -2\pi j\left(\frac{e^{-\omega}}{-2j}\right) = \pi e^{-\omega}. $$ For the pole at \(z=j\): $$ \operatorname*{Res}_{z=j}\left(\frac{e^{-j\omega z}}{(z+j)(z-j)}\righ...