Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ x(t)=t^2 u(t-1) $
$ X(\omega)=\int_{-\infty}^{\infty}t^2 u(t-1) e^{-j\omega t}dt \; = \int_{1}^{\infty}t^2 e^{-j\omega t}dt $
Integration by Parts
$ \int u \; dv = uv - \int v \; du $
$ u=t^2 \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t} $
$ du=2t \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t} $
$ X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t}|_{1}^{\infty} + \frac{2}{j \omega}\int_{1}^{\infty}t^2 e^{-j\omega t}dt $
Integration by Parts
$ u=t \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t} $
$ du=1 \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t} $
$ X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t}|_{1}^{\infty} + \frac{2}{j \omega}[\frac{tj}{\omega}e^{-j\omega t}|_{1}^{\infty}+\frac{1}{j \omega}\int_{1}^{\infty}e^{-j\omega t}dt] $
$ =[\frac{t^2 j}{\omega}e^{-j\omega t} + \frac{2}{j \omega}(\frac{tj}{\omega}e^{-j\omega t}+\frac{1}{\omega ^2}e^{-j\omega t})]_{1}^{\infty} $
$ =(0) - (jt^2 e^{-jt} + 2te^{-jt}+\frac{2}{j}e^{-jt}) $
$ =je^{-jt}(-t^2 + j2t + 2) $
