## 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

$u=t \; \; \; \; \; \; \; \; \; \; \; \; \; dv = e^{-j \omega t}$

$du=1 \; dt \; \; \; \; \; \; \; \; v = \frac{1}{-j\omega}e^{-j \omega t}$

$X(\omega)=\frac{t^2 j}{\omega}e^{-j\omega t}|_{1}^{\infty} + \frac{2}{j \omega}[\frac{tj}{\omega}e^{-j\omega t}|_{1}^{\infty}+\frac{1}{j \omega}\int_{1}^{\infty}e^{-j\omega t}dt]$

$=[\frac{t^2 j}{\omega}e^{-j\omega t} + \frac{2}{j \omega}(\frac{tj}{\omega}e^{-j\omega t}+\frac{1}{\omega ^2}e^{-j\omega t})]_{1}^{\infty}$

$=(0) - (jt^2 e^{-jt} + 2te^{-jt}+\frac{2}{j}e^{-jt})$

$=je^{-jt}(-t^2 + j2t + 2)$

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood