(Fourier Transform of x(t))
(Fourier Transform of x(t))
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\\&= \int_{-\infty}^{\infty} cos(8 \pi t)e^{-t^{2}}e^{-j\omega t}dt  
 
\\&= \int_{-\infty}^{\infty} cos(8 \pi t)e^{-t^{2}}e^{-j\omega t}dt  
 
\\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t}-e^{-j8\pi t}}{2}e^{-t^{2}}e^{-j\omega t}dt
 
\\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t}-e^{-j8\pi t}}{2}e^{-t^{2}}e^{-j\omega t}dt
 +
\\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t-t^2}-e^{-j8\pi t-t^2}}{2}e^{-j\omega t}dt
 +
\\&= \int_{-\infty}^{\infty}\frac{e^{t(j8\pi t-t)}-e^{-t(j8\pi +t)}}{2}e^{-j\omega t}dt
 +
  
 
\end{align}</math>
 
\end{align}</math>

Revision as of 17:00, 8 October 2008

Specify a signal x(t)

$ x(t)=cos(8 \pi t)e^{-t^{2}} $

Fourier Transform of x(t)

$ \begin{align} X(\omega) &=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty} cos(8 \pi t)e^{-t^{2}}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t}-e^{-j8\pi t}}{2}e^{-t^{2}}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{j8\pi t-t^2}-e^{-j8\pi t-t^2}}{2}e^{-j\omega t}dt \\&= \int_{-\infty}^{\infty}\frac{e^{t(j8\pi t-t)}-e^{-t(j8\pi +t)}}{2}e^{-j\omega t}dt \end{align} $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett