Line 27: Line 27:
  
 
<math>
 
<math>
=\frac{1}{3+j\omega}+4*\frac{2}{\omega}*\frac{-e^{-3j\omega} + e^{3j\omega}}{2j}
+
=2\frac{1}{3+j\omega}+4*\frac{2}{\omega}*\frac{-e^{-3j\omega} + e^{3j\omega}}{2j}
</math>
+
</math> and using eulers identity for sin:
  
 
<math>
 
<math>
=\frac{1}{3+j\omega}+\frac{8sin(3\omega)}{\omega}
+
=\frac{2}{3+j\omega}+\frac{8sin(3\omega)}{\omega}
 
</math>
 
</math>

Revision as of 13:23, 8 October 2008

Compute the Fourier Transform of x(t):

$ \,x(t)=2e^{-3t}u(t)+4[u(t+3)-u(t-3)] $

Using the Formula for Fourier Transforms:

$ \mathcal{F}(x(t))= \mathcal{X}(\omega)= \int_{-\infty}^{\infty}x(t)e^{-j\omega t} \,dt $

So the calculation follows as:

$ \mathcal{X}(\omega)= \int_{-\infty}^{\infty}(2e^{-3t}u(t)+4[u(t+3)-u(t-3)])e^{-j\omega t} \,dt $

$ =2\int_{0}^{\infty}e^{-t(3+j\omega)}\,dt + 4\int_{-3}^{3}e^{-j\omega t}\, dt $

$ =2\frac{-e^{-t}}{3+j\omega}\bigg|^{\infty}_{0}+4\frac{-e^{-j\omega t}}{j\omega}\bigg|^{3}_{-3} $

$ =2\frac{1}{3+j\omega}+4*\frac{2}{\omega}*\frac{-e^{-3j\omega} + e^{3j\omega}}{2j} $ and using eulers identity for sin:

$ =\frac{2}{3+j\omega}+\frac{8sin(3\omega)}{\omega} $

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Mu Qiao