(New page: <math> x(t)={e^{-3|t|}, |t|<1}</math><p> <math> x(t)=0, |t|>1 </math>)
 
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<math> x(t)={e^{-3|t|}, |t|<1}</math><p>
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Computer the Fourier Transform of the Following:
          <math> x(t)=0,    |t|>1 </math>
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<math>\, x(t)={e^{-2|t|}, |t|<1}\,</math>
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<math>\, x(t)=0,    |t|>1 \,</math>
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<math>\,\mathcal{X}(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt\,</math>
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<math>\,\mathcal{X}(\omega)=\int_{-1}^{0}e^{2t}e^{-j\omega t}\,dt\ + \int_{0}^{1}e^{-2t}e^{-j\omega t}\,dt\,</math>
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<math>\,\mathcal{X}(\omega)=\int_{-1}^{0}e^{t(2-j\omega)}\,dt\ + \int_{0}^{1}e^{-t(2+j\omega)}\,dt\,</math>
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<math>\,\mathcal{X}(\omega)={\left. \frac{e^{t(2-j\omega )}}{(2-j\omega )}\right]_{-1}^{0} + {\left. \frac{e^{-t(2+j\omega )}}{(2+j\omega )}\right]_0^{1}}}\,</math>
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<math>\,\mathcal{X}(\omega)=\frac{1}{2-j\omega} - \frac{e^{-(2-j\omega)}}{2-j\omega} + \frac{e^{-(2+j\omega)}}{2+j\omega} - \frac{1}{2+j\omega} \,</math>

Revision as of 15:44, 8 October 2008

Computer the Fourier Transform of the Following:

$ \, x(t)={e^{-2|t|}, |t|<1}\, $
$ \, x(t)=0,    |t|>1 \, $

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

$ \,\mathcal{X}(\omega)=\int_{-1}^{0}e^{2t}e^{-j\omega t}\,dt\ + \int_{0}^{1}e^{-2t}e^{-j\omega t}\,dt\, $


$ \,\mathcal{X}(\omega)=\int_{-1}^{0}e^{t(2-j\omega)}\,dt\ + \int_{0}^{1}e^{-t(2+j\omega)}\,dt\, $

$ \,\mathcal{X}(\omega)={\left. \frac{e^{t(2-j\omega )}}{(2-j\omega )}\right]_{-1}^{0} + {\left. \frac{e^{-t(2+j\omega )}}{(2+j\omega )}\right]_0^{1}}}\, $


$ \,\mathcal{X}(\omega)=\frac{1}{2-j\omega} - \frac{e^{-(2-j\omega)}}{2-j\omega} + \frac{e^{-(2+j\omega)}}{2+j\omega} - \frac{1}{2+j\omega} \, $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010