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− | <math></math> | + | <math>\,x(t)=e^{-7t}u(t+1) + e^{23t}u(t-1)\,</math> |
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+ | <math>\,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math> |
Revision as of 07:10, 6 October 2008
Compute the fourier transform of this signal below:
$ \,x(t)=e^{-7t}u(t+1) + e^{23t}u(t-1)\, $
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $