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<math>\,\mathcal{X}(\omega)=\int_{-1}^{\infty} e^{-(7+j\omega )t} dt + \int_{1}^{\infty} e^{-(23+j\omega )t} dt\,</math> | <math>\,\mathcal{X}(\omega)=\int_{-1}^{\infty} e^{-(7+j\omega )t} dt + \int_{1}^{\infty} e^{-(23+j\omega )t} dt\,</math> | ||
| − | <math>\,\mathcal{X}(\omega)=\left. \frac{e^{-(7+j\omega )t}}{-(7+j\omega )}\right]_{-1}^{\infty} \,</math> | + | <math>\,\mathcal{X}(\omega)={\left. \frac{e^{-(7+j\omega )t}}{-(7+j\omega )}\right]_{-1}^{\infty}} + {\left. \frac{e^{-(23+j\omega )t}}{-(23+j\omega )}\right]_{1}^{\infty}}\,</math> |
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\frac{-e^{7+j\omega}}{-(7+j\omega )} - \frac{-e^{-(23+j\omega)}}{-(23+j\omega )} \,</math> | ||
Revision as of 09:42, 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\, $
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{\infty} e^{-7t} u(t+1) e^{-j\omega t} dt + \int_{-\infty}^{\infty} e^{-23t} u(t-1) e^{-j\omega t}dt\, $
$ \,\mathcal{X}(\omega)=\int_{-1}^{\infty} e^{-(7+j\omega )t} dt + \int_{1}^{\infty} e^{-(23+j\omega )t} dt\, $
$ \,\mathcal{X}(\omega)={\left. \frac{e^{-(7+j\omega )t}}{-(7+j\omega )}\right]_{-1}^{\infty}} + {\left. \frac{e^{-(23+j\omega )t}}{-(23+j\omega )}\right]_{1}^{\infty}}\, $
$ \,\mathcal{X}(\omega)=\frac{-e^{7+j\omega}}{-(7+j\omega )} - \frac{-e^{-(23+j\omega)}}{-(23+j\omega )} \, $
