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<math>\,\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\,</math> | <math>\,\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\,</math> | ||
| − | <math>\,\mathcal{X}(\omega)=\int_{-1}^{\infty} e^{-(7+j\omega )t} dt + \int_{1}^{\infty} e^{-(23+j\omega )t} | + | <math>\,\mathcal{X}(\omega)=\int_{-1}^{\infty} e^{-(7+j\omega )t} dt + \int_{1}^{\infty} e^{-(23+j\omega )t} dt\,</math> |
<math>\left. \frac{e^{-15}}{-(j\omega +5)}e^{-(j\omega +5)t}\right]_{1}^{+\infty}</math> | <math>\left. \frac{e^{-15}}{-(j\omega +5)}e^{-(j\omega +5)t}\right]_{1}^{+\infty}</math> | ||
Revision as of 08:17, 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\, $
$ \left. \frac{e^{-15}}{-(j\omega +5)}e^{-(j\omega +5)t}\right]_{1}^{+\infty} $
