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== Answer == | == Answer == | ||
| − | <math>\,\mathcal{X}(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt\,</math> | + | <math>\,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math> |
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}e^{-5(t+3)}u(t-1)e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}e^{-j\pi t}\delta(t-\frac{\pi}{2})e^{-j\omega t}\,dt\,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=\int_{1}^{+\infty}e^{-5t}e^{-15}e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}\delta(t-\frac{\pi}{2})e^{-j(\omega +\pi)t}\,dt\,</math> | ||
| + | |||
| + | <math>\,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-j(5+j\omega)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\,</math> | ||
Revision as of 18:56, 5 October 2008
Compute the Fourier transform of the following CT signal using the integral formula:
$ \,x(t)=e^{-5(t+3)}u(t-1) + e^{-j\pi t}\delta(t-\frac{\pi}{2})\, $
Answer
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}e^{-5(t+3)}u(t-1)e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}e^{-j\pi t}\delta(t-\frac{\pi}{2})e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=\int_{1}^{+\infty}e^{-5t}e^{-15}e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}\delta(t-\frac{\pi}{2})e^{-j(\omega +\pi)t}\,dt\, $
$ \,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-j(5+j\omega)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
