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+ | [[Category:problem solving]] | ||
+ | [[Category:ECE301]] | ||
+ | [[Category:ECE]] | ||
+ | [[Category:Fourier transform]] | ||
+ | [[Category:signals and systems]] | ||
+ | == Example of Computation of Fourier transform of a CT SIGNAL == | ||
+ | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
+ | ---- | ||
+ | |||
Compute the Fourier transform of the following CT signal using the integral formula: | Compute the Fourier transform of the following CT signal using the integral formula: | ||
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<math>\,\mathcal{X}(\omega)=\frac{e^{-15}}{(j\omega +5)}e^{-(j\omega +5)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\,</math> | <math>\,\mathcal{X}(\omega)=\frac{e^{-15}}{(j\omega +5)}e^{-(j\omega +5)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\,</math> | ||
+ | |||
+ | <math>\,\mathcal{X}(\omega)=\frac{1}{(j\omega +5)}e^{-(j\omega +20)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\,</math> | ||
+ | |||
+ | ---- | ||
+ | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] |
Latest revision as of 11:25, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
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\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\left. \frac{e^{-15}}{-(j\omega +5)}e^{-(j\omega +5)t}\right]_{1}^{+\infty} + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\frac{e^{-15}}{-(j\omega +5)}(e^{-j\omega(\infty)}e^{-5(\infty)}-e^{-(j\omega +5)}) + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\frac{e^{-15}}{(j\omega +5)}e^{-(j\omega +5)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\frac{1}{(j\omega +5)}e^{-(j\omega +20)} + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $