Difference between revisions of "HW5.2 Daniel Morris ECE301Fall2008mboutin" - Rhea

Latest revision as of 11:29, 16 September 2013

Example of Computation of Fourier transform of a CT SIGNAL

A practice problem on CT Fourier transform

Fourier Transformations

The Fourier transformation of the signal x(t) is defined as

$ X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt $

Example

$ x(t)=e^{-3|t-2|} $

$ X(\omega)=\int_{-\infty}^{\infty} e^{-3|t-2|} e^{-j\omega t}dt $

$ X(\omega)=\int_{-\infty}^{2} e^{-3(-(t-2))}e^{-j\omega t}dt+\int_{2}^{\infty} e^{-3(t-2)}e^{-j\omega t}dt $

$ X(\omega)=\int_{-\infty}^{2} e^{3t-6-j\omega t}dt+\int_{2}^{\infty} e^{-3t+6-j\omega t}dt $

$ X(\omega)=e^{-6}\int_{-\infty}^{2} e^{(3-j\omega )t}dt+e^6\int_{2}^{\infty} e^{-(3+j\omega )t}dt $

$ X(\omega)=e^{-6}\frac{1}{3-j\omega}e^{(3-j\omega )t}|_{-\infty}^{2}+e^6\frac{1}{-(3+j\omega)}e^{-(3+j\omega)t}|_{2}^{\infty} $

$ X(\omega)=e^{-6}\frac{1}{3-j\omega}(e^{(3-j\omega)(2)}-e^{(3-j\omega)(-\infty)})+e^6\frac{1}{-(3+j\omega)}(e^{-(3+j\omega )(\infty)}-e^{-(3+j\omega)(2)}) $

$ X(\omega)=e^{-6}\frac{1}{3-j\omega}e^{(3-j\omega)(2)}+e^6\frac{1}{3+j\omega}e^{-(3+j\omega)(2)} $

$ X(\omega)=\frac{e^{-2j\omega}}{3-j\omega}+\frac{e^{-2j\omega}}{3+j\omega} $

$ X(\omega)=\frac{6e^{-2j\omega}}{9+\omega^2} $

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