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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]] | ||
| + | ---- | ||
| + | |||
==Fourier Transform== | ==Fourier Transform== | ||
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<font "size"=4> | <font "size"=4> | ||
| − | <math>x(t)= | + | <math>x(t)=(t-1)e^{-6t+6}u(t-1) \,\ </math> |
</font> | </font> | ||
| − | <math>X(\omega)=\int_{-\infty}^{\infty}t^ | + | <math>X(\omega)=\int_{-\infty}^{\infty}x(t)=(t-6)e^{-6t+6}u(t-6) e^{-j\omega t}dt \;</math> |
| + | |||
| + | |||
| + | <math>x(t) \,\ </math>looks like <math>te^{-6t}u(t) \,\ </math> so we evaluate that | ||
| + | |||
| + | the F.T of <math>te^{-6t}u(t) \,\ </math> is | ||
| + | |||
| + | <math>\int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \;</math> | ||
| + | |||
| + | <math>=\int_{0}^{\infty}te^{-6t-j\omega t}dt \;</math> | ||
| + | |||
| + | <math>=\int_{0}^{\infty}te^{-t(6+j\omega t)}dt \;</math> | ||
| + | |||
| + | Do integration by parts | ||
| + | |||
| + | <math> ={\left. \frac{-te^{-t(6+j\omega )}}{6+j\omega }\right]_{0}^{\infty}} - | ||
| + | \int_{0}^{\infty}\frac{-te^{-t(6+j\omega )}}{6+j\omega }dt \;</math> | ||
| + | |||
| + | <math>={\left. \frac{-e^{-t(6+j\omega )}}{(6+j\omega)^2 }\right]_{0}^{\infty}}</math> | ||
| + | |||
| + | <math>= \frac{1}{(6+j\omega)^2}</math> | ||
| + | |||
| + | And now we use the time shift property and get | ||
| + | |||
| + | <math>X(\omega)=\frac{e^{-j\omega}}{(6+j\omega)^2}</math> | ||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:34, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ x(t)=(t-1)e^{-6t+6}u(t-1) \,\ $
$ X(\omega)=\int_{-\infty}^{\infty}x(t)=(t-6)e^{-6t+6}u(t-6) e^{-j\omega t}dt \; $
$ x(t) \,\ $looks like $ te^{-6t}u(t) \,\ $ so we evaluate that
the F.T of $ te^{-6t}u(t) \,\ $ is
$ \int_{-\infty}^{\infty}te^{-6t}u(t) e^{-j\omega t}dt \; $
$ =\int_{0}^{\infty}te^{-6t-j\omega t}dt \; $
$ =\int_{0}^{\infty}te^{-t(6+j\omega t)}dt \; $
Do integration by parts
$ ={\left. \frac{-te^{-t(6+j\omega )}}{6+j\omega }\right]_{0}^{\infty}} - \int_{0}^{\infty}\frac{-te^{-t(6+j\omega )}}{6+j\omega }dt \; $
$ ={\left. \frac{-e^{-t(6+j\omega )}}{(6+j\omega)^2 }\right]_{0}^{\infty}} $
$ = \frac{1}{(6+j\omega)^2} $
And now we use the time shift property and get
$ X(\omega)=\frac{e^{-j\omega}}{(6+j\omega)^2} $
