| (6 intermediate revisions by one other user not shown) | |||
| Line 1: | Line 1: | ||
| + | [[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]] | ||
| + | ---- | ||
| + | |||
Let x(t)= <math>cos(t)</math> | Let x(t)= <math>cos(t)</math> | ||
| Line 10: | Line 19: | ||
<math>X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt</math> | <math>X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt</math> | ||
| − | <math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}+\int_{-\infty}^{\infty}e^{-jt(1+\omega)})</math> | + | <math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math> |
| + | |||
| + | <math>X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt)</math> | ||
| + | |||
| + | <math>X(\omega)={\left. \frac{e^{jt(1-\omega)}}{j(1-\omega)}\right]_{-\infty}^{\infty}} + {\left. \frac{e^{-jt(1+\omega)}}{-j(1+\omega)}\right]_{-\infty}^{\infty}}</math> | ||
| + | |||
| + | <math>X(\omega)={\left.\frac{(1+\omega)e^{jt(1-\omega)}-(1-\omega)e^{-jt(1+\omega)}}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math> | ||
| + | |||
| + | <math>X(\omega)={\left.\frac{2e^{-\omega}(1+\omega)cos(t)}{j(1-\omega^2)}\right]_{-\infty}^{\infty}}</math> | ||
| + | |||
| + | <math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\infty}^{\infty}}</math> | ||
| + | |||
| + | <math>X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\pi}^{\pi}}</math> | ||
| + | |||
| + | <math>X(\omega)=0</math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:35, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
Let x(t)= $ cos(t) $
Then
$ X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $
$ X(\omega) = \int_{-\infty}^{\infty}cos(t)e^{-j\omega t}dt $
$ X(\omega) = \int_{-\infty}^{\infty}\frac{1}{2}(e^{jt}+e^{-jt})e^{-j\omega t}dt $
$ X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt) $
$ X(\omega) = \frac{1}{2}(\int_{-\infty}^{\infty}e^{jt(1-\omega)}dt+\int_{-\infty}^{\infty}e^{-jt(1+\omega)}dt) $
$ X(\omega)={\left. \frac{e^{jt(1-\omega)}}{j(1-\omega)}\right]_{-\infty}^{\infty}} + {\left. \frac{e^{-jt(1+\omega)}}{-j(1+\omega)}\right]_{-\infty}^{\infty}} $
$ X(\omega)={\left.\frac{(1+\omega)e^{jt(1-\omega)}-(1-\omega)e^{-jt(1+\omega)}}{j(1-\omega^2)}\right]_{-\infty}^{\infty}} $
$ X(\omega)={\left.\frac{2e^{-\omega}(1+\omega)cos(t)}{j(1-\omega^2)}\right]_{-\infty}^{\infty}} $
$ X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\infty}^{\infty}} $
$ X(\omega)=\frac{(1+\omega)2e^{-\omega}}{j(1-\omega^2)}{\left.cos(t)\right]_{-\pi}^{\pi}} $
$ X(\omega)=0 $
