| (19 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]] | ||
| + | ---- | ||
| + | |||
| + | ==FOURIER TRANSFORM== | ||
| + | |||
| + | |||
<math> x(t) = e^{-3|t-2|} </math> | <math> x(t) = e^{-3|t-2|} </math> | ||
| Line 5: | Line 17: | ||
When | When | ||
| − | <math> t- | + | <math> t-1 < 0 \rightarrow x_1(t) = e^{3t-3} </math> |
and when, | and when, | ||
| − | <math> t- | + | <math> t-1 >0 \rightarrow x_2(t) = e^{-3t+3} </math> |
So, we can then compute the Fourier series by adding the integrals of each diferent case. | So, we can then compute the Fourier series by adding the integrals of each diferent case. | ||
| Line 15: | Line 27: | ||
<math>\ \mathcal{X}(\omega) = \int_{-\infty}^{\infty} x_1(t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty} x_2(t)e^{-j\omega t} \,dt </math> | <math>\ \mathcal{X}(\omega) = \int_{-\infty}^{\infty} x_1(t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty} x_2(t)e^{-j\omega t} \,dt </math> | ||
| − | <math> \mathcal{X}(\omega) = \int_{-\infty}^{ | + | <math> \mathcal{X}(\omega) = \int_{-\infty}^{1} e^{3t-3}e^{-j\omega t}\,dt + \int_{1}^{\infty} e^{-3t+3}e^{-j\omega t} \,dt </math> |
| + | |||
| + | <math> \mathcal{X}(\omega) = \frac{1}{e^{3}} \int_{-\infty}^{1} e^{3t-j\omega t}\,dt + e^{3} \int_{1}^{\infty} e^{-3t-j\omega t} \,dt </math> | ||
| + | |||
| + | <math> \mathcal{X}(\omega) = \frac{1}{e^{3}} \int_{-\infty}^{1} e^{t(3-j\omega)}\,dt + e^{3} \int_{1}^{\infty} e^{-t(3+j\omega)} \,dt </math> | ||
| + | |||
| + | <math> \mathcal{X}(\omega) = {\left. \frac{e^{t(3-j\omega)}}{3-j\omega} \right]^{1}_{-\infty} } \frac{1}{e^{3}} + {\left. -\frac{e^{-t(3+j\omega)}}{3+j\omega} \right]^{\infty}_1 } e^{3}\,</math> | ||
| − | <math> \mathcal{X}(\omega) = \frac{1}{e^{ | + | <math> \mathcal{X}(\omega) = \frac{1}{e^{3}} \frac{e^{3-j\omega}}{3-j\omega} + e^{3} \frac{e^{-3-j\omega}}{3+j\omega} </math> |
| − | <math> \mathcal{X}(\omega) = \frac | + | <math> \mathcal{X}(\omega) = \frac{e^{-j\omega}}{3-j\omega} + \frac{e^{-j\omega}}{3+j\omega} </math> |
| − | + | ---- | |
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:37, 16 September 2013
Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on CT Fourier transform
FOURIER TRANSFORM
$ x(t) = e^{-3|t-2|} $
Noticing that there is an absolute value, we can proceed to divide in tow cases.
When
$ t-1 < 0 \rightarrow x_1(t) = e^{3t-3} $
and when,
$ t-1 >0 \rightarrow x_2(t) = e^{-3t+3} $
So, we can then compute the Fourier series by adding the integrals of each diferent case.
$ \ \mathcal{X}(\omega) = \int_{-\infty}^{\infty} x_1(t)e^{-j\omega t}\,dt + \int_{-\infty}^{\infty} x_2(t)e^{-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \int_{-\infty}^{1} e^{3t-3}e^{-j\omega t}\,dt + \int_{1}^{\infty} e^{-3t+3}e^{-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \frac{1}{e^{3}} \int_{-\infty}^{1} e^{3t-j\omega t}\,dt + e^{3} \int_{1}^{\infty} e^{-3t-j\omega t} \,dt $
$ \mathcal{X}(\omega) = \frac{1}{e^{3}} \int_{-\infty}^{1} e^{t(3-j\omega)}\,dt + e^{3} \int_{1}^{\infty} e^{-t(3+j\omega)} \,dt $
$ \mathcal{X}(\omega) = {\left. \frac{e^{t(3-j\omega)}}{3-j\omega} \right]^{1}_{-\infty} } \frac{1}{e^{3}} + {\left. -\frac{e^{-t(3+j\omega)}}{3+j\omega} \right]^{\infty}_1 } e^{3}\, $
$ \mathcal{X}(\omega) = \frac{1}{e^{3}} \frac{e^{3-j\omega}}{3-j\omega} + e^{3} \frac{e^{-3-j\omega}}{3+j\omega} $
$ \mathcal{X}(\omega) = \frac{e^{-j\omega}}{3-j\omega} + \frac{e^{-j\omega}}{3+j\omega} $
