(New page: Compute the Fourier Transform of x(t): <math>\,x(t)=2e^{-3t}u(t)+3[u(t+3)-u(t-3)]</math> Using the Formula for Fourier Transforms: <math> \mathcal{F}(x(t))= \mathcal{X}(\omega)= \int_{-...) |
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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 x(t): | Compute the Fourier Transform of x(t): | ||
| − | <math>\,x(t)=2e^{-3t}u(t)+ | + | <math>\,x(t)=2e^{-3t}u(t)+4[u(t+3)-u(t-3)]</math> |
Using the Formula for Fourier Transforms: | Using the Formula for Fourier Transforms: | ||
| Line 12: | Line 20: | ||
So the calculation follows as: | So the calculation follows as: | ||
| + | |||
<math> | <math> | ||
\mathcal{X}(\omega)= | \mathcal{X}(\omega)= | ||
| − | \int_{-\infty}^{\infty}(2e^{-3t}u(t)+ | + | \int_{-\infty}^{\infty}(2e^{-3t}u(t)+4[u(t+3)-u(t-3)])e^{-j\omega t} \,dt |
| + | </math> | ||
| + | <math> | ||
| + | =2\int_{0}^{\infty}e^{-t(3+j\omega)}\,dt + 4\int_{-3}^{3}e^{-j\omega t}\, dt | ||
| + | </math> | ||
| + | <math> | ||
| + | =2\frac{-e^{-t}}{3+j\omega}\bigg|^{\infty}_{0}+4\frac{-e^{-j\omega t}}{j\omega}\bigg|^{3}_{-3} | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | =2\frac{1}{3+j\omega}+4*\frac{2}{\omega}*\frac{-e^{-3j\omega} + e^{3j\omega}}{2j} | ||
| + | </math> and using eulers identity for sin: | ||
| + | |||
| + | <math> | ||
| + | =\frac{2}{3+j\omega}+\frac{8sin(3\omega)}{\omega} | ||
</math> | </math> | ||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:35, 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 x(t):
$ \,x(t)=2e^{-3t}u(t)+4[u(t+3)-u(t-3)] $
Using the Formula for Fourier Transforms:
$ \mathcal{F}(x(t))= \mathcal{X}(\omega)= \int_{-\infty}^{\infty}x(t)e^{-j\omega t} \,dt $
So the calculation follows as:
$ \mathcal{X}(\omega)= \int_{-\infty}^{\infty}(2e^{-3t}u(t)+4[u(t+3)-u(t-3)])e^{-j\omega t} \,dt $
$ =2\int_{0}^{\infty}e^{-t(3+j\omega)}\,dt + 4\int_{-3}^{3}e^{-j\omega t}\, dt $
$ =2\frac{-e^{-t}}{3+j\omega}\bigg|^{\infty}_{0}+4\frac{-e^{-j\omega t}}{j\omega}\bigg|^{3}_{-3} $
$ =2\frac{1}{3+j\omega}+4*\frac{2}{\omega}*\frac{-e^{-3j\omega} + e^{3j\omega}}{2j} $ and using eulers identity for sin:
$ =\frac{2}{3+j\omega}+\frac{8sin(3\omega)}{\omega} $
