(New page: == Signal == <math> X(\omega) = 6\delta (\omega - 5) + 3\pi \delta(\omega - 1) \!</math> == Inverse Fourier Transform == <math> x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega )e...) |
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| + | [[Category:problem solving]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier transform]] | ||
| + | [[Category:inverse Fourier transform]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of inverse Fourier transform (CT signals) == | ||
| + | A [[CT_Fourier_transform_practice_problems_list|practice problem on CT Fourier transform]] | ||
| + | ---- | ||
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== Signal == | == Signal == | ||
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<math> x(t) = \frac{3}{\pi }e^{5jt} + \frac{3}{2}e^{jt} \!</math> | <math> x(t) = \frac{3}{\pi }e^{5jt} + \frac{3}{2}e^{jt} \!</math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 12:51, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
Signal
$ X(\omega) = 6\delta (\omega - 5) + 3\pi \delta(\omega - 1) \! $
Inverse Fourier Transform
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega )e^{j\omega t} d\omega \! $
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} 6\delta (\omega -5)e^{j\omega t} d\omega + \frac{1}{2\pi} \int_{-\infty}^{\infty} 3\pi \delta (\omega -1)e^{j\omega t} d\omega \! $
We know that:
$ \int_{-\infty}^{\infty} \delta (\omega -t_0) e^{jwt} d\omega = e^{jt_0 t} \! $
Therefore:
$ x(t) = \frac{6}{2\pi }e^{5jt} + \frac{3\pi }{2\pi }e^{jt} \! $
$ x(t) = \frac{3}{\pi }e^{5jt} + \frac{3}{2}e^{jt} \! $
