(New page: == Inverse Fourier Transform == <math> X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \!</math> <math> x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega )e^{j\omega t} d\...) |
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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]] | ||
| + | ---- | ||
== Inverse Fourier Transform == | == Inverse Fourier Transform == | ||
<math> X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \!</math> | <math> X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \!</math> | ||
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F(<math> \frac{2}{\pi }e^{3jt} + \frac{5 }{2 }e^{j2t} \!</math>) = <math> X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \!</math> | F(<math> \frac{2}{\pi }e^{3jt} + \frac{5 }{2 }e^{j2t} \!</math>) = <math> X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \!</math> | ||
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| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:47, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
Inverse Fourier Transform
$ X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \! $
$ 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} 4\delta (\omega -3)e^{j\omega t} d\omega + \frac{1}{2\pi} \int_{-\infty}^{\infty} 5\pi \delta (\omega -2)e^{j\omega t} d\omega \! $
Since integrating dirac functions is extremely easy one can easily simplify to the following
$ x(t) = \frac{4}{2\pi }e^{3jt} + \frac{5\pi }{2\pi }e^{j2t} \! $ $ = \frac{2}{\pi }e^{3jt} + \frac{5 }{2 }e^{j2t} \! $
Check:
F($ \frac{2}{\pi }e^{3jt} + \frac{5 }{2 }e^{j2t} \! $) = $ X(\omega) = 4\delta (\omega - 3) + 5\pi \delta(\omega - 2) \! $
