(New page: == Inverse Fourier Transform == let <math>\mathcal{X}(w) = w \times u(-w)</math> then <math>\mathcal{F}^{-1} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathcal{X}(w) e^{jwt}\, dw = \frac...) |
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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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== Inverse Fourier Transform == | == Inverse Fourier Transform == | ||
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<math>\Rightarrow \mathcal{F}^{-1} = \frac{1}{2\pi t}</math> | <math>\Rightarrow \mathcal{F}^{-1} = \frac{1}{2\pi t}</math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:53, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
Inverse Fourier Transform
let $ \mathcal{X}(w) = w \times u(-w) $
then $ \mathcal{F}^{-1} = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathcal{X}(w) e^{jwt}\, dw = \frac{1}{2\pi} \int_{-\infty}^{+\infty} wu(-w) e^{jwt} = \frac{1}{2\pi} \int_{-\infty}^{0} we^{jwt} $
using integration by parts:
let $ u = w, du = dw, dv = e^{jwt} dw, v = \frac{1}{jt}e^{jwt} $
$ \Rightarrow \mathcal{F}^{-1} = \frac{1}{2\pi} \times \left [ uv - \int v \, du \right ] = \left [\frac{w e^{jwt}}{jt} \right ]_{-\infty}^{0} - \int_{-\infty}^{0} \frac{1}{jt}e^{jwt} \, dw $
$ \left [\frac{w e^{jwt}}{jt} \right ]_{-\infty}^{0} = 0 $ and
$ - \int_{-\infty}^{0} \frac{1}{jt}e^{jwt} \, dw = - \frac{1}{jt} \left [\frac{e^{jwt}}{jt} \right ]_{-\infty}^{0} = \frac{1}{t} $
$ \Rightarrow \mathcal{F}^{-1} = \frac{1}{2\pi t} $
