(New page: ==Computing the Inverse Fourier Transform== <math>\ X(\omega)= 8 \pi w \delta(w-9) + 2 \pi w^{3} \delta(w-4 \pi) </math> The inverse Fourier transform is defined as: <math> x(t) = int_{...) |
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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]] | ||
| + | ---- | ||
==Computing the Inverse Fourier Transform== | ==Computing the Inverse Fourier Transform== | ||
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The inverse Fourier transform is defined as: | The inverse Fourier transform is defined as: | ||
| − | <math> x(t) = int_{- | + | <math> x(t) = \int_{-\infty}^{\infty} \frac{X(w)}{2 \pi} e^{jwt} dw </math> |
| + | |||
| + | Using this formula to determine the signal: | ||
| + | |||
| + | <math>\ x(t) = \frac{8 \pi}{2 \pi} \int_{-\infty}^{\infty} w e^{jwt} \delta(w-9) dw + \frac{2}{2 \pi} \int_{-\infty}^{\infty}w^{3} \delta(w-4 \pi) e^{jwt} dw </math> | ||
| + | |||
| + | Now using the sifting property of the delta function we find that the signal is | ||
| + | |||
| + | <math>\ x(t) = 36 e^{j9t} + 64 \pi^{2} e^{j4\pi t} </math> | ||
| + | |||
| + | ---- | ||
| + | [[CT_Fourier_transform_practice_problems_list|Back to Practice Problems on CT Fourier transform]] | ||
Latest revision as of 11:40, 16 September 2013
Example of Computation of inverse Fourier transform (CT signals)
A practice problem on CT Fourier transform
Computing the Inverse Fourier Transform
$ \ X(\omega)= 8 \pi w \delta(w-9) + 2 \pi w^{3} \delta(w-4 \pi) $
The inverse Fourier transform is defined as:
$ x(t) = \int_{-\infty}^{\infty} \frac{X(w)}{2 \pi} e^{jwt} dw $
Using this formula to determine the signal:
$ \ x(t) = \frac{8 \pi}{2 \pi} \int_{-\infty}^{\infty} w e^{jwt} \delta(w-9) dw + \frac{2}{2 \pi} \int_{-\infty}^{\infty}w^{3} \delta(w-4 \pi) e^{jwt} dw $
Now using the sifting property of the delta function we find that the signal is
$ \ x(t) = 36 e^{j9t} + 64 \pi^{2} e^{j4\pi t} $
