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| − | + | =How to obtain the Inverse DT Fourier Transform formula in terms of f in hertz (from the formula in terms of <math>\omega</math>) = | |
| − | + | ||
| − | + | Recall: | |
| − | + | ||
| − | | | + | <math>\, x(t)=\mathcal{F}^{-1}(\mathcal{X}(\omega))=\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{i\omega t} d \omega\,</math> |
| + | |||
| + | To obtain X(f), use the substitution | ||
| + | |||
| + | <math>\omega= 2 \pi f </math>. | ||
| + | |||
| + | More specifically | ||
| + | |||
| + | <div align="left" style="padding-left: 0em;"> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | x(t) &=\mathcal{F}^{-1}(\mathcal{X}(\omega)) \\ | ||
| + | &=\mathcal{F}^{-1}(\mathcal{X}(2\pi f)) \\ | ||
| + | &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(2\pi f)e^{i2\pi ft} d2\pi f \\ | ||
| + | &= \int_{-\infty}^{\infty}X(f)e^{i2\pi ft} df | ||
| + | \end{align} | ||
| + | </math> | ||
| + | </div> | ||
| + | |||
| + | ---- | ||
| + | [[ECE438_HW1_Solution|Back to Table]] | ||
Latest revision as of 10:58, 15 September 2010
How to obtain the Inverse DT Fourier Transform formula in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ \, x(t)=\mathcal{F}^{-1}(\mathcal{X}(\omega))=\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathcal{X}(\omega)e^{i\omega t} d \omega\, $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} x(t) &=\mathcal{F}^{-1}(\mathcal{X}(\omega)) \\ &=\mathcal{F}^{-1}(\mathcal{X}(2\pi f)) \\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{X}(2\pi f)e^{i2\pi ft} d2\pi f \\ &= \int_{-\infty}^{\infty}X(f)e^{i2\pi ft} df \end{align} $
