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| − | + | =How to obtain the CT Fourier transform formula in terms of f in hertz (from the formula in terms of <math>\omega</math>) = | |
| − | + | ||
| − | + | Recall: | |
| − | + | ||
| − | + | <math> \mathcal{X}(\omega )=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt</math> | |
| − | + | ||
| − | | | + | |
| + | To obtain X(f), use the substitution | ||
| + | |||
| + | <math>\omega= 2 \pi f </math>. | ||
| + | |||
| + | More specifically | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(f) &=\mathcal{X}(2\pi f)\\ | ||
| + | &=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | ---- | ||
| + | [[ECE438_HW1_Solution|Back to Table]] | ||
Latest revision as of 09:57, 15 September 2010
How to obtain the CT Fourier transform formula in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ \mathcal{X}(\omega )=\mathcal{F}(x(t))=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} X(f) &=\mathcal{X}(2\pi f)\\ &=\int_{-\infty}^{\infty} x(t) e^{-i2\pi ft} dt \end{align} $
