(New page: A work in progress. == The Continuous Time Fourier Transform (CTFT) == CTFT: <math>X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt</math> Inverse CTFT: <math>x(t) = \fr...) |
m (→The Continuous Time Fourier Transform (CTFT)) |
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'''Example:''' | '''Example:''' | ||
| − | Let x(t) = | + | Let <math>x(t) = <math>\delta (t)</math> |
| − | <math>X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt = \int_{-\infty}^{\infty} \! rect(t)e^{-j \omega t} dt = \int_{-\frac{1}{2}}^{\frac{1}{2}} \! rect(t)e^{-j \omega t} dt = \int_{-\frac{1}{2}}^{\frac{1}{2}} \! e^{-j \omega t} dt</math> | + | <math>\begin{align}X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt \\ |
| + | &= \int_{-\infty}^{\infty} \! rect(t)e^{-j \omega t} dt \\ | ||
| + | &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \! rect(t)e^{-j \omega t} dt \\ | ||
| + | &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \! e^{-j \omega t} dt\end{align}</math> | ||
...because rect(t) has an area of 1 over the limits <math>[-\frac{1}{2}, \frac{1}{2}]</math>. So, | ...because rect(t) has an area of 1 over the limits <math>[-\frac{1}{2}, \frac{1}{2}]</math>. So, | ||
Revision as of 05:37, 23 October 2010
A work in progress.
The Continuous Time Fourier Transform (CTFT)
CTFT:
$ X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt $
Inverse CTFT:
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \! X(\omega)e^{j \omega t} dw $
Example:
Let $ x(t) = <math>\delta (t) $
$ \begin{align}X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt \\ &= \int_{-\infty}^{\infty} \! rect(t)e^{-j \omega t} dt \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \! rect(t)e^{-j \omega t} dt \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \! e^{-j \omega t} dt\end{align} $
...because rect(t) has an area of 1 over the limits $ [-\frac{1}{2}, \frac{1}{2}] $. So,
$ \int_{-\frac{1}{2}}^{\frac{1}{2}} \! e^{-j \omega t} dt = \frac{1}{-j\omega}(e^{-j\frac{\omega}{2}} - e^{j\frac{\omega}{2}}) = \frac{1}{j\omega}(e^{j\frac{\omega}{2}} - e^{-j\frac{\omega}{2}}) = \frac{2}{\omega}(\frac{e^{j\frac{\omega}{2}} - e^{-j\frac{\omega}{2}}}{2j}) = \frac{2}{\omega}sin(\frac{\omega}{2}) $
Therefore,
$ X(\omega) = \frac{2}{\omega}sin(\frac{\omega}{2}) \ \ or \ \ \frac{1}{\omega/2}sin(\frac{\omega}{2}) = sinc(\frac{\omega}{2\pi}) $
