ASN6c - Fourier Transform property: Value at origin: Difference between revisions

Revision as of 13:35, 19 December 2009

This Fourier Transform property says $s(0)$ transforms to $\int _{-\infty }^{\infty }S(f)df$

$s(0)=s(t)|_{t=0}={\mathcal {F}}^{-1}\left[S(f)\right])|_{t=0}$

$s(0)=\int _{-\infty }^{\infty }S(f)e^{j2\pi f(0)}df=\int _{-\infty }^{\infty }S(f)df$

The result is $s(0)$ transforms to $\int _{-\infty }^{\infty }S(f)df$

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 This Fourier Transform property says <math>s(0)</math> transforms to <math>\int_{-\infty}^{\infty} S(f) df</math>
-<math>s(0)= s(t)|_{t=0} = \mathcal{F}^{-1}\left[S(f) \right]</math>
+<math>s(0)= s(t)|_{t=0} = \mathcal{F}^{-1}\left[S(f) \right])|_{t=0}</math>
-<math>s(0)= s(t)|_{t=0} = \int_{-\infty}^{\infty} S(f)e^{ j 2 \pi f(0) } df = \int_{-\infty}^{\infty} S(f) df</math>
+<math>s(0) = \int_{-\infty}^{\infty} S(f)e^{ j 2 \pi f(0) } df = \int_{-\infty}^{\infty} S(f) df</math>
 The result is <math>s(0)</math> transforms to <math>\int_{-\infty}^{\infty} S(f) df</math>