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

Revision as of 13:37, 19 December 2009

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

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

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

This shows how $s(0)\!$ becomes $\int _{-\infty }^{\infty }S(f)df$

@@ Line 3: / Line 3: @@
 == Value at Origin ==
-This Fourier Transform property says <math>s(0)</math> transforms to <math>\int_{-\infty}^{\infty} S(f) df</math>
+This Fourier Transform property says <math>s(0)\!</math> becomes <math>\int_{-\infty}^{\infty} S(f) df</math>
 <math>s(0)= s(t)|_{t=0} = \mathcal{F}^{-1}\left[S(f) \right])|_{t=0}</math>
@@ Line 9: / Line 9: @@
 <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>
+This shows how <math>s(0) \!</math> becomes <math>\int_{-\infty}^{\infty} S(f) df</math>