ASN6c fixing: Difference between revisions

Revision as of 23:31, 13 December 2009

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

$s (0) = S (f) |_{f = 0} = [\int_{- \infty}^{\infty} s (t) e^{- j 2 π f t} d t] |_{f = 0} = \int_{- \infty}^{\infty} s (t) d t$

And $\int_{- \infty}^{\infty} s (t) d t$ in time transforms in frequency to $\int_{- \infty}^{\infty} S (f) d f$

The result is $s (0)$ transforms to $\int_{- \infty}^{\infty} S (f) d f$

@@ Line 5: / Line 5: @@
 This Fourier Transform property says <math>s(0)</math> transforms to <math>\int_{-\infty}^{\infty} S(f) df</math>
-<math>s(0)= S(f)|_{f=0} = \int_{-\infty}^{\infty} s(t)e^{- j 2 \pi f t} dt = \int_{-\infty}^{\infty} s(t) dt</math>
+<math>s(0)= S(f)|_{f=0} = [\int_{-\infty}^{\infty} s(t)e^{- j 2 \pi f t} dt]|_{f=0} = \int_{-\infty}^{\infty} s(t) dt</math>
 And <math> \int_{-\infty}^{\infty} s(t) dt</math> in time transforms in frequency to <math>\int_{-\infty}^{\infty} S(f) df</math>
 The result is <math>s(0)</math> transforms to <math>\int_{-\infty}^{\infty} S(f) df</math>