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 $\int_{- \infty}^{\infty} S (f) d f$

Revision as of 23:31, 13 December 2009 view source Jodi.Hodge (talk \| contribs) 508 edits No edit summary ← Older edit		Revision as of 23:31, 13 December 2009 view source Jodi.Hodge (talk \| contribs) 508 edits No edit summary Newer edit →
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	And <math> \int_{-\infty}^{\infty} s(t) dt</math> in time transforms in frequency to <math>\int_{-\infty}^{\infty} S(f) df</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>		The result is <math>s(0)\!</math> transforms to <math>\int_{-\infty}^{\infty} S(f) df</math>