ASN4 fixing: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
 
No edit summary
Line 1: Line 1:
[[Jodi Hodge| back to my home page]]
[[Jodi Hodge| back to my home page]]

== Parseval's Theorem ==

Parseval's Theorem says that <math>\int_{- \infty}^{\infty} (|s(t)|)^2 dt</math> in time transforms to <math>\int_{- \infty}^{\infty} (|S(f)|)^2 df</math> in frequency

Note that
<math> (|s(t)|)^2 = \frac {s(t)^.s^*(t)}{2}</math>

and also that

<math> s(t)= F ^{-1}[S(f)]=\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} df </math>

Therefore

<math> (|s(t)|)^2 = \frac {1}{2}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} S(f)e^{-j 2 \pi f t} df df</math>

Note that <math> |e^{j 2 \pi f t}|= \sqrt{cos^2(2 \pi f t) + sin^2(2 \pi f t)}=1 </math>

The above equation of <math>|s(t)|</math> simplifies to then <math>|s(t)|= \int_{- \infty}^{\infty}S(f) df= |S(f)|</math>

Therefore,squaring the function and intergrating it in the time domain <math>\int_{- \infty}^{\infty} (|s(t)|)^2 dt</math> is to do the same in the frequency domain <math>\int_{- \infty}^{\infty} (|S(f)|)^2 df</math>

Revision as of 21:43, 13 December 2009

back to my home page

Parseval's Theorem

Parseval's Theorem says that in time transforms to in frequency

Note that

and also that

Therefore

Note that

The above equation of simplifies to then

Therefore,squaring the function and intergrating it in the time domain is to do the same in the frequency domain