ASN4 - Fourier Transform property: Parseval's Theorem: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
 
(4 intermediate revisions by the same user not shown)
Line 1: Line 1:

[[Jodi Hodge|Back to my Home Page]]



== Parseval's Theorem ==
== Parseval's Theorem ==
Line 4: Line 7:
<math>\int_{- \infty}^{\infty} (|s(t)|)^2 dt</math> in time transforms to <math>\int_{- \infty}^{\infty} (|S(f)|)^2 df</math> in frequency
<math>\int_{- \infty}^{\infty} (|s(t)|)^2 dt</math> in time transforms to <math>\int_{- \infty}^{\infty} (|S(f)|)^2 df</math> in frequency


First find the magnitude of <math>s(t)</math> which is also the Inverse Fourier Transform of <math>S(f)</math>.
The magnitude of <math>s(t)</math> is also the Inverse Fourier Transform of <math>S(f)</math>.


<math> |s(t)|= F ^{-1}[S(f)]=|\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} df | </math>
<math> |s(t)|= F ^{-1}[S(f)]=|\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} df | </math>
Line 10: Line 13:
Note that <math> |e^{j 2 \pi f t}|= \sqrt{cos^2(2 \pi f t) + sin^2(2 \pi f t)}=1 </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>


So then <math>|s(t)|= \int_{- \infty}^{\infty}S(f) df</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>

Latest revision as of 08:17, 3 December 2009

Back to my Home Page


Parseval's Theorem

in time transforms to in frequency

The magnitude of is also the Inverse Fourier Transform of .

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