ASN4 fixing: Difference between revisions

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[[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 = s(t)^.s^*(t) \!</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 = \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>



and

<math> \int_{- \infty}^{\infty} (|s(t)|)^2 dt = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty}\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} S(f)e^{-j 2 \pi f' t} df df^'dt</math>






<math> (\int_{- \infty}^{\infty})^5 s(t)e^{-j 2 \pi f t}e^{j 2 \pi f t} s(t)e^{-j 2 \pi f t}e^{-j 2 \pi f' t} df df^'dt\! </math>


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

Latest revision as of 13:30, 16 December 2009

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Parseval's Theorem

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

Note that

and also that

Therefore


and





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