ASN4 fixing: Difference between revisions
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== Parseval's Theorem == |
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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 |
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Note that |
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<math> (|s(t)|)^2 = s(t)^.s^*(t) \!</math> |
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and also that |
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<math> s(t)= F ^{-1}[S(f)]=\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} df \!</math> |
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Therefore |
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<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> |
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and |
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<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> |
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<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> |
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<math> |s(t)|= F ^{-1}[S(f)]=|\int_{- \infty}^{\infty}S(f)e^{j 2 \pi f t} df | </math> |
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Note that <math> |e^{j 2 \pi f t}|= \sqrt{cos^2(2 \pi f t) + sin^2(2 \pi f t)}=1 </math> |
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The above equation of <math>|s(t)|</math> simplifies to then <math>|s(t)|= \int_{- \infty}^{\infty}S(f) df= |S(f)|</math> |
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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
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