FourierIntegral

FourierIntegral - ⌠ δ sin(2 πν t dt A ν = 2  ...

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Colby College Fourier Integrals -5.00E-06 0.00E+00 5.00E-06 1.00E-05 1.50E-05 2.00E-05 -600 -400 -200 0 200 400 600 ν (kHz) RE g( ν ) -1.50E-05 -1.00E-05 -5.00E-06 0.00E+00 5.00E-06 1.00E-05 1.50E-05 -600 -400 -200 0 200 400 600 ν (kHz) IM g( ) Fourier series: f(t) = n=0 A n cos(2 π n ν o t) + n=0 B n sin(2 π n ν o t) ν o = 1/L A n = 2 0 L f(t) cos(2 π n ν o t) dt B n = 2 0 L f(t) sin(2 π n ν o t) dt Let L →∞ then 2 π n ν o t 2 πν t with ν a continuous variable: A( ν ) = 2 0 f(t) cos(2 πν t) dt B( ν ) = 2 0 f(t) sin(2 πν t) dt Square pulse of length δ : f(t) = 1 for t=0 to δ and f(t) = 0 after: A( ν ) = 2 0 δ cos(2 πν t) dt B( ν ) = 2
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Unformatted text preview: ⌠ δ sin(2 πν t) dt A( ν ) = 2    sin(2 πν t) 2 πν δ B( ν ) =2   -cos(2 πν t) 2 πν δ A( ν ) = 2 sin(2 πνδ ) 2 πν B( ν ) =2 1-cos(2 πνδ ) 2 πν A( ν ) = absorption spectrum B( ν ) = dispersion full width to first nulls = 1 δ Combine : e i2 πν t = cos(2 πν t) + i sin(2 πν t) Absorption = RE[g( ν )] f(t) = ⌡ ⌠-∞ ∞ g( ν ) e-i2 πν t d ν g( ν ) = 2 ⌡ ⌠ ∞ f(t) e i2 πν t dt Dispersion = IM[g( ν )] 0 δ 1/ δ t →...
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This document was uploaded on 02/22/2012.

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