fourier_transform_Gaussian

fourier_transform_Gaussian - -ln G(0 = 2 2 2(9 Since the...

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Fourier Transform of the Gaussian Konstantinos G. Derpanis October 20, 2005 In this note we consider the Fourier transform 1 of the Gaussian. The Gaussian function, g ( x ), is deﬁned as, g ( x ) = 1 σ 2 π e - x 2 2 σ 2 , (3) where R -∞ g ( x ) dx = 1 (i.e., normalized). The Fourier transform of the Gaussian function is given by: G ( ω ) = e - ω 2 σ 2 2 . (4) Proof : We begin with diﬀerentiating the Gaussian function: dg ( x ) dx = - x σ 2 g ( x ) (5) Next, applying the Fourier transform to both sides of ( 5 ) yields, iωG ( ω ) = 1 2 dG ( ω ) (6) dG ( ω ) G ( ω ) = - ωσ 2 . (7) Integrating both sides of ( 7 ) yields, ω Z 0 dG ( ω 0 ) 0 G ( ω 0 ) 0 = - ω Z 0 ω 0 σ 2 0 (8) ln G ( ω )
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Unformatted text preview: -ln G (0) = 2 2 2 . (9) Since the Gaussian is normalized, the DC component G (0) = 0, thus ( 9 ) can be rewritten as, ln G ( ) =- 2 2 2 (10) Finally, applying the exponent to each side yields, e ln G ( ) = e- 2 2 2 (11) G ( ) = e- 2 2 2 (12) as desired. 1 The Fourier transform pair is given by: F ( ) = Z- f ( x ) e-ix dx (1) f ( x ) = 1 2 Z- F ( ) e ix d, (2) where i denotes the complex unit. 1...
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This note was uploaded on 02/04/2012 for the course COMPUTER 101 taught by Professor Ahmed during the Spring '11 term at alamo.edu.

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