# sol3 - PHYS 218 SOLUTION TO ASSIGNMENT 3 16 Feb 2007 By...

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PHYS 218 - SOLUTION TO ASSIGNMENT 3 16 Feb 2007 By Chung Koo Kim e-mail : [email protected] 1. Fourier Transform Examples 1 (a) −τ τ h h (b) The integral is the same for complex integrand as long as the integral variable (t here) is real. g 1 ( ω ) = Z -∞ f 1 ( t ) e - iωt dt = Z 0 - τ he - iωt dt + Z τ 0 ( - h ) e - iωt dt = ih ω e - iωt ± ± ± ± 0 - τ - ih ω e - iωt ± ± ± ± τ 0 = ih ω [2 - ( e iωτ + e - iωτ )] = 2 ih ω [1 - cos ωτ ] * Subscript of g is for later use in problem 3. (c) We see that g 1 ( ω ) is pure imaginary, so the real part is zero. Refer to the follow- ing graph. Blue and red curves represent real and imaginary parts, respectively. 4 π ±±±±±±±± τ 2 π τ 2 π τ 4 π τ 4h 2h

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2. Fourier Transform Examples 2 (a) See the ﬁgure below. −τ τ h (b) In the same manner as in problem 1(b), g 2 ( ω ) = Z -∞ f 2 ( t ) e - iωt dt = Z τ - τ b ± 1 - | t | τ ² cos ωt dt = 2 Z τ 0 b ± 1 - t τ ² cos ωt dt = 2 b τω 2 [1 - cos ωτ ] Note that f 2 ( t ) is an even function of t. (c) Now g 2 ( ω ) is real, so the imaginary part is zero. Refer to the following graph. Blue and red curves represent real and imaginary parts, respectively. 6 π ±±±±±±±± τ 4 π τ 2 π τ 2 π τ 4 π τ 6 π τ b τ 3. Fourier Transform and diﬀerential equations (a)-(b) The integrals can be evaluated as before : G ( ω ) = Z -∞ F ( t ) e - iωt dt = - 1 F ( t ) e - iωt ³ ³ ³ ³ -∞ + Z -∞ f ( t ) e - iωt = g ( ω )
g ( ω ) = Z -∞ f ( t ) e - iωt dt = F ( t ) e - iωt ± ± ± ± -∞ + Z -∞ F ( t ) e - iωt = iωG ( ω ) Here F ( t ) e - iωt ± ± -∞ = 0 because the magnitude of e - iωt is at most one but F(t) approaches zero fast enough. Although vague, mathematically strict deﬁnition can be given to ’fast enough’. (c) From the answers to Problem (1) and (2), we ﬁnd that g 1 ( ω ) = iωg 2 ( ω ) provided that b/ τ =h, or f 1 ( t ) is the derivative of f 2 ( t ). We can check that the latter is true if b/ τ =h. Further note - we have just learned that diﬀerential relation in time domain is converted into simple algebraic manipulation (multiplication by in this case) in the frequency domain when Fourier transformed. Combined with the linearity of Fourier transform 1 , this simpliﬁes the task of solving diﬀerential equations spectacularly. For example, let’s solve equation of motion for a driven simple harmonic oscillator, whose motion is described by a diﬀerential equation ¨ x ( t ) + ω 2 0 x ( t ) = cos ω d t, with driving frequency ω diﬀerent from the natural frequency ω d . Then the Fourier transform(FT) of the equation is ( ) 2 X

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## This note was uploaded on 03/13/2008 for the course PHYS 2218 taught by Professor Wittich,p during the Spring '08 term at Cornell.

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sol3 - PHYS 218 SOLUTION TO ASSIGNMENT 3 16 Feb 2007 By...

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