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 : ck269@cornell.edu 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 )
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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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sol3 - PHYS 218 - SOLUTION TO ASSIGNMENT 3 16 Feb 2007 By...

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