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ps04soln

# ps04soln - PHYSICS 218 SOLUTION TO HW 4 Created 8:05am Last...

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PHYSICS 218 SOLUTION TO HW 4 Created: September 18, 2004 8:05am Last updated: September 26, 2004 11:24am 1. (a) Recall that integration by parts is b a u d v = u v b a - b a v d u . So for this problem, it is of our interest to let u = F ( t ) and d v = e - iωt d t . Starting with G ( ω ), we integrate by parts to get G ( ω ) = -∞ F ( t ) e - iωt d t = F ( t ) e - iωt - i ω -∞ - -∞ d F d t e - iωt - i ω d t = 1 i ω -∞ f ( t ) e - iωt d t = g ( ω ) i ω , where the first term vanishes due to boundary conditions lim t →±∞ F ( t ) = 0. (b) Here, we let u = e - iωt d t and d v = f ( t ) = d F d t to get g ( ω ) = -∞ d F d t e - iωt d t = F ( t ) e - iωt -∞ - -∞ F ( t ) ( - i ω ) e - iωt d t = i ω -∞ F ( t ) e - iωt d t = i ω G ( ω ) . 2. (a) The function f ( t ) = 0 for | t | > τ , h for - τ < t < 0 , - h for 0 < t < τ , (1) is sketched in figure (1). (b) For piecewise continuous function like f ( t ), we can separate the domain of integration into several regions. So the Fourier transform of f ( t ) is g ( ω ) = -∞ f ( t ) e - iωt d t = - τ -∞ + 0 - τ + τ 0 + τ f ( t ) e - iωt d t = 0 - τ h e - iωt d t + τ 0 ( - h ) e - iωt d t = h e - iωt - i ω 0 - τ + - h e - iωt - τ 0 = h i ω ( e iωτ + e - iωτ - 2 ) = 2 h i ω cos( ω τ ) - 1 = 4 h i ω sin 2 ω τ 2 . (c) The graph of g ( ω ) is shown in figure (2). 3. (a) Note that the function f ( t ) actually has four regions. To be more precise, we can write it as f ( t ) = 0 for | t | > τ , b ( 1 + t/τ ) for - τ t 0 , b ( 1 - t/τ ) for 0 t τ . (2) The sketch of it is shown in figure (3). 1

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PHYSICS 218 SOLUTION TO HW 4 0 - τ τ t f ( t ) - h h Figure 1: Sketch of f ( t ). -1.5 -1 -0.5 0 0.5 1 1.5 -6 -4 -2 0 2 4 6 g ( ω ) ω (in units of π ) Figure 2: Graph of g ( w ) with h = 1. Note that we have left out the imaginary number i . (b) Similar to problem (2), the Fourier transform of f ( t ) is g ( ω ) = 0 - τ b ( 1 + t/τ ) e - iωt d t + τ 0 b ( 1 - t/τ ) e - iωt d t = b τ - τ e - iωt d t + b τ 0 - τ t e - iωt d t - b τ τ 0 t e - iωt d t . (3) The first integral is easily evaluated to be τ - τ e - iωt d t = 2 ω sin( ωτ ). The second and third integral can be evaluated by employing the trick of partial differentiation under the integral 2
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ps04soln - PHYSICS 218 SOLUTION TO HW 4 Created 8:05am Last...

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