PS-5-2007-Solutions

PS-5-2007-Solutions - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2007 Solutions to Problem Set Five 1. (30 points) Evaluating integrals with the help of Fourier transforms Evaluate the following integrals using Parseval’s Theorem and one other method. (Yes, we expect you to evaluate the integral twice, and if you do it right you should get the same answer for both approaches (obviously)): (a) ± −∞ sinc 4 ( t ) dt (b) ± −∞ 2 1+(2 πt ) 2 sinc(2 t ) dt (c) ± −∞ t 2 sinc 4 ( t ) dt Solution: (a) We Frst note that ± −∞ sinc 4 ( t ) dt = ± −∞ | sinc 2 ( t ) | 2 dt, since sinc( t ) is a real-valued function. Using Parseval’s theorem we have, ± −∞ | f ( t ) | 2 dt = ± −∞ |F f ( s ) | 2 ds. 1
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We set f ( t )=sinc 2 ( t )andus ing F ( s )=Λ( s ), we obtain ± −∞ sinc 4 ( t ) dt = ± −∞ | Λ( s ) | 2 ds == ± 1 1 Λ 2 ( s ) ds =2 ± 1 0 ( s +1) 2 ds (since Λ is an even function) ± 1 0 s 2 2 s +1 ds ² s 3 3 s 2 + s ³ 1 0 = 2 3 . For the other approach, we recognize that sinc 4 ( t )trans formsto(Λ Λ) ( s ). Hence, the integral ´ −∞ sinc 4 ( t ) dt would simply be (Λ Λ) (0). Writing this out as an equation yields: ± −∞ sinc 4 ( t ) dt = ± −∞ Λ( s )Λ( s ) ds = ± −∞ Λ 2 ( s ) ds (since Λ is an even function) = ± 1 1 Λ 2 ( s ) ds This is the same integral as the Parseval approach and hence we get 2 3 again. (b) Here we have a multiplication of two functions whose Fourier transforms are known. The given integral can therefore be evaluated using Parseval’s theorem in its general form: ± −∞ f ( t ) g ( t ) dt = ± −∞ F ( s ) G ( s ) ds. For the functions we have, this becomes: ± −∞ 2 1+(2 πt ) 2 sinc(2 t ) dt = ± −∞ 2 ) 2 sinc(2 t ) dt = ± −∞ e −| s | 1 2 Π µ s 2 ds = ± 1 1 1 2 e −| s | ds ± 1 0 1 2 e s ds = ± 1 0 e s ds =1 1 e . 2
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For the other approach, we recognize that 2 1+(2 πt ) 2 sinc(2 t )trans formsto e −| s | 1 2 Π 2 ( s ). Hence, the integral ± −∞ 2 1+(2 ) 2 sinc(2 t ) dt would be the convolution of e −| s | and 1 2 Π 2 ( s ) evaluated at zero. ² −∞ 2 1+(2 ) 2 sinc(2 t ) dt = ² −∞ e −| s | 1 2 Π 2 ( s ) ds = ² 1 1 1 2 e −| s | ds =2 ² 1 0 1 2 e s ds = ² 1 0 e s ds =1 1 e . (c) The integral can be written as: ² −∞ t 2 sinc 4 ( t ) dt = ² −∞ ( t sinc 2 ( t ) )( t sinc 2 ( t ) ) dt Using the derivative property, which states: F{ t n g ( t ) } = ³ i 2 π ´ n G ( n ) ( s ) . we can deduce that t sinc 2 ( t ) transforms to i 2 π Λ ± ( s ). Here we take the derivative of Λ( s ) in a piecewise manner. The graph of Λ ± ( s )isshownbe low . Hence, from Parseval’s we have: ² −∞ t 2 sinc 4 ( t ) dt = ² −∞ i 2 π Λ ± ( s ) i 2 π Λ ± ( s ) ds = 1 4 π 2 ² −∞ Λ ± ( s ± ( s ) ds = 1 4 π 2 (2) = 1 2 π 2 3
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An alternative method would be to again use the derivative property in conjunction with convoution. Writing the left hand side as an integral, ± −∞ t n g ( t ) e i 2 πst dt = ² i 2 π ³ n G ( n ) ( s ) .
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PS-5-2007-Solutions - EE 261 The Fourier Transform and its...

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