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**Unformatted text preview: **EE 261 The Fourier Transform and its Applications Fall 2009 Solutions to Problem Set Three 1. (25 points) Piecewise linear approximations and Fourier transforms. (a) The stretched triangle function is defined by a ( t ) = ( t/a ) = braceleftBigg a | t | a , | t | a , | t | > a Find F a ( s ). (b) Find the Fourier transform of the following signal. 1 2 2 2 . 5 4 6 t Hint: Think s. (c) Consider a signal f ( t ) defined on an interval from 0 to D with f (0) = 0 and f ( D ) = 0. We get a uniform, piecewise linear approximation to f ( t ) by dividing the inter- val into n equal subintervals of length T = D/n , and then joining the values 0 = f (0) , f ( T ) , f (2 T ) , . . . , f ( nT ) = f ( D ) = 0 by consecutive line segments. Let g ( t ) be the linear approximation of a signal f ( t ), obtained in this manner, as illustrated in the following figure where T = 1 and D = 6. 1 1 1 2 2 3 3 4 5 6 t f ( t ) g ( t ) Find F g ( s ) for the general problem ( not for the example given in the figure above) using any necessary information about the signal f ( t ) or its Fourier transform F f ( s ). Think s, again. Solution: (a) From a ( t ) = ( t/a ) = braceleftBigg a | t | a , | t | a , | t | > a and the stretch theorem we have F a ( s ) = 1 1 /a sinc 2 ( s/ (1 /a )) = a sinc 2 as. (b) The function in part (b) is given by the sum of two scaled and shifted triangle functions. 2 2 ( t 2) + 2 . 5 2 ( t 4) . 1 2 2 2 . 5 4 6 t Thus, using the shift and stretch theorems we find the following Fourier transform, 2 2 ( t 2) + 2 . 5 2 ( t 4) 2 e 2 is (2) 2sinc 2 (2 s ) + 2 . 5 e 2 is (4) 2sinc 2 (2 s ) sinc 2 (2 s ) ( 4 e 4 is + 5 e 8 is ) . 2 (c) This is an extension of the previous part (a). The piecewise linear approximation g ( t ) is the sum of shifted and stretched triangles that are scaled by values of function f ( t ) at their center points: 1 1 2 2 3 3 4 5 6 t f ( t ) g ( t ) We compute F f ( s ) by F f ( s ) = F parenleftBigg n 1 summationdisplay k =1 f ( kT ) T ( t kT ) parenrightBigg = n 1 summationdisplay k =1 f ( kT ) F T ( t kT ) = n 1 summationdisplay k =1 f ( kT ) T sinc 2 ( Ts ) e 2 is ( kT ) = T sinc 2 ( Ts ) n 1 summationdisplay k =1 f ( kT ) e 2 iskT . 2. (15 points) Hubbard-Stratonovich Formula Show that integraldisplay e x 2 e 2 sx dx = e s 2 . In Physics, this equation and higher dimensional generalizations are called Hubbard-Stratonovich formulas; do a Google search at your own peril. (You might recognize it as the two-sided Laplace transform of e x 2 . Note that theres no i in the e 2 sx . ) Solution: We argue as we did for finding the Fourier transform of the Gaussian. Let F ( s ) = integraldisplay e x 2 e 2 sx dx ....

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