# HW00 - University of Illinois Spring 2011 ECE 361: Problem...

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Unformatted text preview: University of Illinois Spring 2011 ECE 361: Problem Set 0: Solutions Linear Systems and Probability Review Note: In this course, we use Fourier transforms with respect to the frequency variable f (measured in Hertz) rather than the radian frequency variable ω = 2 πf (radians per second) that you used in ECE 210. Thus, X ( f ) = Z ∞-∞ x ( t )exp(- j 2 πft ) dt, x ( t ) = Z ∞-∞ X ( f )exp( j 2 πft ) df, and Z ∞-∞ | x ( t ) | 2 dt = Z ∞-∞ | X ( f ) | 2 df. Note also that the unit rectangular pulse rect( · ) and the sinc function sinc( · ) are defined as rect( t ) = ( 1 , | t | < 1 2 , , | t | > 1 2 , and sinc( t ) = sin( πt ) πt , t 6 = 0 , 1 , t = 0 . The sinc function has the useful property that sinc(0) = 1 while sinc( n ) = 0 for nonzero integers n . 1. [Drill exercises on Fourier transforms and linear systems] (a) Z ∞-∞ rect( t/T )exp(- j 2 πft ) dt = Z T/ 2- T/ 2 exp(- j 2 πft ) dt = exp(- j 2 πfT/ 2)- exp( j 2 πfT/ 2)- j 2 πf = sin( πfT ) πf = T · sinc( fT ). Note that narrow pulses in the time domain (small T ) correspond to broad main lobes (from- 1 /T Hz to 1 /T Hz) in the frequency domain. (b) From the duality theorem x ( t ) ↔ X ( f ) ⇔ X ( t ) ↔ x (- f ) and the result of part (a), we have that W · sinc( Wt ) ↔ rect(- f/W ) = rect( f/W ). Setting W = T- 1 , we deduce the transform pair sinc( t/T ) ↔ T · rect( fT ) . Note that the Fourier transform of sinc( t/T ) is nonzero only for | f | < 1 2 T ....
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HW00 - University of Illinois Spring 2011 ECE 361: Problem...

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