HW1-S.pdf - ECE 318 Communication Systems Winter 2018 HW1...

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ECE 318 - Communication Systems Winter 2018 HW1; Signals and Systems : Solutions Problem 1 (Haykin 2.4) Be careful that the figure in the book uses f [Hz] as its frequency variable, whereas we conventionally use ω [rad/s] in this course. If we want to use the angular frequency version of the inverse Fourier transform, then we should first convert the units: B = 2 πW Using the inverse Fourier transform formula, we have g ( t ) = 1 2 π Z 0 - B exp( jπ/ 2) exp( jωt ) + 1 2 π Z B 0 exp( - jπ/ 2) exp( jωt ) = j 2 π 1 - exp( - jBt ) jt - j 2 π exp( jBt ) - 1 jt = 1 - exp( - jBt ) 2 πt + 1 - exp( jBt ) 2 πt = 1 - cos( Bt ) πt = 1 - cos(2 πWt ) πt (Where we used the fact that exp( ± jπ/ 2) = cos( π/ 2) ± j sin( π/ 2) = ± j .) Problem 2 (Haykin 2.10) A property of Fourier transform is g 1 ( t ) g 2 ( t ) ←→ G 1 ( f ) * G 2 ( f ). This means that a multi- plication in the time domain results in a convolution in the frequency domain. Since y ( t ) = x 2 ( t ), its Fourier transform is Y ( f ) = X ( f ) * X ( f ) = Z ν = ν = -∞ X ( ν ) X ( f - ν ) dν. Note that Y ( f ) is nonzero only for those f for which there exists a ν such that X ( ν ) X ( f - ν ) is nonzero. Also note that | X ( ν ) | 6 = 0 when | ν | ≤ W , and | X ( f - ν ) | 6 = 0 when | f - ν | ≤ W . For a given f , when you visualize the convolution of X ( f ) with itself as the sum (inte- gration) of the product of X ( ν ) with its time-reversed copy that slides past X ( ν ), it is easy to see that such a ν exists for all f in the interval f [ - 2 W, 2 W ] (see Fig. P-2 in page 7). Therefore, for every f [ - 2 W, 2 W ], Y ( f ) is nonzero, but for all f ( -∞ , - 2 W ) and f (2 W, ), Y ( f ) is zero. Fig. P-2 illustrates the convolution process.

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