HW2sol - ECEN 314: Signals and Systems Solutions to HW 2...

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ECEN 314: Signals and Systems Solutions to HW 2 Problem 1.21 -0.5 0 -1 1 2 3 -1 1 2 0 1 2 3 4 x(t-1) x(2-t) 1 2 -1 -1.5 -1 -0.5 0 0.5 1 2 -1 x(2t+1) 4 6 8 10 12 2 1 0 1 x(4-t/2) [x(t)+x(-t)]u(t) 3 -1.5 1.5 2 1 -1 -0.5 Problem 1.23 -3t/2 -2 -1 2 1/2 1/2 1/2 -1/2 -2 -1 0 1 2 e x (t) o x (t) (a) -1 -2 1 0 1 0 2 -2 -1 0 1 2 x (t) e (b) 1/2 -1/2 1/2 1 1 1/2 x (t) o 0 x (t) e 3t/2 (c) x (t) o -t/2 1
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Problem 1.25 (a) x ( t ) = 3 cos(4 t + π 3 ) is periodic with period: π 2 . To see why this is true, note that we need to find a T so that x ( t + T ) = x ( t ) Note that x ( t ) = 3 cos(4 t + π 3 ) = 3cos ( 4( t + π 2 ) + π 3 ) , which yields the desired answer. (b) x ( t ) = e j ( πt - 1) is periodic with period = 2 π π = 2. (c) x ( t ) = [cos(2 t - π 3 )] 2 [cos(2 t - π 3 )] 2 = 1 + cos(4 t - 2 π 3 ) 2 which is periodic with period T = π 2 . (d) x ( t ) = E (cos(4 πt ) u ( t )) = cos(4 πt ) u ( t ) + cos(4 πt ) u ( - t ) 2 = 1 2 cos(4 πt ) - ∞ < t ≤ ∞ which is periodic with period 1 2 . (Here, E is the even part). (e)
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HW2sol - ECEN 314: Signals and Systems Solutions to HW 2...

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