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228_w06_q5_sol

228_w06_q5_sol - [1/6(t’ – 3 tri[1/3(t’ tri(t’ Use...

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EE 228 Continuous-Time Signals and Systems Cal Poly, slo Quiz 5 John Saghri Solution 1 - 1 1 tri (t) Part a: (5 pints) Express the function x(t) shown below in terms of rect and/or tri functions shown on the right. t 0 1 6 4 2 3 2 5 1 x (t) t 0.5 rect (t) 1 - 0.5 t Answer: x (t) =___ ___ ________________________________________ 2 rect [1/6 (t - 3)] 3 tri [1/3 (t - 3)] + tri (t - 3) Part b: (5 pints) Given the Fourier transform pairs: rect (t) sinc ( f ) and tri (t) sinc 2 ( f ), evaluate the Fourier transform X ( f ) in terms of sinc functions Do change of variable t’ = t - 3 to obtain: r (t’) =2 rect
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Unformatted text preview: [1/6 (t’)] – 3 tri [1/3 (t’)] + tri (t’) Use rect (t) ⇔ sinc ( f ) and tri (t) ⇔ sinc 2 ( f ) and the scaling property of FT to obtain: r ( f ) = 12 sinc (6 f ) - 9 sinc 2 ( f ) + sinc 2 ( f ) To find x ( f ), since t’ = t-3, use the shift property of FT to obtain: x ( f ) = [12 sinc (6 f ) - 9 sinc 2 ( f ) + sinc 2 ( f )] e –j2 π f (3) = [12 sinc (6 f ) -9 sinc 2 ( f ) + sinc 2 ( f )] e –j6 π f Answer: X ( f ) =________________ ______________________ ____ [12 sinc (6 f ) -9 sinc 2 ( f ) + sinc 2 ( f )] e –j6 π f...
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