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3. (50 points] Consider the signal (t= cos(27 (100)+]: 1) Lets take samples of x(t) at a sampling rate fs = 180 Hz. Sketch t

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© xlt) = Cos 21 (100)+) alt) FT Xlf) patitet - il-fo) - entot (fit) X16= Clos (217 (100) 4) = { [e]278.00€ jarrxo0 +7 - canx1NO -8. S. 100 18.26.0 280 360 460 f (H) 90 -360-286 - 160 n=1 n=2 90 olp of titter - so I so X, (+) = { $(4-80) +5 (+80)] ${}neo 120 (1-48° / -340 -24. -140 100 100 -145 246 340 380 480 58. .) -58 58. Pass Low pass filter - 120 100 - 100 f. x (+) = 1

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