Consider a signal v(t) = b(t)cos² (wot + 0). As in the figure, b(t) = bx for t E 5. 1,-1,-1,1 for k = 1,2, 3,4 with 1/T, « fo. v(t) passes through an integrator that takes integration on v(t) for every T, seconds and outputs vo(kT,) for k = 1,2, 3, 4. Determine vo(kTb) for k = 1,2,3,4 with your derivations. b(t) 4 1 -1 -1 1 2 3 4 t(Tp) 1

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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b(t)cos2 (@ot + 0). As in the figure, b(t) = bx for t E ((k – 1)Tp,kTp] and bg =
5. Consider a signaf v(t) =
1,-1,-1,1 for k = 1,2, 3,4 with 1/T, « fo. v(t) passes through an integrator that takes integration on v(t) for
every T, seconds and outputs vo(kTp) for k = 1,2, 3, 4. Determine v(kTp) for k = 1,2, 3, 4 with your derivations.
b(t) A
1
-1
-1
1
4
t(Tp)
1
Transcribed Image Text:b(t)cos2 (@ot + 0). As in the figure, b(t) = bx for t E ((k – 1)Tp,kTp] and bg = 5. Consider a signaf v(t) = 1,-1,-1,1 for k = 1,2, 3,4 with 1/T, « fo. v(t) passes through an integrator that takes integration on v(t) for every T, seconds and outputs vo(kTp) for k = 1,2, 3, 4. Determine v(kTp) for k = 1,2, 3, 4 with your derivations. b(t) A 1 -1 -1 1 4 t(Tp) 1
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