Question

Problem 5. For the sinusoidal voltage signal given by: v(t) = Vmcos(wt+o)V (-0.333,6) - (-1.333,0) (0.667,0) (2.667,0) (1.667,-6) a) Maximum Amplitude of v(t)= Vm b) Time period: T c) Frequency: f d) Angular frequency : 0

Answer 1

Sola given, wt)= Vie Cos(wtto).V. 0,333,6) 1 U13 (2.667,0) 160.667,0) t(sec) . (1.667,-6) . M M fox simplicity we can write (-1333,0) = (-44,0) 1 (6667,0) =( 23.;0) (-0.333, b) = (-43, 6) 112.667,0) = (3/310) . 1(1.667,-6)=(5131-6). Lay from above signal plot, we can see signal ist attaining a maximum value of b at t= -0.333. so we can say maximum Amplitude of Ult) = Nam= 6v. <b> Let, time period of sinusoidal signal be. I. we know that time difference between consecutive - Crest and is trough is half of time period. cresta upper peak] Ata trough texest trough: dower Peak. = 11667–601333) = 15+ 1 = sec 2020/5/12 19 Now,

.. >= At =) T=2ot = 2x2 = 4 sec T= 4 sec time beriod The required i s 4 sec. . (c) we know that, frequency (0) = * = 0.25 #7 . angular frequency (W) = zit = 298 0.25 = Trad/sec. 2020/5/12