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hwsolB03

# hwsolB03 - [0.160 0.180 i.e at t = 0.170 By periodicity the...

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Sheet1 Page 1 HW 3 Solution _____________ (i) We know that the value 2.5 occurs at t=0, 0.160 and 0.180 sec and at no other intermediate times. Since a sinusoid crosses any given level (except its maximum and minimum) twice in each period, it follows that the period T equals exactly 0.180 - 0 = 0.180 sec. W = 2*pi/T = 100*pi/9 and f = 1/0.180 = 5.556 Hz (ii) Since 2.5 is positive and every sinusoid is symmetric about any one of its peaks, it follows that a peak occurs at the midpoint of
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Unformatted text preview: [0.160, 0.180], i.e., at t = 0.170. By periodicity, the first peak to the left of the origin (t=0) occurs at t=-0.010. Thus the initial phase of the sinusoid corresponds to 0.010/0.180 = 1/18 of a cycle, i.e., q = (1/18)*2*pi = 0.3491 rad which lies in the required range. (iii) At t=0, A*cos(q) = 2.5, therefore A = 2.5/cos(0.3491) = 2.6604 (iv) t = (0 : 1/150 : 2-(1/150))*0.180 x = 2.6604*cos(100*pi*t/9 + pi/9) plot(t,x) p...
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