elg3155-solution6.pdf - a n2 = 16 r\/s 2n = 3 Therefore = 0.375 n = 4 Ts = 4 = 2.667 s TP = n 2 n 1 = 0.8472 s%OS = e 2 1 x 100 = 28.06 nT =(1.763 0.4172

elg3155-solution6.pdf - a n2 = 16 r/s 2n = 3 Therefore =...

This preview shows page 1 - 10 out of 10 pages.

Image of page 1
a. n 2 = 16 r/s, 2  n = 3. Therefore = 0.375, n = 4. T s = 4  n = 2.667 s; T P = n 1- 2 = 0.8472 s; %OS = e - / 1 - 2 x 100 = 28.06 %; n T r = (1.76 3 - 0.417 2 + 1.039 + 1) = 1.4238; therefore, T r = 0.356 s . Program: '(a)' clf numa=16; dena=[1 3 16]; Ta=tf(numa,dena) omegana=sqrt(dena(3)) zetaa=dena(2)/(2*omegana) Tsa=4/(zetaa*omegana) Tpa=pi/(omegana*sqrt(1-zetaa^2)) Tra=(1.76*zetaa^3 - 0.417*zetaa^2 + 1.039*zetaa + 1)/omegana percenta=exp(-zetaa*pi/sqrt(1-zetaa^2))*100 subplot(221) step(Ta) title('(a)') Computer response: ans =
Image of page 2
(a) Transfer function: 16 -------------- s^2 + 3 s + 16 omegana = 4 zetaa = 0.3750 Tsa = 2.6667 Tpa = 0.8472 Tra = 0.3559 percenta = 28.0597
Image of page 3
Image of page 4
a. Measuring the time constant from the graph, T = 0.0244 seconds. Estimating a first-order system, G(s) = K s+a . But, a = 1/T = 40.984, and K a = 2. Hence, K = 81.967. Thus, G(s) =
Image of page 5
Image of page 6
Image of page 7
Image of page 8
Image of page 9
Image of page 10

You've reached the end of your free preview.

Want to read all 10 pages?

  • Summer '15
  • TRA, n2, Percenta, 2n

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes