Question

If the initial cone A E Re has a root locus plot started in Figure P1. Determine the following about the root locus determine a) the transfer f a) Of points A, B & C indicated on the real axis which are on the root locus? Ans b) the DC gain of b) How many zeros are there at infinity? Ans c) What angles do the infinity zero asymptote(s) make with the positive real axis? Ans d) Where do the infinity zero asymptotes intersect the real axis? c) the value of y a Figure PIsT Ans e) What range of K, is required for stability? Use the Routh Hurwitz criterion Ans: K(s + 4) Problem 2: A system has the loop transfer function Which of the s plane points listed below is on the root locus for this system at a point where the a) damping ratio is 0.707? (Circle the correct root location & show your verification calculations.) 0.25 For this system, use root locus analysis to determine: Circle one answer -1+1-34i-2 21-14.3-1+0.707 em 3 (15 points) l performance crit Ans: What is the gain, K, associated with that point? b) applies.t Problem 3: A system is characterized by the following differential equation y+8y +16y 2x+x If the input is x » sinSt, use the frequency response method to determine the steady-state response of the output, yalt). In showing your work, clearly identify expressions for G(s)-Y(s)/X(s), Gjo), IGljo)) and o) tate error requency me ratio: ce rejection: ershoot

Answer 1

Roots of a second order characteristic equation are:

$\small s = -\zeta w_n\pm i w_d$

For any complex root,

$\small a\pm ib = -\zeta w_n\pm i w_d$

6 bwd

2 7t 1

(b)2 = (に 6 )2

$\small b^2\zeta^2 =a^2(1-\zeta^2)$

$\small \zeta^2 = \frac{a^2}{a^2+b^2}$

a2 + b2

option 1 or 3 is correct

$\small s= -1+i \;or\;s = -2+2i$

$\small \zeta = \sqrt{\frac{1}{1+1}} = 0.707$

$\small \zeta = \sqrt{\frac{2^2}{2^2+2^2}} = 0.707$

$\small L(s) = \frac{K(s+4)}{s(s+2)}$

characteristic equation is:

K(s + 4) s(s+2

$\small s^2+2s+Ks+4K = 0$

$\small s^2+(2+K)s+4K = 0$

solution of this option 1: $\small -1\pm i$

$\small (s+1-i)(s+1+i) = 0$

$\small (s+1)^2-i^2 = 0$

$\small s^2+2s+2 = 0$

compare with

$\small s^2+2s+Ks+4K = 0$

No Solution.

This is not valid. So, Option 1 is not correct.

solution of this option 3: $\small -2\pm2i$

$\small (s+2-2i)(s+2+2i) = 0$

$\small (s+2)^2-(2i)^2 = 0$

$\small s^2+4s+4+4 = 0$

$\small s^2+4s+8 = 0$

compare it with

$\small s^2+2s+Ks+4K = 0$

K = 2 is the answer.

So, Option 3 is correct. $\small -2\pm2i$