Question

9. Consider a negative unity-feedback control system with the loop transfer function s +8 D(s) G(8)=K- s+1) ((s + 1)2 + 22 (s + 94 + 793 + 1932 +33s + 20 (a) Determine the asymptotes of the root-locus diagram for K > 0, if any. (06pts) Answer: The real-axis crossing of the asymptote(s), a = The angle(s) of the asymptote(s), 0q = _ (b) Determine the break-away and the break-in points of the root-locus diagram for K > 0, if any. (O6pts) Answer: The break-away point(s) is (are) at The break-in point(s) is (are) at (c) Determine the angle of departure of the root-locus branch from the loop-gain pole with the positive imaginary part for K > 0. (06pts) Answer: The angle of departure from the pole with the positive imaginary part is (d) Determine the imaginary-axis crossings of the root-locus diagram for K > 0, if any. (06pts) Answer: The imaginary-axis crossing(s) is (are) at

Answer 1

DISG(S) = Kisto) - (5+1) f($+4)2 +9] [5+42 oles : s=-1 s=-4 - and 6+ 1)² = - 4 - St1= 121 = = -1 + 2 il Is=-1+2j - 4 Poles S=-8 > I zero 15=-1-2 j Istal - in P = 4 wrses z = 1 - ② - 3 ples will terminate at Infinity : : Angle of Asymptates 9 - (2k-1) 180° > D= (2x1-1)x180° 4-1 4-1 -D = 60° 02 = (2x2-1) x180° 4-1 > 102 = 180°

DATE PAGE D 3 = (2x3-1) X180° => 0 3 = 300 Real axis crossing of Asymptates se centrsis = X ALL X-EP-EZ P-Z where EP is the Summation and Ez is the Summation of of real parts of all Poly real parts of all woy E TTA atat EP = (-1)+(-4) + (-1) + (-1) EP = -7 EZ -8 > X = -7-7-8 3 2 1 4-1 6 Break in Points points and to to fi find Break away

ito(s) G(s) = 0 characteristic equation = 1 kls+8) 54+753+19528335 +20 » K= [54+753 +195² +335+20) 15+8) -- For Break away points and break in dk =0 -dk= as [518] [493+2152+385 +337-15"+753 +1932 1335 +20) - 15+8) 2 -- - 45+2183 +385 +335 +3253+ 16852+3045 +264 = 54+753 +1952+335 +20 - = 354 +4653 + 1875²+ 3045 +244=0 = S = -10.05 S = -3.1458 5=-1-065 +51-198 IS =-1-065- 11.198

DATE PAGE t21 2 *'--+-23 Angle of departure for Pole P1 = Od od = 180°-[op - oz . o 02 = 84 = tan "[2] Oz - 15-94° Qp = 0 + O2 +93 0-90° 0 3 = 90° 4 = tam l 3 01 - 33-69° > Op = 33-69° +90'+90° $P = 213-69 => Od = 180°- 213.69°-15.94] => Da = 19.950)

@ characteristic equation I+D(S) Gils =0 + K (+6) 54+753+1992+335+20 => 54 +253 +1952 + s (k+33) +20 +8K=0 S4 S3 52 S' 1 19 2018K Z K+330 19x7)-{\C+33) 20+8k o (K +33)A-7(20+86) o Ear gold 20+84 Marginal Stability rou Should be wo and the Poles will system is lie on the imaginary aris, when Marginally Stable. UK+33) A = 7 (2018k) A = 1947 (+33) = A= 100-K 3 (KH33)(100-K) = 7(2048k) 7 look-K² +330033K =49(2018k) = - -K² +678 +3300 = 980+ 392 K.

→ K2+36 K2+3 25k - 2 320 = 0 DATE, PAGE = K= 6.988 I K=-331-988 as Ko k= 6.988 ~ 7 Ikut Tsees an Imag any given by - - AS² + 20+8K) = 0 A= _ 10o-k t A = 10o-t = 13-285 7 .. 13-2852 + (20 + 8x 7) = 0 . 13- 2852 = -76 → 522-5-722 - - S = 1 ; 2.392