Question

Exercise 10 (8 Marks) Given the open loop transfer function of a system: KH(S) = K s(s +3Xs? +2s +2) Draw manually the root locus plot for the system and determine: a) The number of branches. b) The starting and ending points of all the branches. c) The location of the centroid d) The range of K to keep this and angles of asymplotes. e) The intersections of the root loci with the imaginary axis and the corresponding value of K. 6) The system's oscillating frequency associated with the gain K found above. 9) The angle of departure and angle of arrival (if any). h) Given that the breakaway point is at s = -2.3. find the value of at that point Verify your answers with that generated using MATLAB. Click on any of the locus and move your cursor accordingly You will see, among the 6 items appeared on th 1. 3 of them are 'Gain', Pole' and 'Frequency (rad/sec)' "Gain' refers to the value of K. "Pole' refers to the value of poles (the roots of your denominator) "Frequency (rad/sec)' refers to w.

Answer 1

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open loop trama fer fenchon- K KHOS) = No. of polex 4 =P S(5+3) (5?+25+2) Noig zeron = 0 = 2 Polen S=0,-3,-1 & j (a) No. of branch in the soot locous in the same as no. oof the pole No. of branch = No. of potent=4) Am open loop Zeros b) (1) Root locaus will always inticute from open Loop poles of the system which is the condition for k=0 (in Root locores will always terminate at which in the condition for ko (C) Cembriod = x EP-Ez EP- Sum of the location of poles. P-2 {z- sum of the location of resor P = Total no. of poles 2= Total no. of zeron {P: (0-3-1+/- 2-5 {zuo P = 4 X 2=0 x= EP-{z . - 5-0 4 5 P-2 4 x-1.25 Answer

(ii) Angle of asymptotes) k=0, 1, 2, ... IP-261) Ok = 244] X 180 P-2 K=0,1,2,3 P-2-154-001 - 3 On : 2x0+1 X 186 4-o Oi? (2x1)+|x 180' O2 = (2x2)+1 101=135" U-O x 180 uro Do - 45 O2 = 22501 0,= (2x3) +1 x 180 u-o 103 = 315 Angle of asymptotes 45', 135,225,315' Ay (d) Range of K to keep syitem stable KHCS) S(5+39 (25+3) chara. Equation. IAKH(S)0 K 1 + s(3+ 232435+ 35'+65+9) S9+258 +35² +38² +68² +95 +1 =0 S4 & 5 5 ² 9 95² + 9st kan RH Table to be stable 4k no sign change in fat +53 5 coloum of RH table k meam for system + SY 1 o g 0 0 +S² 7.2 64.8-5k so 0 0 64.8.5k 2.2 ts 7.2 64.8-5k>o 5k < 64.8 K <12.96

for keeping system stable sange of k in less than 12.96 IK< 12.96 Am (e) Intersection of rool locou al with ju axis when noot locous cut the jwaxis, at that point system in marginally stable. for rystem to be marginaly dtable as' now of RH Table Auxilary equation : 7-252+K=0 7.25²+12.960 SE 12.90) 7.2 (S. tj 1:3416 murt be zero 64.8-5k 7-2 K = 12.96 Interection of soot locous with jw axin in tj 1:3416 and value of corresponding value of k in 12.96. of The rystem Oscillating frequency. System oscillating frequeny come with auxilary equation. S = +1.3416 jre = J 1.3416 W = 1.3416 roed Isec

(9) only angle of departure in present because in 8.2.7f only complex polen are prevent. Angle of arivol in not prevent becoure in Ol.Tif there in complex teron are prevent. Angle o de porteve! dd = 180°-(0p-da) dp- Total angle rubrtended by all the polen to the ref pole d2 - Total angle subertanded by all the zeros to the ref pole. 0 = 180 - 0 0 = tan't 0, 5180-45 0.-135 d2 = tanilt 1900-037.j d2= 26.561 الم Od 5-71:56 A-141 tj 0 = 45 - 3 &d=+91-56' 09 = 90 Now, p= 135'+ 26.56+ 90 dp = 251.56 der=o Od: 180-(&p-dz) = 180-251.76 - 71.56° dd = + 71.56 Answer

(h) break point in S:- 23 So, It KHC) = 0 K 614 S(5+3) ($?+25+2) S(5+3) (5?+ 2S+2)+K-O Kers(5+3) (3 ²4 25+2) :-(-2.3) (-23 +3)(C-2-3)*4 26-2-2) +2) kel IS:-2-3 'S=-2.3 = 4.33 ki so g ke at break point in 4.33 · Anwer The value Drowing root locows Ok=no K 500 Lico 3 1.3416 K= 12.96 -14) Break point 5 = 4.33 - 1-25 K=0 k=o T -3 1 Centried (od -1) - 11:3416 KE 12.96 아 ok=00 k=00

Im 1.5 5 X 0.5 -3.0 -2.5 -2.0 -1.5 -1.0 1 Re 0.5 -0.5 -0.5 X. -1.a -1.5