Question

1- [a] For positive values of K, plot the root locus for a unity negative feedback control system having the following open-loop transfer function: K G(s)= (5 + 1)(8 + 4)(8 + 7) For what values of gain K does the system become unstable? Find also the value of k at which the damping ratio is 0.5 and the closed loop poles. (25%) [b] The characteristic equations of linear control systems are given below. Apply Routh-Hurwitz criterion to determine the root distribution and the system stability. (15%) 1) s? +386 +35+ 84+ S +352 +35 + 1 = 0 2) s? +85+2+45 +2s+4=0 3) S+s+2s+ + 25 + 35² + 25 + 4 = 0

Answer 1

Ginen. k : دراما کیا ر (5+1)(5+) ($+73 No. of poles= 3 (P) No of zeros = 0(2) Rut psz Total no. of Branches of Root Locy=p=3 to -1 (D, IL Root locus will lie inre[wa, -1) and reto, -7 m real axiçi since, p-z=3 . There will be three asymptotes with centroid Relp) - 2 Relz) 67-4 - 1) P-Z -4 3 & angles, Zq+1) (@q+)st P-7 50 (300°) .: 01= 1/3, 0, - it Qq = gt, 93=517 Break away and Break in point calculation It Kulss o (Stilsta) (347) =) ks (5+1)(5+2)(8+7) Now News - .dk ds 3+1252&398+ 28 28 [Not lie on -5•732 RL 11 !" 352 +2es+39 s2+85+13=0C sa 2.268, Boople

1 مفهومیلی (= intersection of root loay with jew axis :- method rowth table in method- I+his) = (1 + (-))=0 CS+US+4) LS+ 7) +K=O (+ju + 12[w]?139 () +be+k) => S3+12527395+28+K=O >>-j63-1262 +j3960 +2+K=0 -> (28+K-1204]+jw[ 25.09)-- Comparing lead & imaginary tering; s² 12 s' (12235 – 626 ++] 39 کی 1 28th 62 =39 = o 0 & 28+ k-12 we ka126224 K=12*39-28 so 28tk K = 440 for system to be marginally stable | roots lasty] i [12x25 +23+ y) = 5) 28+k's 12x39 k = 440 i point point of intersection 1232 +68+440) = 0 se + j 13g = 5; 6.295

3 حياز oaxis RL diagram:- Step=2 -jwnoxio steps j6.9287 16248 (K=440 (ماز ره ) P ما را kao CoN + + -4 2 - 2•268 It sy clear from RED RLD thad for (k>440) system will become unstable. --; 6.245 (K=440) -;6.928 & st o Tom for f= 0.5 O = Cos's = 60° Tom 10 - w=o Tom 6o aos Hence, point P. (-0, jois) must line on the roof loay Itkh(s=-6t; 873) OB It must satisfy angle condition 1 go 2 fr tjóvg +1) ( 5+j8v3 +4-) (-stjors +7)

J3 = + [ Tant + :-( .( م + ( 1 + 3 |-6 =) Ton ( 3 / ک ژله + 3 + Tan ک - 1 که - 4 از - (3) 4-)4-) :: 24+ Tah7 و 15 = - 75,1/4+s کے Tant! + Tannl هم |t-ي ++ - ه - ۱ 2( = وااه + 87 -6 Tant - ] احمد + (2 -۱۱اردی = )C ی 8- 10 -28 (- |- (3) 11 - 27) 3 و ہے Tan که - ۱ 118- 28 از - | - (2) (1-2) همع که با ها -11- 24 (2- 11 اردی) 3 C کا ما کا 1-8 2 ای 8 (دل سے) « وی 8 - 116-28 + (1 - 6) 11-26) -" (ء 7 ه + سعی 8- 1-T =

5 J 39 - 245 -62_ -) : 1.2+ ) + ) -S24 P : (-1-24 + 12.1477) 1에 S = P : ( -1-24 +] 2. 1477) (S+JL+4) Ce+ | 124 + 2-144) k 46046 Any.

ъ 0 913): 57 +356+355 +54 + s 6 3 +352 +35+1=0 7 3 One s 1 3 Ore so Det 613) [plore] 4] L) o ㅗ o orest 1 Avel Ove s3 ore s² ose 1 0. o Ques' - E Two sign changed love so 1 FSO,1,43 since s3 Row is zers. Hence, forming Auxiliary eqn j2K+1) It Als)= se + 0.52+ + E) e da(s) 4S3+ o ds Ist element of s2_row is zero. Hence, we replaced will a very small one € 14+eto) There are two sign changes below Auxillary equation Hence, two roots of auxillary eqn will be in the right & No soot f on jw-axit half

ㅋ Hence, The system is unstable will following root distribution :- Left half با ۵۸ - حبان Right half 5 2 bi(2) 968): s5+854, 253 4 452+25tt S 1 2 2 o re s I o rest 2 به I 3 ore s3 3 Two Gres? od ㅗ Sign chayel Ove s' + 2 ove so 1 UNCTABLE Left halt jw-axis Right half 2 3

b(3) 9ls): 56 +95 + 259 +233 +352+25+4 so o rest I 2 3 2 2 1 کیر co E 4 그 ve s o o & Eve) s É (26-1) Ć (24-4) s² le - 1) [_124-1) = (2-a)] : 4 s! (26-1)(862-20) 0 re sign charp o o o do Gre (66-261 0 ore s 8: (26-2)(=) they re] > It 66-262-1 262+46 = 1 ore 26-1 s?. 2E-1 t 2E-1 [24 es':- Gare / : اي - Ć (26-2).(4-1) 66-2 56-2€21) (26-1) (26-12 26-4) - 416€ 4166-26?, € (66-26?-1)2 (26-12 [-10€ +8€]= 4 (24-1) (84?20). 166=2€_,) (66-22-1 [2€ 118 = ive

UUSTABLE 9 Left Half jw-axis Righ Half 4