Question

A2. (a) Explain how the open-loop polar plot can be used to assess closed-loop stability by applying Nyquist's stability criterion. Apply Nyquist's stability criterion to determine the stability condition for a closed-loop system that is unstable in the open-loop. [30%] = K (b) An unstable system has transfer function given by G(S) in which the gain K is S(S-2) positive. A derivative compensator H(s) = 0.5s + 1 is inserted in the negative feedback path to form a control loop. Derive the expression for any asymptotes and axis crossing points and the frequency at which they occur, as function of K. Draw the open-loop polar frequency response for all positive and negative frequencies. [45%] (c) For the control loop in part (b), determine: (i) (ii) The value of K for marginal stability The values of K for which the system is unstable and the number of unstable poles in this case The range of positive values of Kfor which the system is stable The gain margin for K = 10 [25%] (iv)

Answer 1

stability of the s from dood co to on ito drow centro Rat screens, the complete urod ou SL states if the 2 w enclosed the (a) Nyquist plast one the continuation of polare plot for friding an Soup regelemes dy verying Nygust Ploti сероли, отеть. А доо. ogen loop trausfer function - The Nyquist stability cuterion works the thore principle of consument p poles and ley plone corresponding Gas He flone must Dirgin write [N=P-Z] byoumple of unstable opun lopen transfer function and opou loop deusfer fruchom. (ores-arius) (22+2)(4+2) Poles of Oltr, (1²-24+2 6+2) 4 + 1 je closed poll then encude the p. z Times. So encuciment k Minus sigui + 2 + 14-8 os GH(g) = K 2 s-2 " Complex poles I Ij2 lies Height holf of s-plan which makes OLTE uustable

now, Nyquist plat for quien OLTt will be drown as Wo Now if Nyquist plat encircles Itjo them from given diogram here no of of cou conclude os N = P-z -tjo enciô clement is a ie N=2 a = 3-3 Z=11 closed loop systém since ut implies zfo is also unstable. (6) 613) - 1 8(3-2) 0LTE , q (5) HA) : compensator M(s) : 0.5811 negative flo (0.58 +1) 1(3-2) Now GH(s) : K s(3-2) I sehe nenus alt(s) - K(3+2) k(st2) 20 (392) 20(3-2) sign 2 poly Now the point from where call symptolis crosses akis, quien (+2) {-(0+2)} is os, 4 c = 2-1 reapritude where ist exists as 4, ...

4 w 1/3 Deas the oqun bloc frequency response chts) k(3+2) 25(3-2) when wroj 1 an(s)) 2w utwe lansilo ao futwa kutwa ie к на =o.o lauls) go + 2ton el At W-O 44] -90 MWO, lanlal +90 (c) The value of k for marginal stedelity CE بن given as 1+ 418)-H(2) Sune GHIO) K(8+2) 25(3-2) New characteristic equation 1+ k(8+2) 28(3-2) 28²-45+k8+2x = 0 234 + s(K-4) CE +2ko

Nooi by Routh Hurwitz Criteria 2. 2k K-4 Qk stable k-y=0 for moriginally system, [K-4) (ii) The value of k for which system unstable there will be sign change put column of Routh Huritz criterion which represents unletetly Now if if kah then مل of systers [k <4 slm will be unstable e then 2 the Number of sign change K-4 the (forke4) 2k - the then there will be no clu a sign changes : 2 UNSTABLE POLES. (1) Rouge of k for stable system when k> 4 sign Change column of R4 Toble. for [K341 system will be stable (iv) Goir margin for k= to .

Since Grain Pergin in celulated at please comover for phase and over 180 - got a low we +(-180") 180' Lilou we ilou" w 225" tou (225) itou" (225) ton 180 tus) fonus = 1 - 10 2x2 Moquitida at cop: 2 for Kato I Gr(s) | 2.5 2 op Joperty log { lantall} de 20 . Gain Plonginn 2o lung () Geir margin- Sime negative value of Gain margin represents unstable system - 7.95