Solution 7.9.9.13 ΣΣ + + R C K r G GG c 1 2 θ Figure 1: Maglev block diagram For the system of Figure 1 wehave G 1 (s)= 1 Js and G 2 (s)= 1 s : We rst note that the proposed design turns a type 2 system into a type 1. The reason is that if we nd the equivalenttransfer function for the subsystem consisting of G 1 and K r wehave G eqiv (s) = 1=J s + K r =J : Then the new blockdiagarm is shown in Figure 2 Wehavetwochoices. One is to movethepole at s = ;K r =J to the left using K r and let G c = K c .In that case wehaveanequivalentforward loop transfer function G c G p (s)= K c K r =J s(s + K r =J) = K s(s + p) ;; where K = K c K r =J and p = K r =J: Toachieve the speed of reponse we require  = 1 p  0:15 s: The rationale is that the time to reach90% of nal value is speci ed to be 00.2 s. If welet =0:15 then for critical damping we should reach63%of 1 Σ + R C GG c2 G equiv Figure 2: Equivalentblock diagram X X -p -p 2 Im(s) Re(s) Figure 3: Calculation of gain for critical damping nal value in about 0.15 s, whichshould assure that wereach90%of nal value in 0.2 s. However, this will not be sucienttogiveusthe required steady state accuracy to a ramp input. Figure 3 shows why.Theoverall gain required top achievecritical damping is K = p 2  p 2 = p 2 4 : The K v =lim s!0 sK s(s + p) = p 4 : Suppose welet p = 20 so that the compensated system has twopoles at s = ;10, clearly meeting the speci cation on rise time. Then K v =5;; 2 XX X -p Im(s) Re(s)O -a Figure 4: Root locus for K r =J small XX X -p Im(s) Re(s)O -a p-a -c -b a - b a - c p Figure 5: Calculation of gain for critical damping K r =J small and we certainly do no meet the speci cation on steady state accuracy. Therefore, we consider making the ratio K r =J smnall, and adding a lead compensator as shown in Figure 4.The idea is to approximate an integrator. As pointed out earlier this is somewhat contradictory since byadding rate feedback, weeliminated an integrator whichwenoware trying to approx- imate. In any case, we can now nd the gain that will critically damp the system, namely K = (p; a) a (a; b) a;c ;; as shown in Figure 4. Since weareplacing both the pole at ;K r =J and the zero of the lead close to the origin, wecangetaroughtidea of the 3 magnitude of p by simply letting a; c = p and a;b = pand a  p 2 : Then, K = p 2 4 : If wenowletp=20, wehave K v = (p 2 =4) c pb =  p 4 )  c d  If we let p =20thenforK v =50,wemust have 5  c d  =50;; or  c d  =10 Thus, for convenience wecould make b =0:1andc =1. Then, letting p = 24, we can now use the short MATLAB program: gcgp = zpk([-1],[0 -0.1 -24],1) rlocus(gcgp) print -deps rl79913.eps to generate the root locus shown in Figure 6. As can be seen from the gure the break-out is very closetos = ;12. Wetherefore assume that a =12, can compute the gain K c to place twopoles at s = ;12. K c = jsjjs +0:1jjs+24j js +1j j s=;12 = 155:78;; and K v = K c  1 24 0:1 =64:9: Thus, wemeet our speci cations. The following MATLAB program gener- ates the unit step response shown in Figure 7. Weseethat we are close to meeting the speci cations speci cations. Therefore, weletp = 50, and then try to nd the gain that yields critical damping with a little more precision. That is we solve the equation K c = jsjjs +0:1jjs+50j js +1j 4 -25 -20 -15 -10 -5 0 5 -15 -10 -5 0 5 10 15 Real Axis Imag Axis Figure 6: Calculation of gain for critical damping K r =J small Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 7: Unit step response of compensated system 5 s ;24:7 ;24:9 ;25 ;24:8 ;24:85 K c 629.50 629.545 629.536 629.533 629.541 Table 1: Locating break-out pointonreal axis Time (sec.) Amplitude Step Response 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 8: Unit step response of compensated system for values around s = ;25 to nd the break-out point. Table 1 summarizes that search. Thus the gain for critical damping is K c = 630. Then we can generate the step response of the compensated system, shown in Figure 8, using the MATLAB program: gcgp = zpk([-0.2],[0 -0.02 -50],630) tc = feedback(gcgp,1) step(tc,2) print -deps sr79913b.eps Wecheck K v = 6300:2 0:02 50 =126: By examining this last 6