Solution 7.9.7.1 R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback For the system of Figure 1 wehave G p = 1 s(s+0:5) and G c = k pd + k d s: Part a. The PD compensator can be put in the form: G c = k d s+ k pd = k d (s+k pd =k d )=k d (s+a);; whichisapurezero. The calculation of k d and the ratio k pd =k d is based on Figure 2. Eachofthe vectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;3+j2. That is the vector V 1 is the polar representation of the factor s in the denominator of G c G p , V 2 the polar representation of the factor s+0:5, and V 3 the polar representation of s +a.Thus, G c G p j s=;3+j2 = K c jV 3 j 6 (jV 1 j 6  1 )(jV 2 j 6  2 ) = K c jV 3 j jV 1 jjV 2 j 6 ( ; 1 ; 2 ) The evaluation of G c G p at s = ;3+j2hasnowbeen broken down into acomposite magnitude and a composite angle. For the root locus to pass through this point, wemust have ; 1 ; 2 = ;180  : Since all the angles except are known, wecanwrite =  1 +  2 ;180  = h 180  ;tan ;1 (2=3) i + h 180  ;tan ;1 (2=2:5) i ;180  = 146:31  +141:34  ;180  =107:65  1 θ 1 θ 2 -0.5 α -3 -a Im(s) Re(s) 2 V 3 = s + a V 2 = s + 0.5 V 1 = s Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles Then a = 3; 2 tan(180  ;107:65  ) = 3; 2 tan(72:35  ) =2:364 The gain that places a closed loop pole at s = ;3+j2, and another at s = ;3;j2isobtained by solving K c jV 3 j jV 1 jjV 2 j =1;; (1) or K = jV 1 jjV 2 j jV 3 j : (2) The gain to place the poles at this location is then K = p 13 p 10:25 p 4:46 =5:5: The PD compensator is then G c =5:5(s+2:364): Since k d =5:5, and k pd k d =2:364;; 2 wehave k pd = 2:364k d =2:3645:5=13: Part b. K v = lim s!0 sG c (s)G p (s)=lim s!0 s  5:5(s+2:36 s(s+0:5)  = 5:52:36 0:5 =26 Then the steady state error to a ramp input is e ss = 1 K v = 1 26 =0:0385;; or less than four percent. Part c. If we add a PI compensator k pi + k i s = k pi s +k i s = k pi (s+k i =k pi ) s and place the zero close to the pole at the origin, then the gain to place the poles at s = ;3j2isstill approximately 5.5, if wemake k pi =1.We then choose k i =0:1. This is an arbitrary choice. The nal compensator is G c = 5:5(s+2:364)(s+0:1) s = 5:5(s 2 +2:464s+0:2364) s = 13+ 1:3 s +5:5s: Part d. K P =13 K I =1:3 K D =5:5: The acceleration error constantis K a = lim s!0 s 2  5:5(s+2:364)(s+0:1) s 2 (s+0:5)  = 5:52:360:1 0:5 =2:596: Then e ss = 1 K a = 1 2:596 =0:385 Thus, if we build the conventional PD/PI, the steady state error to a parabolic input is very large. 3 Im(s) -0.5 -3 -2.36 2 2 Poles Figure 3: Root Locus of Compensated System Part e. The damping ratio of the dominantclosedlooppolesis  =cos[tan ;1 (2=3)] = cos[33:69  ]=0:83: The damped frequency is ! d =2rad/s., making t p =3:1412=2=1:57 s. Thus, the system should reachitspeak in about 1.5 s. and haveovershoot of less than 2%. To nd the time response we rst nd the closed loop system in factored form. The system has three closed loop poles, twoofwhichweknow, namely the closed loop poles at approximately s = ;3 j2. The root locus of the compensated system is shown in Figure 3. One might suspect that there are break-in and break-out points between the zero at s = ;0:1 and the pole at s = ;0:5. However, as the root locus shows there are not. One could verify this by building Table 1 and noting that the gain along the real axis rises steadily from near the pole to near the zero. The following MATLAB dialogue veri es these computations. EDU>gcgp = zpk([-2.364 -0.1],[0 0 -0.5],5.5) Zero/pole/gain: 4 s -0.4 -0.35 -0.3 -0.2 -0.15 -0.11 -0.1002 1.00322 K 0.027 0.037 0.044 0.055 0.0712 0.21 8.87 5.519 Table 1: Searchfor Closed Loop Pole 5.5 (s+2.364) (s+0.1) --------------------- s^2 (s+0.5) EDU>tc = feedback(gcgp,1) Zero/pole/gain: 5.5 (s+2.364) (s+0.1) ------------------------------- (s+0.1003) (s^2 + 5.9s + 12.96) EDU> We see that one closed loop pole will be almost on top of the zero at s = ;1:002 The other twoclosed loop poles will be at approximately s = ;3j2. Thus the closed loop system in factored form is: T c (s)= 5:5(s+2:36)(s+0:1) (s+3;j2)(s+3+j2)(s+0:1002) This is approximate. When weadded the PI compensator wemoved the poles o slightly from s = ;3  j2. We are not worried about the exact location of the third pole because as wewill see shortly,ithasvery little e ect on the system response. The output C(s)forastepis C(s) = 1 s T c (s) = 5:5(s+2:36)(s+0:1) s(s +2:95;j2:06)(s+2:95+j2:06)(s+0:100322) At this pointweknowthegeneral form of c(t). That is, c(t)= h 1+Be ;0:100322t +2jMje ;2:95t cos(2:06t+ ) i 1(t);; where  = 6 M.Soweknow quite a bit. What remains to be done is to nd the constants B and M. B = (s+0:100322)C(s) j s=;0:100322 5 Time (sec.) Amplitude Step Response 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 4: Unit step response of compensated system = (s+0:100322)  5:5(s+2:36)(s+0:1002) s(s+2:95;j2:06)(s+2:95+ j2:06)(s+0:100322)  j s=;0:100322 = 0:0032 M = (s+2:95;j2:06)C c (s) j s=;2:95+j2:06 = (s+2:95;j2:06)  5:5(s+2:36)(s+0:1) s(s+2:05;j2:06)(s+2:95+ j2:06)(s+0:100322)  j s=;2:96+j2:06 = 0:794 6 ;2:25 rad: Thus, nally wehave c(t)= h 1+0:0032e ;0:100322t +1:588e ;2:95t cos(2:06t;2:25) i 1(t) The unit step response is shown in Figure 4. 6