G pG c + - R C Solution: 7.9.3.10 For the system shown abovewehave G p (s)= 1 s 2 and G c (s)= K c (s + a) s + b Todetermine ! n we rstsolve t p =  ! d =  ! n p 1; 2 ;; with t p  1. Then ! d  : Aconvenientchoice is ! d =3:6 r./s. This should more than do the job in terms of keeping the time to peak under one second. Tokeep the overshoot down requires a large damping ratio. The choice  =0:8willprobably do the trick. Then ! n = ! d p 1; 2 = 3:6 0:6 =6: Then the real part of the desired dominantclosed loop poles is  = ;! n = ;0:86=4:8: The dominantclosedlooppoles are then at s = ;4:8j3:6: For T N2 (s)= ! 2 n s 2 +2! n s + ! 2 n The choice of  =0:8 means the system is nearly critically damped. How- ever, the actual closed loop system will be third order with a zero and for systems of this sort the percentovershoot predicted from the formula for T N2 (s)isalmost always too optimistic. 1 -b θ 1θ 2 α Re(s) Im(s) -a 3.6 -4.8 2 Figure 1: Vector Diagram to Find Pole of Lead Compensator Figure 1 shows the vector diagram necessary to nd the location of the pole and zero of the lead compensator. Wemust have ;2 1 ; 2 = ;180  ;; or ; 2 =2 1 ;180  : It is clear from this last expression than there are an in nite number of choices for the lead compensator. One choice would be to place the zero close to the origin. This will move the pole of the lead to the left and result in a root locus like that sketched in Figure 2. The advantage of this design is that the near cancellation of one of the poles at the origin makes the resulting open loop system look very nearly like G c (s)G p (s)= K s(s + b) : Since this last transfer function leads to exactly T N2 (s) under feedback, the formula for per centovershoot is quite accurate. A second approachistomove the zero of the lead to the left, resulting in the root locus shown in Figure 3. This design will lead to more overshoot, but note that the damping ratio remains nearly constantover a large range of gain. This had its advantages. For the rst approach, assume that the zero is placed at s = ;0:1. Then  2 = + 180  ;2 1 2 -b Re(s) Im(s) -a 3.6 -4.8 2 Figure 2: Vector Diagram to Find Pole of Lead Compensator -b Re(s) j Im(s) 3.6 -a 2 Figure 3: Root Locus for Second Lead Compensator 3 = [180  ;tan ;1 (3:6=4:7)]+ 180  ;2[180  ;tan ;1 (3:6=4:8) = 142:55  +180  ;2[143:13  ] = 36:29  Then by simple trigonometry b = 4:8+ 3:6 tan(36:29  ) = 9:7: The gain is K c = p 3:6 2 +4:8 2  p 3:6 2 +4:8 2  p 3:6 2 +4:9 2 p 3:6 2 +4:7 2 = 36:97 ! 37 Then G c (s)G p (s) = 37(s+0:1) s 2 (s +9:7) For the second approachtheprocedure is the same except that weneed to either pickazero or a pole of the lead compensator. If we pick the zero to be at s = ;3, then  2 = +180  ;2 1 = [180  ;tan ;1 (3:6=1:8)]+ 180;2[180  ;tan ;1 (3:6=4:8)] = 116:57  + 180  ;286:26 = 10:3  Then b = 4:8+ 3:6 tan(10:3  ) = 24:6 The gain is K c = p 3:6 2 +4:8 2  p 3:6 2 +4:8 2  p 3:6 2 +19:8 2 p 3:6 2 +1:8 2 = 180 4 -b Re(s) Im(s) -a s + a s s + b Figure 4: Computation of Gain Along Real Axis to Find Third Pole s ;0:106 ;0:1045 ;0:103 ;0:1025 ;0:10274 K 17.97 23.39 33.93 40.33 36.97 Table 1: Gain Versus s Along Real Axis Then G c (s)G p (s) = 180(s+3) s 2 (s +24:6) The second design requires considerably more gain to implementthan the rst. In manycases the amountofgain, sometimes called `control e ort' is limited. In suchacasethe rst design would probably be better. The third closed loop pole can be found byavarietyofmethods. One is to simply look for the same gain, i.e. 36.97(180) in the vicinityofs = ;0:1 (s = ;3). Figure 4 shows how that calculation is made. K c = jsj 2 js + bj js + aj This same formula applies to either of the proposed designs. Table 1 sum- marizes the search for the location of the pole in the rst design, while Table 2 does the same for the second design. Thus, for the rst design The closed loop transfer function is T c (s)= 36:97(s+0:1) (s +0:10274)(s+4:8;j3:6)(s+4:8+j3:6) 5 s ;18 ;17 ;16 ;15 ;14 K 142 156.9 169.4 180 188.9 Table 2: Gain Versus s Along Real Axis 0 0.5 1 Closed Loop Response 00.511.5 2 2.5 3 Time In Seconds Figure 5: Step Response of First Lead Design and for the second design T c (s)= 180(s +3) (s +24:6)(s+4:8;j3:6)(s+4:8+j3:6) The step response for the rst design is shown in Figure 5. As can be seen, the speci cations are more than met. The time response for the second design is shown in Figure 6. As pre- dicted there is more overshoot in this case, and, in fact, wedonot meet the speci cation. 6 0 0.5 1 1.5 Closed Loop Response 00.511.5 2 2.5 3 Time in Seconds Figure 6: Step Response of Second Lead Design 7