Solution: 7.9.6.2 For the system of Figure 1 let R + C G c G p Σ - Figure 1: Cascade Compensation with UnityFeedback G p = 1 (s +1)(s +2) and G c (s)= K c (s +2)(s + z 2 ) s(s +10) : Figure 2 shows the evaluation of G c G p (s)ats = ;3+j2incomponent form. That is, eachofthe vectors in Figure 2 is the polar representation of one of the factors in G c G p evaluated at s = ;3+j2. The vector V 1 , for instance, 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+1,and so on. The pole at s = ;2has been cancelled by the zero at s = ;2 and so those twoterms do not enter into the angle and magnitude calculations to be made next. 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 )(jV 4 j 6  3 ) = K c jV 3 j jV 1 jjV 2 jjV 4 j 6 ( ; 1 ; 2 ; 3 ) 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 ; 3 = ;180  : Since all the angles except are known, wecanwrite =  1 +  2 +  3 ;180  = h 180  ;tan ;1 (2=3) i + h 180  ;tan ;1 (2=1) i +tan ;1 (2=7);180  = 146:31  +135  +15:95  ;180  =117:26  1 θ 1 θ 2 -1 α -3 -a Im(s) Re(s) 2 V 3 = s + a V 2 = s + 1 V 1 = s θ 3 V 4 = s + 10 -10 Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles Then a = 3; 2 tan(180  ;117:26  ) = 3; 2 tan(62:74  ) =1:97 The gain that places closed loop poles at s = ;3j2isobtained bysolving K c jV 3 j jV 1 jjV 2 jjV 4 j =1;; (1) or K = jV 1 jjV 2 jjV 4 j jV 3 j : (2) The gain to place the poles at this location is then K = p 13 p 8 p 53 p 5:06 =33: The lead/pi compensator is then G c (s)= 33(s+2)(s+1:97) s(s +10) 2 -10 -3 -1.97 -1 2 -2 Im(s) Re(s) Figure 3: Root Locus of Compensated System Part (b) K v = lim s!0 sG c (s)G p (s)=lim s!0 s  33(s+1:97) s(s +1)((s+10)  = 331:97 10 =6:5 Then the steady state error to a ramp input is e ss = 1 K v = 1 6:5 =0:1538;; or roughly 15%. Part (c) The system has three closed loop poles, under the assumption of perfect cancellation. Weknow where twoofthethree closed loop poles are, namely at s = ;3  j3. The next task is to nd the third closed loop pole. The root locus in Figure 3 shows where that third closed loop pole must be. One wayto nd the pole is to searchalongthe real axis between s = ;10 and 3 V 4 = s + 10 V 3 = s + 1.97 V 2 = s + 1 V 1 = s Re(s) Im(s) -10 -1.97 -1 Figure 4: Calculation of Gain Along Real Axis s -8 -7.5 -7 -6.5 -6 -5.5 -5 K 18.6 22 25 27.6 29.8 31.5 33 Table 1: SearchforClosed Loop Pole s = ;1:97 until we nd a value of s that yields a gain of 33. Figure 4 shows how this calculation is made. That is K c = jV 1 jV 2 jjV 4 j jV 3 j Table 1 summarizes the searchforthe third closed loop pole. Thus, the closed loop transfer function is T c (s)= 33(s+1:97) (s +5)(s +3;j2)(s+3+j2) Part (d) The damped frequency is 2 r./s. and the damping ratio is  = cos[tan ;1 (2=3)] = cos(33:69  )=0:83 Thus, a good guess at the time response is shown in Figure 5. The formula for overshoot predicts less than 2% overshoot, but we knowfrom experience that the formula is often too optimistic, so a good guess is 10%. The time to peak should be t p =  ! d  1:5s.: 4 1.5 s. Over shoot = 10% t c(t) Figure 5: Estimated Time Response C(s) = 33(s +1:97) s(s +5)(s+3;j2)(s+3+j2) = 1 s + B s +6 + M s +3;j2 + M  s +3+j2 : B = (s +5)C(s) j s=;6 = (s +5)  33(s +1:97) s(s +3;j2)(s+3+j2)(s+6)  j s=;5 = 2:5 M = (s +3;j2)C c (s) j s=;3+j2 = (s +3;j2)  33(s +1:97) s(s +3;j2)(s+3+j2)(s+5)  j s=;3+j2 = ;1:75;0:5 = 1:82 6 ;2:8633 r. Thus, nally wehave c(t)= h 1+2:56e ;5t +3:64e ;3t cos(2t;2:8633) i 1(t) The time response is shown in Figure 6. The actual per centovershoot is under 10 per cent, and the time to peak is 0.8 s. 5 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 0.00 0.40 0.80 1.20 1.60 2.00 2.40 2.80 Time in Seconds Total Response 3.6378exp(-3t)cos(2t-2.8635) 2.4986exp(-5t) Figure 6: Step Response of Compensated System 6