Solution 4.6.5.3 (a) The closed loop transfer function is T c (s) = K c s +2 1+ K c s +2 = K c s +(K c +2) Thus, to place the closed loop pole as s = ;5wesimply havetochoose K c =3: Then the closed loop system is T c (s)= K c s +(K c +2) : The unit step response is generated bytheMATLAB dialogue EDU>tc = zpk([],[-5],3) Zero/pole/gain: 3 ----- (s+5) EDU>step(tc) EDU>print -deps sr4653a.eps EDU> The step response is shown in Figure 1. As can be seen the response is ne except that it only reaches a nal value of 0.6 or 60% of the requested nal value. (b) he closed loop transfer function is T c (s) = K c (s +2:1) s(s +2) 1+ K c (s +1) s(s +2) = K c (s +2:1) s 2 +(2+K c )s +2:1K c 1 Time (sec.) A mp li tu d e Step Response 0 0.2 0.4 0.6 0.8 1 1.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Figure 1: Step Response for K =3 2 Wewishthe characteristic equation to be p(s)=(s +5)(s+ q)=s 2 +(5+q)s +5q: But we also knowthatthe characteristic equation is p(s)=s 2 +(2+K c )s +2:1K c : Equating the coecients of likepowers of s weobtain the two equations 5+q = 2+K c 5q = 2:1K c From the second equation we obtain q = 2:1K c 5 : subsituting this value of q into the rst equation weobtain 5+ 2:1K c 5 =2+K c ;; or 25+ 2:1K c =10+5K c or K c = 15 2:9 : Then the closed loop system is T c (s) = (15=2:9)(s+2:1) s 2 +(2+15=2:9)s+2:1(15=2:9) = 5:1724(s+2:1) (s +2:1724)(s+5) : The unit step response is generated bytheMATLAB dialogue EDU>tc = zpk([-2.1],[-2.1724 -5],5.1724) Zero/pole/gain: 5.1724 (s+2.1) --------------- (s+2.172) (s+5) EDU>step(tc) EDU>print -deps sr4653b.eps EDU> The step response is shown in Figure 2. 3 Time (sec.) A mp li tu d e Step Response 0 0.2 0.4 0.6 0.8 1 1.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step Response for K =5:1724 4