Problem: 7.9.9.7 It is a trivial matter to solve this problem with a second order compensator. The compensator G c (s)= 2:5(s0:1)(s+5) s(s +10) leads to the root locus shown in Figure 1 and the step response shown in Figure 2. If this design has a drawbackitisthat it settles rather slowly, -15 -10 -5 0 5 -8 -6 -4 -2 0 2 4 6 8 Real Axis Imag Axis Figure 1: Root locus for second order compensator although it is within 5% of its nal value in less than twoseconds. A somewhat more challenging problem is to meet the speci cations using a rst order lead. Suppose wecancelthe plantpole at s = ;5withazero. Then the problem of nding the pole of the lead compensator is that shown in Figure 3. We pickadamping ratio  and a natural frequency ! n . Then we knowthat ! c = ! n q 1; 2  = ! n : We can then show that to satisfy the angle condition at the point s = (;;;! d ), wemust have  2 =tan ;1 (! d =);tan ;1 [! d =(15;]: 1 Time (sec.) Amplitude Step Response 0 5 10 15 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Figure 2: Step response for second order compensator, Then the pole of the lead compensator is located at s = ;b with b =  + ! d tan( 2 ) : The gain to place the poles at the desired location is K c = q ! 2 d +  2  q ! 2 d +(b;) 2  q ! 2 d +(15;) 2 200 : Finally,wehave K v =lim s!0 sG c G p (s)= 200K c 15b : Wearenowinaposition to write a short MATLAB program a=15 zeta = 0.6 wn = 3 knt = 0 while wn < 10 knt = knt + 1;; 2 XX X θ 1 θ 2 θ 3 Re(s) Im(s) σ? ω d -15 -b Figure 3: Satisfying Angle Condition Along LIne of ConstantDamping sigma = wn * zeta;; wd = wn * sqrt(1 - zeta^2);; wdk(knt) = wd;; theta2 = atan(wd/sigma) - atan(wd/(a-sigma));; x=wd/tan(theta2);; b(knt) = sigma + x;; gain(knt) = sqrt(sigma^2 + wd^2)*sqrt(wd^2 + (a- sigma)^2)*sqrt(wd^2 + x^2);; kv(knt) = gain(knt)/(a * b(knt));; wn = wn+ 0.1;; end plot(wdk,kv) save kv6.dat kv /ascii save wdk6.dat wdk /ascii that compares K v to ! d for three di erentdamping ratios as shown in Fig- ure 4 If wechoose  =0:5andK v =5:1, then the compensator is G c (s)= 4:77(s+5) s +12:44 If wechoose  =0:6, G c (s)= 7:49(s+5) s +19:53 The step responses for the twocompensators are shown in Figure 5. These designs havealittle more overshoot than the second order compensator, but somewhat better settling characteristics. 3 1 2 3 4 5 6 7 8 246810 ζ = 0.5 ζ = 0.6 ζ = 0.7 ω d K v Figure 4: K v versus  0 0.5 1 1.5 Step Response 0 0.5 1 1.5 2 Time in Seconds ζ = 0.5 ζ = 0.6 Figure 5: Step Responses for  =0:5 and  =0:6 4