Solution 8.8.3mtr4cx The full MATLAB program used to nd the transfer function is: % %load the data % load mtr4step t=mtr4step(1:500,1);; y=mtr4step(1:500,2);; plot(t,y) print -deps sr883mtr4xa.eps y=y(51:500);; t=t(51:500);; y=y-y(1);; y(1) = 0;; t(1) = 0;; plot(t,y) print -deps sr883mtr4xb.eps y(180:450) = mean(y(180:450));; y(175:179) = mean(y(175:179));; y(170:174) = mean(y(170:174));; y(165:169) = mean(y(165:169));; y(160:164) = mean(y(160:164));; y(155:159) = mean(y(155:159));; y(150:154) = mean(y(150:154));; y(145:149) = mean(y(145:149));; y(140:144) = mean(y(140:144));; y(135:139) = mean(y(135:139));; y(130:134) = mean(y(130:134));; y(125:129) = mean(y(125:129));; y(120:124) = mean(y(120:124));; y(115:119) = mean(y(115:119));; y(110:114) = mean(y(110:114));; y(105:109) = mean(y(105:109));; y(105:109) = mean(y(105:109));; y(100:104) = mean(y(100:104));; y(95:99) = mean(y(95:99));; y(90:94) = mean(y(90:94));; y(85:89) = mean(y(85:89));; 1 y(80:84) = mean(y(80:84));; y(75:79) = mean(y(75:79));; y(70:74) = mean(y(70:74));; y(65:69) = mean(y(65:69));; y(60:64) = mean(y(60:64));; plot(t,y) print -deps sr883mtr4xc.eps %zero the summer % asum = 0.0;; % %Find the length of the array y(t) Note:this the step response. It has to be %pulled into Matlab before this program is run. % top = size(y);; top = top(1,1) % %Initialize counter. % k=1;; % % %Find maximum value of step response. % K0 = max(y) % % %Form function K0 - y(t). Note y(t) is the same as yu(t) in the write up. % ftemp = K0 - y;; % fsave1 = ftemp;; % %Reset counter % k=2;; % %Integrate K0 - y(t). and form the function y1(t). % 2 while ( k <= top) l= k-1;; asum = asum + ( ( (ftemp(k)+ftemp(l))/2 ) *( t(k) - t(l) ) );; t1(k) = t(k);; f1(k) = asum;; k=k+1;; end % %Save data for plotting. % save time.dat t1 -ascii save f1.dat f1 -ascii save ftempa.dat ftemp -ascii % %SetK1 = to max(y1(t)), that is its final steady state value. % K1 = max(f1) % % %form function: K1 - y1(t). % ftemp1 = K1 - f1;; % %Save function for later plotting % save ftempb.data ftemp1 -ascii % %reset summer % asum = 0.0;; % %Reset counter % k=2;; % %Integrate K1 - y1(t) to get y2(t) (called f2 in program) % while ( k <= top) l= k-1;; asum = asum + ( ( (ftemp1(k) + ftemp1(l) ) /2 ) *( t(k) - t(l) ) );; 3 f2(k) = asum;; k=k+1;; end % %SetK2 equal to maximum value of y2(t). % K2 = asum % %Save function y2(t) for later plotting. % save f2.data f2 -ascii % %Find poles of transfer function. % a1 = K1/K0 a2 = ( a1*K1 - K2 )/K0 cpol = [1, a1/a2 , 1/a2] roots(cpol) % %Find gain of transfer function % Kplant = K0/a2 g=zpk([],[-33.6165+j*5.1687 -33.6165-j*5.1687],Kplant) T=linspace(0,0.5,402);; u=ones(402,1);; [yhat,T] = lsim(g,u,T);; plot(t,y,'k-',T,yhat,'k--') print -deps sr883mtr4xd.eps load mtr4av topm = size(mtr4av) top = topm(1,1) t=mtr4av(1:top,1);; av = mtr4av(1:top,2);; plot(t,av) print -deps 883mtr4xav.eps The MATLAB statements y(180:450) = mean(y(180:450));; y(175:179) = mean(y(175:179));; y(170:174) = mean(y(170:174));; 4 y(165:169) = mean(y(165:169));; y(160:164) = mean(y(160:164));; y(155:159) = mean(y(155:159));; y(150:154) = mean(y(150:154));; y(145:149) = mean(y(145:149));; y(140:144) = mean(y(140:144));; y(135:139) = mean(y(135:139));; y(130:134) = mean(y(130:134));; y(125:129) = mean(y(125:129));; y(120:124) = mean(y(120:124));; y(115:119) = mean(y(115:119));; y(110:114) = mean(y(110:114));; y(105:109) = mean(y(105:109));; y(105:109) = mean(y(105:109));; y(100:104) = mean(y(100:104));; y(95:99) = mean(y(95:99));; y(90:94) = mean(y(90:94));; y(85:89) = mean(y(85:89));; y(80:84) = mean(y(80:84));; y(75:79) = mean(y(75:79));; y(70:74) = mean(y(70:74));; y(65:69) = mean(y(65:69));; y(60:64) = mean(y(60:64));; plot(t,y) print -deps sr883mtr4xc.eps are a crude smoothing program designed to help the algorith converge. The smoothed response is shown in Figure 1 For this problem the alogrithm gives the answer G(s)= 81:2667 (s +2:3507)(s+9:2081) : This is unadjusted for the size of the input voltage, about 35 V, as shown inf Figure 2. So we nally arriveat G(s)= 2:319 (s +33:6165;j5:1687)(s+33:6165+ j5:1687) : The step response of the model is compared to the actual recorded step response in Figure 3 The matchisprettygood. 5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 3.5 4 Figure 1: Smoth step response data 6 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 5 10 15 20 25 30 35 40 Figure 2: Armature voltage applied to motor 7 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 0 0.5 1 1.5 2 2.5 3 3.5 4 Figure 3: Smoth step response data 8