Solution 8.8.8.a The full MATLAB program used to nd the transfer function is: % %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. % load p2zid1 topm = size(p2zid1) top = topm(1,1) t=p2zid1(1:top,1);; y=p2zid1(1:top,2);; % % asum = 0.0;; plot(t,y) print -deps sr888a.a.ep t=t+0.004 y=y(78:top);; t=t(78:top);; t(1)=0;; y(1) = 0 plot(t,y) print -deps sr888a.b.eps topm = size(y) top = topm(1,1) % % k0 = y(top) % f0 = k0-y;; % %Reset counter % k=2;; % %Integrate f0 = k0 - y(t). and form the function y1(t). % while ( k <= top) 1 l=k-1;; asum = asum + ( ( (f0(k)+f0(l))/2 ) *( t(k) - t(l) ) );; y1(k) = asum;; k=k+1;; end % %Save data for plotting. % save time.dat t -ascii save y.dat y -ascii save y1.dat y1 -ascii save f0.dat f0 -ascii % %SetK1 = to max(y1(t)), that is its final steady state value. % k1 = asum % %form function:f1 = k1 - y1(t). % f1 = k1-y1;; % %Save function for later plotting % save f1.dat f1 -ascii % %reset summer % asum = 0.0;; % %Reset counter % k=2;; % %Integrate f1(k) = k1 - y1(t) to get y2(t) % while ( k <= top) l= k-1;; asum = asum + ( ( (f1(k)+f1(l))/2 ) *( t(k) - t(l) ) );; y2(k) = asum;; k=k+1;; 2 end % %Setk2 equal to maximum value of y2(t). % k2 = asum % %Save function y2(t) for later plotting. % save y2.data y2 -ascii % % %form function: f2 = k2 - y2(t). % f2 = k2-y2;; % %Save function for later plotting % save f2.data f2 -ascii % %reset summer % asum = 0.0;; % %Reset counter % k=2;; % %Integrate f2 = k2 - y2(t) to get y3(t) % while ( k <= top) l= k-1;; asum = asum + ( ( (f2(k)+f2(l))/2 ) *( t(k) - t(l) ) );; y3(k) = asum;; k=k+1;; end % %SetK3 equal to maximum value of y3(t). % k3 = asum A=[k0,0,-1;;k1,-k0,0;;k2,-k1,0] 3 b=[k1;;k2;;k3] x=inv(A)*b a1 = x(1) a2 = x(2) b1 = x(3) polyd=[a2 a1 1] p=roots(polyd) Kplant = b1/a2 z=-k0/b1 T=linspace(0,1,200);; g=zpk([z],[p(1) p(2)],Kplant) [yhat,T] = step(g,T);; plot(t,y,'k-',T,yhat,'k--') print -deps sr888a.c.eps The MATLAB statements load p2zid1 topm = size(p2zid1) top = topm(1,1) t=p2zid1(1:top,1);; y=p2zid1(1:top,2);; % % asum = 0.0;; plot(t,y) print -deps sr888a.a.ep t=t+0.004 y=y(78:top);; t=t(78:top);; t(1)=0;; y(1) = 0 plot(t,y) print -deps sr888a.b.eps topm = size(y) top = topm(1,1) adjust the data so that the response begins at zero for t =0. The rest of the program is the integration routine. The rawdataisshown in Figure 1 The adjusted data is shown in Figure 2 4 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 0.5 1 1.5 2 2.5 3 Figure 1: Unadjusted step response 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.5 1 1.5 2 2.5 3 Figure 2: Adjusted step response 5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.5 1 1.5 2 2.5 3 Figure 3: Comparison of measured response to model response For this problem the alogrithm gives the answer G(s)= 78:6152(s+1:718) (s +15:33)(s+9:274) : The step response of the model is compared to the actual recorded step response in Figure 3 The matchisprettygood. 6