Solution 8.8.7 The MATLAB progam load mtr6step size(mtr6step) t=mtr6step(1:500,1);; y=mtr6step(1:500,2);; plot(t,y) print -deps sr881a.eps y34 =y(34) y35 =y(35) t34 =t(34) t=t-t(34);; y(34) = 0;; t=t(34:500);; y=y(34:500);; plot(t,y) print -deps sr881b.eps maxy = max(y) y1 = 3.2566 -y;; w=log(y1);; plot(t,w);; s=-(1.3 -(-0.52))/0.073 print -deps 881lny1.eps Bv = (y - max(y))./exp(s*t);; plot(t(100:150),Bv(100:150)) print -deps 881B.eps B=mean(Bv(100:150)) y2 = y-(max(y) +B*exp(s*t));; plot(t,log(y2)) print -deps 881ln2.eps s1 = -(2.5)/0.0225 C=-(max(y) + B) yhat = max(y) + B*exp(s*t) + C*exp(s1*t);; plot(t,y,'k-',t,yhat,'k--') print -deps sr881c.eps K=max(y)*abs(s)*abs(s1) load mtr6av tav = mtr6av(1:500,1);; 1 av = mtr6av(1:500,2);; plot(tav,av) print -deps 881av.eps Was written to solveproblem 8.8.1. The statements load mtr6step size(mtr6step) t=mtr6step(1:500,1);; y=mtr6step(1:500,2);; plot(t,y) print -deps sr881a.eps y34 =y(34) y35 =y(35) t34 =t(34) t=t-t(34);; y(34) = 0;; t=t(34:500);; y=y(34:500);; plot(t,y) print -deps sr881b.eps s=-(1.3 -(-0.52))/0.073 maxy = max(y) y1 = 3.2566 -y;; w=log(y1);; plot(t,w);; print -deps 881lny1.eps laod the data, adjust the data so the graph begins at y(t)=0fort =0, nds the function y 1 (t)=y ss ;y(t);; and plotlny 1 (t)versus t. The slope of this line is the location of the pole with the largest time constant. The part of the code that weareinterested in comes immediately after this. Once wehavefound the location of the rst pole the wecan nd B  y(t);A e ;p 1 t : As discussed in the chapter, B should become relatively constantinthe time period just before wereachsteady state. The MATLAB statements 2 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0 0.5 1 1.5 2 2.5 3 3.5 Region for evaluating B Figure 1: Root locus s=-(1.3 -(-0.52))/0.073 print -deps 881lny1.eps Bv = (y - max(y))./exp(s*t);; plot(t(100:150),Bv(100:150)) print -deps 881B.eps B=mean(Bv(100:150)) evaluate B over a range of time values. Figure 1 Shows the adjusted time response, and the approximate region for evaluating B. The statement Bv = (y - max(y))./exp(s*t);; cancluates a vector of values of B using the ./ whichallows us to divide elements in one vector bythose in another on an entry byentry basis. The statement 3 0.035 0.04 0.045 0.05 0.055 0.06 -4.1 -4.05 -4 -3.95 -3.9 -3.85 -3.8 -3.75 -3.7 -3.65 -3.6 Figure 2: Plot of B for 0:04 <t<0:06 plot(t(100:150),Bv(100:150)) Then plots B versus t for 0:04 <t<0:06 s. The plot is shown in Figure 2. The plot is jumping all over the place, because of sensor noise, but wecan use the statement B=mean(Bv(100:150)) to nd the mean value. We use this mean value for B in problem 8.8.1. 4