how can i find the intersection point between the two curves and the minimum point to the other curve?

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mohamed asran 2019-10-26

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回答： mohamed asran 2020-11-9

k1=10^-4;k2=2*10^5
k2 = 200000
d=0.08;
a=50*10^-6:1*10^-6:100*10^-6;
cr    =k1./a;
cf   =k2*d.*a;
ctot   =cr+cf;
plot(a,cr,a,cf,a,ctot)
 title('optimal')
xlabel('cross section area')
ylabel('costs')
legend('cr','ctot','cf')

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Answer 1

Image Analyst 2019-10-26

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在 MATLAB Online 中打开

Do you want the (harder) analytical answer (like from the formula) or the (easier) digital answer from the digitized vectors, like

distances = abs(cf-cr)
[minDistance, indexAtMin] = min(distances);
y1AtMin = cf(indexAtMin)
y2AtMin = cr(indexAtMin)
aAtMin = a(indexAtMin)
hold on;
line([aAtMin, aAtMin], ylim);

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mohamed asran 2019-10-26

no i'm just starting to use matlab

Image Analyst 2019-10-26

MATLAB is NOT needed for an analytical solution. Haven't you taken a calculus course yet?

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Answer 2

mohamed asran 2020-11-9

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在 MATLAB Online 中打开

clc
clear all
r=0.05;
l=0.01;
st=0.0001;
v=220;
Kf=18;
j=3;
Tl=60;
i=0;
w=0;
I=[];
W=[];
t=[];
for dt=0:0.0001:1
    I=[I i];
    t=[t dt];
    W=[W w];
    i=i+(((v-r*i)-(Kf*w)/l)*st);
     w=w+((((Kf*i)-Tl)/j)*st);
end
plot(t,W,'linewidth',4)
xlabel('time (sec)','fontsize','18','fontweight','b');
ylabel('SPEED (rpm)','fontsize','22','fontweight','b');
title('Dynamic model of separately excited dc motor under constant excitation');
axis([0 0.1 0.5])
gri;d
plot(t,I,'linewidth',4)
xlabel('time (sec)','fontsize','18','fontweight','b');
ylabel('current (A)','fontsize','22','fontweight','b');
title('current response of rl circuit');
axis([0 0.1 0.5])