How to plot a matrix with loop of input Variables
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Muhammed Ajmal Pallithadathil Nazer
am 22 Nov. 2017
Bearbeitet: Joy Mondal
am 23 Nov. 2017
I want to run the below code from the value of Bx 0 to 0.0118 with an interval of .00092 and plot it in e vs Bx, where e is a 4*1 matrix. I found some problem while plotting and I am very poor in working with loops.
Bx=0:.00092:0.0115;
B=0.5*1.978*0.467*30*Bx;
D=-7.31;
E=-1.71;
Bz=0.5*1.969*0.467*30*0;
A=[2*D+3*Bz,sqrt(3)*B,sqrt(3)*E,0;sqrt(3)*B,Bz,2*B,sqrt(3)*E;sqrt(3)*E,2*B,-Bz,sqrt(3)*B;0,sqrt(3)*E,sqrt(3)*B,2*D-3*Bz]
e=eig(A)
D=eig(A,'matrix')
plot(Bx,e)
Here I wrote the code straight away and may not be correct,
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Antworten (4)
Walter Roberson
am 22 Nov. 2017
syms Bx
B=0.5*1.978*0.467*30*Bx;
D=-7.31;
E=-1.71;
Bz=0.5*1.969*0.467*30*0;
A=[2*D+3*Bz,sqrt(3)*B,sqrt(3)*E,0;sqrt(3)*B,Bz,2*B,sqrt(3)*E;sqrt(3)*E,2*B,-Bz,sqrt(3)*B;0,sqrt(3)*E,sqrt(3)*B,2*D-3*Bz]
e=eig(A)
fplot(e, [0, 0.0115])
1 Kommentar
Joy Mondal
am 22 Nov. 2017
- clear all
- clc
- for Bx=0:.00092:0.0115
- i=0;
- B=0.5*1.978*0.467*30*Bx;
- D=-7.31;
- E=-1.71;
- Bz=0.5*1.969*0.467*30*0;
- A=[2*D+3*Bz,sqrt(3)*B,sqrt(3)*E,0;
- sqrt(3)*B,Bz,2*B,sqrt(3)*E;
- sqrt(3)*E,2*B,-Bz,sqrt(3)*B;
- 0,sqrt(3)*E,sqrt(3)*B,2*D-3*Bz];
- e=eig(A);
- j=0;
- for i=1:4
- j=j+1;
- ei=e(i,1);
- plot(Bx,ei,'ok')
- hold on
- end
- end
2 Kommentare
Stephen23
am 22 Nov. 2017
@Joy Mondal: please format your code correctly. It is very simple: first select the code text, then click the {} Code button above the textbox.
Joy Mondal
am 23 Nov. 2017
Bearbeitet: Joy Mondal
am 23 Nov. 2017
clc
clear all
for Bx=0:.00092:0.0115
i=0;
B=0.5*1.978*0.467*30*Bx;
D=-7.31;
E=-1.71;
Bz=0.5*1.969*0.467*30*0;
A=[2*D+3*Bz,sqrt(3)*B,sqrt(3)*E,0;
sqrt(3)*B,Bz,2*B,sqrt(3)*E;
sqrt(3)*E,2*B,-Bz,sqrt(3)*B;
0,sqrt(3)*E,sqrt(3)*B,2*D-3*Bz];
e=eig(A);
j=0;
for i=1:4
j=j+1;
ei=e(i,1);
if j==1
plot(Bx,ei,'dk')
elseif j==2
plot(Bx,ei,'og')
elseif j==3
plot(Bx,ei,'sb')
elseif j==4
plot(Bx,ei,'xm')
end
hold on
end
end
end
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