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Why do i get a blank graph?

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jpmontoya
jpmontoya on 1 Jun 2019
Commented: jpmontoya on 2 Jun 2019
Hello, as you can see in the code below I am trying to make two plots but the problem is that the subplot (2,1,2) goes blank. I got nothing. I think maybe there is something wrong with the variables k1 and v1 since they depend on X(1) and X(3) which come from the solution of an ODE. Can somebody help me? Thanks in advance.
function [T,X] = call_osc()
tspan = [0 2000];
x1_0=0.05;
x2_0=0.05;
x3_0=298.15;
x4_0=0.1;
odeset('RelTol', 1e-10, 'AbsTol', 1e-12);
[T,X]=ode15s(@osc,tspan,[x1_0 x2_0 x3_0 x4_0]);
R=8.314;
Ea1=80000;
alpha=11;
A=1;
k1= 0.8*exp(-Ea1/(alpha*R*X(3)));
v1=k1*A*X(1);
subplot(2,1,1)
plot(T,X(:,3))
grid on
xlabel('tiempo(s)','Fontsize',13,'FontWeight','Bold')
ylabel('Temperatura (K)','Fontsize',13,'FontWeight','Bold')
subplot(2,1,2)
plot(X(:,1),v1)
grid on
end

  1 Comment

Walter Roberson
Walter Roberson on 2 Jun 2019
It would help to have your osc function

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Accepted Answer

gonzalo Mier
gonzalo Mier on 2 Jun 2019
Edited: gonzalo Mier on 2 Jun 2019
You are computing v1 as
k1= 0.8*exp(-Ea1/(alpha*R*X(3)));
v1=k1*A*X(1);
v1 is a matriz of size 1x1. So when you do plot( "vector of size 1xn" , "cte" ) the output is an empty plot.
Check if the way to compute v1 is correct or you have to use X(:,1).
PD: You also can check this doing:
plot(X(:,1),v1,'*')

  3 Comments

jpmontoya
jpmontoya on 2 Jun 2019
Gonzalo, I checked by doing:
plot(X(:,1),v1
And I got something like this:
graphv1.jpg
As you can see all of the values of v1 are zero but they shouldn't be.
Walter Roberson
Walter Roberson on 2 Jun 2019
tspan = [0 2000];
[T,X]=ode15s(@osc,tspan,[x1_0 x2_0 x3_0 x4_0]);
X will be output as something with an unpredictable number of rows, and 4 columns.
k1= 0.8*exp(-Ea1/(alpha*R*X(3)));
v1=k1*A*X(1);
X is a 2D array. X(1) and X(3) are linear indexing, so X(3) is X(3,1) ans X(1) is X(1,1) . X(1,1) should be the same as x1_0 but X(3,1) will be the x1 output at the third time point, whatever the third time point happens to be.
jpmontoya
jpmontoya on 2 Jun 2019
Thank you so much. I did a lot of things considering the answers of Gonzalo and Walter and finally everything works perfect.

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