Why am I getting imaginary values with ode45 ??

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Sai charan Bandi
Sai charan Bandi 2014-12-29
编辑： Shoaibur Rahman 2014-12-31
%This is my program R=1;
L=10^-2;
C=400/62505000010;
Is=10^-12;
Vdo=25*10^-3;
f=@(t,y) [(-R*y(1)/L-(y(2)/(C*L)-(Vdo/L)*log(1+(y(1)/Is))));y(1)];
close all;
tic
options = odeset('RelTol',10^-12,'AbsTol',10^-8);
[t,y]=ode45(f,0:10^-3:0.01,[10^-6 -10^-6],options);
toc
plot(t,y)
xlabel('time "t"')
ylabel('output "y"')
figure(2)
plot(y(:,2),y(:,1),'r')
xlabel('current I');
ylabel('charge Q')
I am simulating a simple circuit, which is governed by two differential equations
differential (Q)=I;
differential(I)=(-R/L)*I-(Vdo/L)*log(1+(I/Is))-Q/(C*L);
I,Q are my variables. I am getting the simulated output but the problem is I am getting imaginary values
for y.
As the only possible cause could be 'log' term in the second differential equation,
from that 1+I/Is>0 ==> I>-Is 'to get real values'
so I used odeset to put obsolute tolerence limit as 10^-8.
But I am still getting imaginary values for 'y'!! Why??
Please let me know what I am doing wrong !!

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回答（3 个）

Shoaibur Rahman
Shoaibur Rahman 2014-12-29
Your I has both negative and positive values, which are in the range of 10^(-7), but Is is in 10^(-12). So, I/Is gives high negatives for some negative values of I. In that case, log is getting a negative input, and thus resulting imaginary parts.
I is the current, and negative indicates the direction. So, when using inside log, you can use both I and Is in same direction, hence same signs. This is also equivalent to making I as abs(I) inside the log.
Sai charan Bandi
Sai charan Bandi 2014-12-29
Thank You for replying Shoaibur Rahman,
My question how you are saying I is taking negative values also ?? Could you elaborate a little please!! and using abs(I) is the option that I know but I want I to take values > -Is with out using any absolute values just by changing the options in odeset
and
'AbsTol' is not taking negative values, even then when I am using AbsTol as 10^-8 how can I take negative values ??
as AbsTol > - Is i.e 10^-8>-10^-12
maybe I am missing some understanding of ode45 here! please explain me if you can understand my question !!
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Jan
Jan 2014-12-30
@Sai charan Bandi: Please use the section for comments and not the one for answers, if you post a comment. Thanks.
Sai charan Bandi
Sai charan Bandi 2014-12-30
Sorry! That's my bad to use answer rather than comment section. @Jan Simon
@Shoaibur Rahman I am using a simple series circuit with R,L,C and one diode without any voltage source.Anyway thanks for the reply.
I could not understand how to control ode45 from not taking any imaginary values.

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Jan
Jan 2014-12-30
If you create a function instead of using an anonymous function, you can add a short test for imaginary values:
function dy = fun(t,y)
dy = [(-R*y(1)/L-(y(2)/(C*L)-(Vdo/L)*log(1+(y(1)/Is)))); ...
y(1)];
if any(~isreal(dy))
keyboard
end
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Sai charan Bandi
Sai charan Bandi 2014-12-31
编辑：Sai charan Bandi 2014-12-31
I tried using break points. If I put break point on ode45 line the next step matlab will perform the entire ode calculation and goes to next step. I am asking, is it possible to see , updation of values for each single time-step ??
I cannot tell ode45 to ignore imaginary values , then can I plot the ode output values with imaginary values included . Why my plot function is neglecting the imaginary values of output.??
And finally in my problem I am taking two parameters Q ,I as ode outputs.In One column, I am getting imaginary values and for other column all zeros,which is wrong!! That is why I want to see how ode executing output values in each time step. Is it possible??
-Thanks for the reply
Shoaibur Rahman
Shoaibur Rahman 2014-12-31
编辑：Shoaibur Rahman 2014-12-31
You can have a look at the ode45 function itself to see how the function is written. Type edit ode45 on your command window, and see that. It is a complex algorithm though.
Regarding the ignoring of imaginary values, the plot function ignores those, not the ode45. This is the way how plot function works. If you are interested to plot the imaginary part then use imag function. However, in RLC circuit analysis, the most effective way will be to see the magnitude and phase, which can be computed using abs and angle functions, respectively.
Two columns in output have some values; none of them is entirely zero, but one is very small comparing to another. This is not surprising since, I = dQ/dt, your first equation. So, the relation between I and Q vastly depends on the time step.

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