if time is so small in series so how plot the figure
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a(1)=1;
b(1)=1;
c(1)=1;
s1=0;
s2=0;
s3=0;
k3=2.8;
k1=0.33;
k2=1;
T=0.1;
t=0:0.01:0.1;
tau=0:0.1:1
tau=zeros(length(t));
for i=1:20
a(i+1)=(1/(i+1)).*(b(i)-a(i))*k2;
b(i+1)=(1/(i+1)).*(a(i).*(k3-c(i))-b(i));
c(i+1)=(1/(i+1))*(a(i).*b(i)-k1*c(i));
s1=s1+(a(i).*tau.^i);
s2=s2+(b(i).*tau.^i);
s3=s3+(c(i).*tau.^i);
disp([s1 s2 s3])
end
tau=0:0.1:1;
so how to plot the graph between tau =t/T where range of tau will be 0 to 1 ie tau=[0 1]; so the range of t=0 to 0.1
so pl plot the graph between plot(t,s1,t,s2,t,s3)
Accepted Answer
a(1)=1;
b(1)=1;
c(1)=1;
s1=0;
s2=0;
s3=0;
k3=2.8;
k1=0.33;
k2=1;
T=0.1;
t=0:0.01:0.1;
tau = t ./ T;
%tau=zeros(length(t));
for i=1:20
a(i+1)=(1/(i+1)).*(b(i)-a(i))*k2;
b(i+1)=(1/(i+1)).*(a(i).*(k3-c(i))-b(i));
c(i+1)=(1/(i+1))*(a(i).*b(i)-k1*c(i));
s1=s1+(a(i).*tau.^i);
s2=s2+(b(i).*tau.^i);
s3=s3+(c(i).*tau.^i);
disp([s1;s2;s3])
end
plot(t, s1, t, s2, t, s3)
6 Comments
shiv gaur
on 2 Mar 2022
Edited: Walter Roberson
on 3 Mar 2022
shiv gaur
on 2 Mar 2022
Walter Roberson
on 2 Mar 2022
I have been up all night, and I am feeling ill. I will not be resolving this problem.
Steven Lord
on 2 Mar 2022
shiv gaur, looking at your question history you've been asking a lot of questions that could probably be solved if you knew how to use the debugging tools included in MATLAB. In this particular situation I recommend using an error breakpoint to pause execution when MATLAB generates a NaN or Inf value and then look at the operation you performed on that line and the data on which you performed that operation. Determine if that data matches what you expect it to be. If it does not, rerun your code with other standard, conditional, and/or error breakpoints set at earlier points to determine where the problem started.
If you haven't already taken them perhaps the MATLAB Onramp and/or MATLAB Fundamentals courses on the tutorials page would give you a good foundation of MATLAB knowledge.
Your values are exploding.
The graphs show log10 of the absolute values. (The values are alternating signs in practice.)
Notice that the values are headed up past 10^200. At i = 18 b and c both become infinite (that is, they overflow.)
format long g
a(1)=0;
b(1)=1;
c(1)=0;
s1=0;
s2=0;
s3=0;
k3=28;
k1=8/3;
k2=10;
T=0.1;
t=0:0.01:1;
tau = t ./ T;
%tau=zeros(length(t));
for i=1:20
a(i+1)=(1/(i+1)).*(b(i)-a(i))*k2;
b(i+1)=(1/(i+1)).*(a(i).*T*(k3-c(i))-b(i));
c(i+1)=(1/(i+1))*(T*a(i).*b(i)-k1*c(i));
if ~all(isfinite([a,b,c]),'all')
subplot(3,1,1); plot(abs(log10(a))); title('log10(abs(a))')
subplot(3,1,2); plot(abs(log10(b))); title('log10(abs(b))');
subplot(3,1,3); plot(abs(log10(c))); title('log10(abs(c))');
disp(a)
disp(b)
disp(c)
error('first loop broke at i = %d', i);
end
end
for i=1:20
s1=s1+(a(i).*tau.^i);
s2=s2+(b(i).*tau.^i);
s3=s3+(c(i).*tau.^i);
disp([s1;s2;s3])
if any(any(isnan([s1,s2,s3])))
error('second loop broke at i = %d', i);
end
end
plot(tau, s1, tau, s2, tau, s3)
And your s values are over 10^300 and 10^500
format long g
a(1) = sym(0);
b(1) = sym(1);
c(1) = sym(0);
s1 = sym(0);
s2 = sym(0);
s3 = sym(0);
k3=28;
k1=8/3;
k2=10;
T=0.1;
t=0:0.01:1;
tau = t ./ T;
%tau=zeros(length(t));
for i=1:20
a(i+1)=(1/sym(i+1)).*(b(i)-a(i))*k2;
b(i+1)=(1/sym(i+1)).*(a(i).*T*(k3-c(i))-b(i));
c(i+1)=(1/sym(i+1))*(T*a(i).*b(i)-k1*c(i));
if ~all(isfinite([a,b,c]),'all')
subplot(3,1,1); plot(abs(log10(a))); title('log10(abs(a))')
subplot(3,1,2); plot(abs(log10(b))); title('log10(abs(b))');
subplot(3,1,3); plot(abs(log10(c))); title('log10(abs(c))');
disp(a)
disp(b)
disp(c)
error('first loop broke at i = %d', i);
end
end
for i=1:20
s1=s1+(a(i).*tau.^i);
s2=s2+(b(i).*tau.^i);
s3=s3+(c(i).*tau.^i);
if ~all(isfinite([s1,s2,s3]),'all')
disp(s1)
disp(s2)
disp(s3)
error('second loop broke at i = %d', i);
end
end
plot(tau, s1, tau, s2, tau, s3);
title('original');
legend({'s1', 's2', 's3'});
plot(tau, log10(abs(s1)), tau, log10(abs(s2)), tau, log10(abs(s3)));
title('log10(abs(s))');
legend({'s1', 's2', 's3'});
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