Same Function but Different Result (Basic Function)

1 view (last 30 days)
Hi all, can anybody help me ? I have 2 same functions here, but when I evaluate them, I got a slightly different result. Perhaps someone could help me to check it ? Thank you
The first version is
t = [4:1:14];
A0 = 23
rho = 1.2
Ton = 1.2
Toff = 6
r = 0.6
del_a = 3.8
del_m = 0.1
p = t>=Ton;
q = t <Toff;
s = t>=Toff;
A = (A0*exp(rho.*(t - Ton)).*p.*q) + (A0*exp(rho*(Toff - Ton))*exp(-(r+del_a).*(t - Toff))).*s;
M = ((r/(r+del_a - del_m))*(A0*exp(rho*(Toff - Ton))*exp(-del_m.*(t-Toff)) - A)).*s;
y = A+M
plot(t,y)
and the second version is :
t = [4:1:14];
Par(1) = 23
Par(2) = 1.2
Par(3) = 1.2
Par(4) = 6
Par(5) = 0.6
Par(6) = 3.8
Par(7) = 0.1
model = @(Par,t) ( ( Par(1)*exp(Par(2).*(t - Par(3))).*(t>=Par(3)).*(t < Par(4)) )...
+ ( Par(1)*exp(Par(4) - Par(3))*exp(-(Par(5)+Par(6)).*(t - Par(4))).*(t >= Par(:,4)))...
+ ( (Par(5)/(Par(5) + Par(6) - Par(7))) *( (Par(1)*exp(Par(2)*(Par(4) - Par(3))))* ...
exp(-Par(7).*(t - Par(4))) - (( Par(1)*exp(Par(2).*(t - Par(3))).*(t>=Par(3)).*(t < Par(4)) )...
+ ( Par(1)*exp(Par(4) - Par(3))*exp(-(Par(5)+Par(6)).*(t - Par(4))).*(t >= Par(4)))...
) ) ).*(t>=Par(4)) ...
)
y = model(Par,time);
plot(t,y)
the problem is I have a slightly different result, but both of them are a same function. Can anybody help ? What I want to do is to convert my function in the first version into a function handle like in the second version
Thanks very much
  1 Comment
Jan
Jan on 30 Sep 2013
Edited: Jan on 30 Sep 2013
I get the error message "??? Undefined function or variable 'time'." for the second version. Do you mean "t" instead?

Sign in to comment.

Accepted Answer

Jan
Jan on 30 Sep 2013
Edited: Jan on 30 Sep 2013
If I replace "time" by "t", I get this:
y1 = A+M;
y2 = model(Par, t);
y1 - y2
% >> 0, 0, 3875.769e, 47.58413, 0.5842066, 7.172503e-3, 8.805925e-5, 1.081133e-6, 1.327351e-8, 1.629701e-10, 1.989520e-12
This seems to show, that the functions are different.
The 2nd version is very hard to read. I'd really avoid such ugly code, because, as you already see, it is nearly impossible to debug it. Do you have any good reasons to prefer this kind of code?
  3 Comments
Jan
Jan on 1 Oct 2013
You can simply write the code into a function and create a function handle either by setting "@" before the name of the function, or by using str2func.

Sign in to comment.

More Answers (0)

Categories

Find more on Interactive Control and Callbacks in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!