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CVR

Same Function but Different Result (Basic Function)

Asked by CVR
on 30 Sep 2013
Latest activity Commented on by CVR
on 1 Oct 2013

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 Simon
on 30 Sep 2013

I get the error message "??? Undefined function or variable 'time'." for the second version. Do you mean "t" instead?

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1 Answer

Answer by Jan Simon
on 30 Sep 2013
Edited by Jan Simon
on 30 Sep 2013
 Accepted answer

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

CVR
on 1 Oct 2013

Yes you are right. I would prefer the first version now. The reason is because I need a function handle, not sure if a matlab function could be converted into a function handle

Jan Simon
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.

CVR
on 1 Oct 2013

wow, great.. It works Thx


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