hey guys I'm trying to do an if statement inside an embedded matlab function in simulink, but it is not working, I don't know why.
------------------------------------------------
This is the matlab function
function [Wv_membr,lambda_m] = fcn(Ist,T_st,phi_ca,phi_an)
Mv=18.02E-3;
n=381;
Afc=280;
F=96485;
tm=0.01275;
%phi_an=0.5;
%phi_ca=0.80;
am=(phi_ca+phi_an)/2;
if (am>0) && (am<=1)
lambda_m=0.043+17.81*am-39.85*am^2+36*am^3;
elseif (am>1) && (am<3)
lambda_m=14+1.4*(am-1);
end
i=Ist/Afc;
if (lambda_m<2)
Dy=1E-6;
elseif (lambda_m>=2) && (lambda_m<=3)
Dy=1E-6*(1+2*(lambda_m-2));
elseif (lambda_m>3) && (lambda_m<4.5)
Dy=1E-6*(3-1.67*(lambda_m-3));
elseif (lambda_m>=4.5)
Dy=1.25E-6;
end
if (phi_an>0) && (phi_an<=1)
lambda_an=0.043+17.81*phi_an-39.85*phi_an^2+36*phi_an^3;
elseif (phi_an>1) && (phi_an<3)
lambda_an=14+1.4*(phi_an-1);
end
if (phi_ca>0) && (phi_ca<=1)
lambda_ca=0.043+17.81*phi_ca-39.85*phi_ca^2+36*phi_ca^3;
elseif (phi_ca>1) && (phi_ca<3)
lambda_ca=14+1.4*(phi_ca-1);
end
pm_dry=0.002;
Mm_dry=1.1;
cv_an=pm_dry/Mm_dry * lambda_an;
cv_ca=pm_dry/Mm_dry * lambda_ca;
Dw=Dy*exp(2416*(1/303-1/T_st));
nd=0.0029*lambda_m^2+0.05*lambda_m-3.4E-19;
Nv_membr=nd*i/F * Dw*(cv_ca-cv_an)/tm;
Wv_membr=Nv_membr*Mv*Afc*n;
end
---------------------------------------------------
when I run it, it says that lamda_m is not a defined variable, but this is impossible. It is very weird to me. I wonder for some help.

1 commentaire

bym
bym le 20 Déc 2011
please format your code to make it readable

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 Réponse acceptée

Jan
Jan le 20 Déc 2011

0 votes

When I uncomment the lines as you mentioned, I still do not get an error:
phi_an=0.5;
phi_ca=0.80;
am=(phi_ca+phi_an)/2;
if (am>0) && (am<=1)
lambda_m=0.043+17.81*am-39.85*am^2+36*am^3;
elseif (am>1) && (am<3)
lambda_m=14+1.4*(am-1);
end
if (lambda_m<2)
Dy=1E-6;
elseif (lambda_m>=2) && (lambda_m<=3)
Dy=1E-6*(1+2*(lambda_m-2));
elseif (lambda_m>3) && (lambda_m<4.5)
Dy=1E-6*(3-1.67*(lambda_m-3));
elseif (lambda_m>=4.5)
Dy=1.25E-6;
end
All is fine.

7 commentaires

C.J. Harris
C.J. Harris le 20 Déc 2011
Keep in mind this is an embedded matlab function within Simulink. Different rules apply, hence all is not fine.
Danilo NASCIMENTO
Danilo NASCIMENTO le 20 Déc 2011
So I don't know what is going on. I made attempts into differents computers with different matlabs. When I do the same you did, it gives me this message "Variable 'Dy' is undefined on some execution paths."
It can't happen because it is independent of the inputs of the function.
C.J. Harris
C.J. Harris le 20 Déc 2011
See my answer above. 'am' is dependent on inputs phi_ca and phi_an. 'lambda_m' is dependent on 'am', and 'Dy' is depenent on 'lambda_m'. Therefore it is undefined on some execution paths, and this is no allowed for embedded matlab.
You must give this variable a default value, or add 'else' conditions to your 'if-else if' statements.
Danilo NASCIMENTO
Danilo NASCIMENTO le 20 Déc 2011
I cannot change if statement. What I can do is set an initial value to the variables 'phi_ca' and 'phi_an. But I don't know if it is correct the way I did above, just setting it within the function. Because at every iteration of simulink it must update this values. And doing this within the function I presuppose that 'phi_ca' and 'phi_an' will always stay with the initial values.
C.J. Harris
C.J. Harris le 20 Déc 2011
No, you can't set your inputs to have default values, only the internal parameters (such as Dy) and outputs (such as lambda_m).
As stated, please initialise all your outputs and parameters to default values.
Danilo NASCIMENTO
Danilo NASCIMENTO le 20 Déc 2011
Ok. Now it is working man. Thanks.
C.J. Harris
C.J. Harris le 20 Déc 2011
No problem. Mark answer as accepted.

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Plus de réponses (3)

C.J. Harris
C.J. Harris le 20 Déc 2011

1 vote

Within an embedded matlab function you have to ensure all variables are set independent of execution path through the code.
While you might be able to assure us the values of 'lambda_m' and 'Dy' are always within the specified range, Matlab cannot assure this, as the values of 'lambda_m' and 'Dy' are dependent on the function inputs.
In order to fix this error just assign 'lambda_m' and 'Dy' a value at the top of your code, and if they are within the specified ranges defined by your 'if-else if' statements these 'initial' values will just be overwritten.

1 commentaire

Jan
Jan le 20 Déc 2011
As far as I understand, this answer solves the problem. And in addition initializing all used variables is a good programming habit.

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Walter Roberson
Walter Roberson le 20 Déc 2011

0 votes

Consider your code
if (am>0) && (am<=1)
lambda_m=0.043+17.81*am-39.85*am^2+36*am^3;
elseif (am>1) && (am<3)
lambda_m=14+1.4*(am-1);
end
What happens in your code if am is not in the range 0 < am < 3 ? If, that is, am <= 0 or am >= 3 ?

2 commentaires

Danilo NASCIMENTO
Danilo NASCIMENTO le 20 Déc 2011
No man.. I assure to you that am is in this interval. You can, for example, take off % of lines 12 and 13 and you'll see that there is another mistake. Matlab will give you that variable Dy is not defined. But this is also impossible.
Jan
Jan le 20 Déc 2011
@Danilo: If "am" is not in the expected ranges, the error message you have posted will appear. Without doubtm there can be a further problem also. But Wlater's correct statement is affect by this.

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arief hidayat
arief hidayat le 24 Nov 2017
Modifié(e) : per isakson le 24 Nov 2017

0 votes

Hi anyone help me for my script, when i running script, i got error "not enough statement" script :
if ((rate ~= rate_tx || (Nbpsc ~= Nbpsc_tx) || (psdu_byte ~= psdu_byte_tx)))
Percounter = 1;
noviterbi_Y = [];
PSDU = [];
return ;
else
Percounter = 0;

1 commentaire

Jan
Jan le 24 Nov 2017
Modifié(e) : Jan le 24 Nov 2017
Do not hijack an existing thread by inserting a new question in the section for answers. Open a new thread instead and delete this one. Otherwise it is confusing, to which question an answer belong and you cannot accept the answer, which solves your problem. Thanks.
Include the complete error message, not just a part of it. Format your code using the "{} Code" button to make it readable.

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