if statement

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Danilo NASCIMENTO
Danilo NASCIMENTO 2011-12-20
编辑： Jan 2017-11-24
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 个评论
bym
bym 2011-12-20
please format your code to make it readable

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采纳的回答

Jan
Jan 2011-12-20
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 个评论
显示 5更早的评论
Danilo NASCIMENTO
Danilo NASCIMENTO 2011-12-20
Ok. Now it is working man. Thanks.
C.J. Harris
C.J. Harris 2011-12-20
No problem. Mark answer as accepted.

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更多回答（3 个）

C.J. Harris
C.J. Harris 2011-12-20
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 个评论
Jan
Jan 2011-12-20
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 2011-12-20
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 个评论
Danilo NASCIMENTO
Danilo NASCIMENTO 2011-12-20
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 2011-12-20
@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 2017-11-24
编辑：per isakson 2017-11-24
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 个评论
Jan
Jan 2017-11-24
编辑：Jan 2017-11-24
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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