if statement

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Danilo NASCIMENTO
Danilo NASCIMENTO am 20 Dez. 2011
Bearbeitet: Jan am 24 Nov. 2017
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 Kommentar
bym
bym am 20 Dez. 2011
please format your code to make it readable

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Akzeptierte Antwort

Jan
Jan am 20 Dez. 2011
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 Kommentare
Danilo NASCIMENTO
Danilo NASCIMENTO am 20 Dez. 2011
Ok. Now it is working man. Thanks.
C.J. Harris
C.J. Harris am 20 Dez. 2011
No problem. Mark answer as accepted.

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Weitere Antworten (3)

C.J. Harris
C.J. Harris am 20 Dez. 2011
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 Kommentar
Jan
Jan am 20 Dez. 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 am 20 Dez. 2011
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 Kommentare
Danilo NASCIMENTO
Danilo NASCIMENTO am 20 Dez. 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 am 20 Dez. 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 am 24 Nov. 2017
Bearbeitet: per isakson am 24 Nov. 2017
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 Kommentar
Jan
Jan am 24 Nov. 2017
Bearbeitet: Jan am 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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