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where is the mistake in this code?
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v=0.25;
x= 60;
D=0.01;
kmax=3.1624E-5;
X=1;
Ks=8.9550E-5;
a0=1;
t=1:56;
for tt= 1:1:length(t)
ka=kmax*X/Ks;
B=(((v^2)/(4*D^2) +(ka/D)) ^ 0.5);
C(tt)=(a0/2)*exp(-B*x)*erfc((x-(v^2 +4*ka*D)^0.5*t)/(2*(D*t)^0.5)) + exp(B*x)*erfc((x-(v^2 +4*ka*D)^0.5 *t)/(2*(D*t)^0.5));
end
plot(tt,C)
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답변 (3개)
Chad Greene
2014년 9월 5일
I'm not sure if this is a coincidence, but you've bolded two of the mistakes. There's an 0.5 t and a 2(D*t). Did you mean 0.5*t and a 2*(D*t)? Or perhaps plus or minus or divided by or to the power of?
Another thing: put a . before the ^ signs to make them .^ if you're doing element-wise raising to a power.
And finally, should the plot be plot(t,C)?
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per isakson
2014년 9월 5일
Don't write longer lines than you can handle. Introduce temporary variables like
meaningful_name = (2*(D*t)^0.5);
that will make the error messages easier to interpret.
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Star Strider
2014년 9월 5일
First, consider whether you should replace t by tt in your C(tt) assignment. Otherwise, C(tt) never changes.
Second, even with that change, if you break C(tt) down into its components, the problem becomes quickly apparent. Add this line before C(tt):
Cv(tt,:) = [exp(-B*x) erfc((x-(v^2 +4*ka*D)^0.5*tt)/(2*(D*tt)^0.5)) exp(B*x) erfc((x-(v^2 +4*ka*D)^0.5 *tt)/(2*(D*tt)^0.5))];
For all values of tt:
Cv(tt,:) = [0 0 Inf 0]
so: C(tt) = NaN everywhere.
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Star Strider
2014년 9월 5일
I actually did not suggest any parts for you to change — you have to figure that out, since I do not know what you are doing.
I only suggested a way for you to figure out what is wrong with your code.
The arguments to erfc are going to be zero as you calculate them, so you first need to understand exactly what you are doing with erfc. Once you have that problem solved, you can use logarithms if necessary, replacing exp(±B*x) with (±B*x) for your interim calculations, Then take antilogs at the end.
You may have problems with the units of your variables, so look closely at those to be sure all the units of your variables are correct and the magnitudes are correct.
per isakson
2014년 9월 5일
편집: per isakson
2014년 9월 5일
Set Stop on Errors:
dbstop on errors
Run the function. At line 17 select f(tt,:), right click, chose Evaluate Selection and you will find that it is not a  <1x4 double>.
Possibly, you have to pre-allocate f with something like
f = nan( length(t), 4 );
nan is good because it will make a fuzz if not all element are assigned values.
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