How to plot this code correctly?

6 vues (au cours des 30 derniers jours)
Jay
Jay le 21 Nov 2012
tt=1000;
B=1700;
b=0.0013;
sigma=0.0099;
t = 0 : 1 : tt;
for K = 1 : length(t)
p(K) = (B-(b*t(K)))/(sigma);
end
p(K)= roundn(p,-1);
p(K)=p;
if p==0.0
l=.5*k;
elseif p==0.1
l=.5478*k;
elseif p==0.2
l=.5793*k;
elseif p==0.3
l=.6255*k;
elseif p==0.4
l=.6554*k;
elseif p==0.5
l=.6985*k;
elseif p==0.6
l=.7257*k;
elseif p==0.7
l=.7642*k;
elseif p==0.8
l=.7881*k;
elseif p==0.9
l=.8212*k;
elseif p==1.0
l=.8413*k;
elseif p==1.1
l=.8686*k;
elseif p==1.2
l=.8849*k;
elseif p==1.3
l=.9066*k;
elseif p==1.4
l=.9192*k;
elseif p==1.5
l=.9357*k;
elseif p==1.6
l=.9452*k;
elseif p==1.7
l=.9573*k;
elseif p==1.8
l=.9641*k;
elseif p==1.9
l=.9726*k;
elseif p==2.0
l=.9772*k;
elseif p==2.1
l=.9830*k;
elseif p==2.2
l=.9861*k;
elseif p==2.3
l=.9898*k;
elseif p==2.4
l=.9918*k;
elseif p==2.5
l=.9941*k;
elseif p==2.6
l=.9953*k;
elseif p==2.7
l=.9967*k;
elseif p==2.8
l=.9974*k;
elseif p==2.9
l=.9982*k;
elseif p==3.0
l=.9987*k;
elseif p>3.0
l=1*k;
elseif p==-0.1
l=.4522*k;
elseif p==-0.2
l=.4207*k;
elseif p==-0.3
l=.3745*k;
elseif p==-0.4
l=.3446*k;
elseif p==-0.5
l=.3015*k;
elseif p==-0.6
l=.2743*k;
elseif p==-0.7
l=.2358*k;
elseif p==-0.8
l=.2119*k;
elseif p==-0.9
l=.1788*k;
elseif p==-1.0
l=.1587*k;
elseif p==-1.1
l=.1314*k;
elseif p==-1.2
l=.1151*k;
elseif p==-1.3
l=.0934*k;
elseif p==-1.4
l=.0808*k;
elseif p==-1.5
l=.0643*k;
elseif p==-1.6
l=.0548*k;
elseif p==-1.7
l=.0427*k;
elseif p==-1.8
l=.0359*k;
elseif p==-1.9
l=.0274*k;
elseif p==-2.0
l=.0228*k;
elseif p==-2.1
l=.0170*k;
elseif p==-2.2
l=.0139*k;
elseif p==-2.3
l=.0102*k;
elseif p==-2.4
l=.0082*k;
elseif p==-2.5
l=.0059*k;
elseif p==-2.6
l=.0047*k;
elseif p==-2.7
l=.0033*k;
elseif p==-2.8
l=.0026*k;
elseif p==-2.9
l=.0018*k;
elseif p==-3.0
l=.0013*k;
else
l=0*k;
end
plot(t, p)
  1 commentaire
Matt Fig
Matt Fig le 21 Nov 2012
Modifié(e) : Matt Fig le 21 Nov 2012
That code errors, so there is nothing to plot!
Also, you are not going to get the results you expect from that massive IF block. You are comparing floating points for equality, which is a bad idea. I also wonder if you are hoping the IF block will pick out each element of p one at a time... it won't.

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Réponse acceptée

Walter Roberson
Walter Roberson le 21 Nov 2012
Read the documentation on histc() and in particular pay attention to the multiple-output version of it.

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