How to plot this code correctly?
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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 Kommentar
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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