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 个评论
Matt Fig
2012-11-21
编辑:Matt Fig
2012-11-21
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.
采纳的回答
Walter Roberson
2012-11-21
Read the documentation on histc() and in particular pay attention to the multiple-output version of it.
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