Info
This question is closed. Reopen it to edit or answer.
Surface Plot or Mesh Plot
1 view (last 30 days)
Show older comments
I am quite naive to MATLAB, so i beg your apology for asking simple questions. Here is the code:
A=50;
tau=1;
P_f=0.02;
for i=1:100
v0(i)=i;
v(i)=v0(i).*5/18;
Kf=tan(atan((v(i).*tau)/sqrt((4*A^2)-((v(i)^2)*(tau^2))))-((pi*P_f)/2));
M=(2*A*Kf)/(v(i).*sqrt(1+Kf^2));
Phf(i)=(2/pi)*(atan((v(i).*tau)/sqrt((4*A^2)-(((v(i))^2)*tau^2)))-atan((v(i).*M)/sqrt((4*A^2)-(((v(i))^2)*M^2))));
end
plot (v0,Phf)
Now, how to make a Surface Plot or Mesh Plot of this ?
6 Comments
Star Strider
on 13 Jul 2014
When I calculated your data, the range (max(Phf)-min(Phf)) = 69.3889e-018. The variations in Phf are on the order of machine precision.
Do this calculation (dPhf/dv0) and plot to illustrate that:
dPhfdv0 = diff([Phf])./diff([v0]);
plot(dPhfdv0)
Answers (1)
Star Strider
on 14 Jul 2014
Probabilities are by definition always positive.
I used dPhfdV0 to illustrate the fact that your Phf array has very little variation. If you want to calculate the statistics, I would calculate those on Phf itself. Calculating the mean and standard deviation are easy enough:
Phfmn = mean(Phf)
Phfsd = std(Phf)
If you want the values that will define the range that Phf will be found within with 95% probability, you can calculate them as:
PhfRng95 = [Phfmn-1.96*Phfsd Phfmn+1.96*Phfsd]
You have to use format long e to see the details of PhfRng95. The value of 1.96 is the inverse normal distribution for the probabilities of 0.025 and 0.975, encompassing a total probability of 0.95.
0 Comments
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!