Surface Plot or Mesh Plot

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 comentarios

Azzi Abdelmalek
Azzi Abdelmalek el 13 de Jul. de 2014
Editada: Azzi Abdelmalek el 13 de Jul. de 2014
You need three variables. What are they?
Adnan
Adnan el 13 de Jul. de 2014
I just have two variables here ~ v0 and Phf. How to define this third matrix or vector
Star Strider
Star Strider el 13 de Jul. de 2014
Which other one do you want to plot: Kf, M, or something else?
Adnan
Adnan el 13 de Jul. de 2014
Can you just help me to adjust the scale of this plot (as per the above code). The plot comes around 0.02, but i can't figure out the exact values on the y axis with respect to the x axis. I need something where i can clearly see the rising or the declining points
Star Strider
Star Strider el 13 de Jul. de 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)
Adnan
Adnan el 14 de Jul. de 2014
Thanks Star Strider. I m calculating probability and it can't be negative. Is there any method to avoid negative probabilities in this plot, plot(dPhfdv0)

Respuestas (1)

Star Strider
Star Strider el 14 de Jul. de 2014

0 votos

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.

La pregunta está cerrada.

Preguntada:

el 13 de Jul. de 2014

Cerrada:

el 20 de Ag. de 2021

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