Azzera filtri
Azzera filtri

Plotting problem

1 visualizzazione (ultimi 30 giorni)
J
J il 22 Feb 2012
Modificato: sony il 20 Ott 2013
Hi. I've written this code to plot the graph it produces. However, the top three lines have complex parts to them. How can I plot only the real parts? Code follows:
%Matlab code for "MATLAB
%Question 1
%
%13/02/12
function [lmda, Ct] = problem
%QUESTION 1%
rs=0.05; %rotor solidity
Cl=2*pi; %specific lift coefficient
pa1=0.1745329250; %pitch angle 1
pa2=0.0872664625; %pitch angle 2
pa3=-0.0017453292; %pitch angle 3
pa4=-0.0349065850; %pitch angle 4
pa5=-0.0872664625; %pitch angle 5
%--------------------%
i=1;
for lmda=0:0.1:20;
a2(i)=(((((rs*lmda*Cl)/16)+0.5)^2)-((rs*lmda*Cl*(1-(lmda*pa1)))/8));
if a2(i) < 0
break
end
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
a1(i)=(((rs*lmda*Cl)/16)+0.5);
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
at(i)=a1(i)-sqrt(a2(i));
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
Ct1(i)=((rs*(lmda^2)*Cl)/2)*(((1-at(i))/lmda)-pa1);
i=i+1;
end
%----------------------------------------------------------------%
i=1;
for lmda=0:0.1:20;
a2(i)=(((((rs*lmda*Cl)/16)+0.5)^2)-((rs*lmda*Cl*(1-(lmda*pa2)))/8));
if a2(i) < 0
break
end
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
a1(i)=(((rs*lmda*Cl)/16)+0.5);
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
at(i)=a1(i)-sqrt(a2(i));
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
Ct2(i)=((rs*(lmda^2)*Cl)/2)*(((1-at(i))/lmda)-pa2);
i=i+1;
end
%----------------------------------------------------------------%
i=1;
for lmda=0:0.1:20;
a2(i)=(((((rs*lmda*Cl)/16)+0.5)^2)-((rs*lmda*Cl*(1-(lmda*pa3)))/8));
if a2(i) < 0
break
end
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
a1(i)=(((rs*lmda*Cl)/16)+0.5);
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
at(i)=a1(i)-sqrt(a2(i));
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
Ct3(i)=((rs*(lmda^2)*Cl)/2)*(((1-at(i))/lmda)-pa3);
i=i+1;
end
%----------------------------------------------------------------%
i=1;
for lmda=0:0.1:20;
a2(i)=(((((rs*lmda*Cl)/16)+0.5)^2)-((rs*lmda*Cl*(1-(lmda*pa4)))/8));
if a2(i) < 0
break
end
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
a1(i)=(((rs*lmda*Cl)/16)+0.5);
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
at(i)=a1(i)-sqrt(a2(i));
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
Ct4(i)=((rs*(lmda^2)*Cl)/2)*(((1-at(i))/lmda)-pa4);
i=i+1;
end
%----------------------------------------------------------------%
i=1;
for lmda=0:0.1:20;
a2(i)=(((((rs*lmda*Cl)/16)+0.5)^2)-((rs*lmda*Cl*(1-(lmda*pa5)))/8));
if a2(i) < 0
break
end
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
a1(i)=(((rs*lmda*Cl)/16)+0.5);
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
at(i)=a1(i)-sqrt(a2(i));
i=i+1;
end
%--------------------%
i=1;
for lmda=(0:0.1:20);
Ct5(i)=((rs*(lmda^2)*Cl)/2)*(((1-at(i))/lmda)-pa5);
i=i+1;
end
%----------------------------------------------------------------%
lmda=(0:0.1:20);
figure,plot(lmda,real(Ct1),'k-','Linewidth',2),grid on, hold on
plot(lmda,real(Ct2),'k--','Linewidth',2)
plot(lmda,real(Ct3),'k-.','Linewidth',2)
plot(lmda,real(Ct4),'r-','Linewidth',2)
plot(lmda,real(Ct5),'r--','Linewidth',2)
axis([0 20 0 1])
xlabel('\lambda');ylabel('C_t');title('Thrust coefficient against tip speed ratio.');
legend('\theta_T=10^{o}','\theta_T=5^{o}','\theta_T=-0.1^{o}','\theta_T=-2^{o}','\theta_T=-5^{o}');
end
%QUESTION 2%

Risposta accettata

Matt Tearle
Matt Tearle il 22 Feb 2012
You are plotting only the real parts. I think the problem is coming with how you're calculating these functions. You have break statements to terminate your for loops when a2 goes negative. However, the last value of a2 (the negative value) is still there.
Furthermore, you're reusing a2, but you stop calculating after a2 goes negative. This means that the rest of a2 is still there from the previous calculation. Do you really want that?
If you zoom out, so the y-axis goes higher, you will see that the function carries on doing something else. My suspicion is that this comes from calculating at based on old a2 values.
Finally, you really don't need these loops. This code is really nasty to read and debug.
lmda=0:0.1:20;
a1=(((rs*lmda*Cl)/16)+0.5);
Nice and easy. Just remember to use .* and ./ when multiplying/dividing arrays.
  1 Commento
J
J il 22 Feb 2012
Great, thank you very much! I'm new to Matlab so this style was really helpful.

Accedi per commentare.

Più risposte (2)

J
J il 22 Feb 2012
Ok, I've tried to edit the code and get rid of loops.
rs = 0.05;
Cl = 2*pi;
pa = [10*(pi/180) 5*(pi/180) -0.1*(pi/180) -2*(pi/180) -5*(pi/180)];
lmda=1:0.1:20;
for v = 1:5
for i=1:length(lmda)
a(i) = (((rs*lmda(i)*(Cl))/16)+0.5)-sqrt((((rs*lmda(i)*(Cl))/16)+0.5).^2)-((rs*lmda(i)*(Cl).*(1-lmda(i).*pa(v)))/8);
Ct(i) = ((rs*lmda(i).^2*Cl)/2).*(((1-a(i))./1-pa(v)));
end
Z = imag(Ct);
for d = 1:1:201
if Z(v,d) ~= 0
Ct(v,d) = NaN;
end
end
end
plot (lmda,Ct)
legend('10','5','-0.1','-2','-5')
axis([0 20 0 1])
xlabel('\lambda');ylabel('C_t');title('Thrust coefficient against tip speed ratio.');
legend('\theta_T=10^{o}','\theta_T=5^{o}','\theta_T=-0.1^{o}','\theta_T=-2^{o}','\theta_T=-5^{o}')

J
J il 22 Feb 2012
I changed it again, thanks so much for your help!
clear all
clc
rs=0.05;
Cl=2*pi;
pa=[(pi/18) (pi/36) -(pi/1800) -(pi/90) -(pi/36)];
for v=1:5
lmda=0:0.1:20;
Ct(v,:) = ((rs*lmda.^2*Cl)/2).*(((1-((rs*lmda*Cl/16+1/2)-...
((rs*lmda*Cl/16+1/2).^2-(rs*lmda*Cl.*(1-lmda*pa(v)))/8).^(1/2))))./lmda-pa(v));
Z= imag(Ct);
for d = 1:1:201
if Z(v,d) ~= 0
Ct(v,d) = NaN;
end
end
end
figure,plot(lmda,Ct)
grid on
axis([0 20 0 1]);
xlabel('\lambda');ylabel('C_t');title('Thrust coefficient against tip speed ratio.');
legend('\theta_T=10^{o}','\theta_T=5^{o}','\theta_T=-0.1^{o}','\theta_T=-2^{o}','\theta_T=-5^{o}');
  3 Commenti
J
J il 24 Feb 2012
<3 haha teach me more! Why couldn't you be my lecturer!
Matt Tearle
Matt Tearle il 24 Feb 2012
Because MathWorks pays more :)
Maybe some of these suggestions might help you:
http://www.mathworks.com/matlabcentral/answers/8026-best-way-s-to-master-matlab

Accedi per commentare.

Categorie

Scopri di più su Loops and Conditional Statements in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by