Legend Color doesn't Match the Curves Color
6 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Good Day;
Here my question is happened often to occur as I see when I searched to solve my problem,but each case it depend on their code. I faced this problem as shown in the attached figure, and I hope to find the solution from you. I used 'hold on'but nothing changed, the code as shown below:
Z=100:250:5000; %Distance Between The Transmitter and Reciever in m
pointing_Err_angle=1*10^-5; %Pointing Error Angle in mrad
lambda=1550e-9; %Wavelength in nm
wo=5*10^-2; %Beam Waist at Z=0 in m
Cn=sqrt(5e-15); %Refractive Index Structure Parameter
% --------------------Calculation Section----------------------------------
k=2*pi/lambda; %The number of optical wave
po=(0.55*(Cn)^2*k^2*Z).^(-3/5); %The Coherence Length
E=(1+2*wo^2./po.^2);
wz=wo*(1+E.*((lambda.*Z)./(pi*wo^2)).^2).^(1/2); %The Beam Waist
r=Z*(pointing_Err_angle); %The Radial Displacement B
%----------------------------------------Calculate Hp-----------------------------
v=(sqrt(pi)*a)./(sqrt(2)*wz);
a0=(erf(v)).^2; %Maximal Fraction of Collected Power at r=0
Wzeq=sqrt(wz.^2*((sqrt(pi).*erf(v))/(2*v.*exp(-v.^2))));
Hp=a0.*exp((-2*r.^2)./((Wzeq).^2));
%----------------------------------Intensity Distribution-------------------------
Rytov_var=[0.1,0.2,0.5,0.8]; %Log irradiance variance (Roytov variance)
for i=1:length(Rytov_var)
for j=1:5:length(Hp)
A1(i,j)=1./(Hp(j).*sqrt(2*pi*Rytov_var(i)));
A2(i,j)=((log(Hp(j))+0.5.*Rytov_var(i)).^2)./2.*Rytov_var(i).^2;
PDF_Hp(i,j)=A1(i,j).*exp(-(A2(i,j)));
end
end
%--------------------------Plot Section------------------------------------
z=1250:1250:5000; %Distance for Plot Purpose
figure(1)
semilogy(z,PDF_Hp,'LineWidth',4);
xlabel('The Distance Z')
ylabel('The Probability Density Function ')
title('The PDF of Hp Vs The Distance')
legend('σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')
grid on
figure(2)
semilogy(Rytov_var,PDF_Hp,'LineWidth',3);
xlabel('Roytov variance')
ylabel('The Probability Density Function ')
title('The PDF of Hp Vs Rytove Variance')
legend('σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')
grid on
0 Kommentare
Akzeptierte Antwort
Adam Danz
am 15 Sep. 2020
Bearbeitet: Adam Danz
am 15 Sep. 2020
I see the problem. Look at the values you're plotting. Assuming a=0.5e-2;
semilogy(z,PDF_Hp,'LineWidth',4);
PDF_Hp =
Columns 1 through 12
59.204 0 0 0 0 74.572 0 0 0 0 130.59 0
33.684 0 0 0 0 41.243 0 0 0 0 66.92 0
5.2821 0 0 0 0 5.3479 0 0 0 0 5.1881 0
0.42448 0 0 0 0 0.30818 0 0 0 0 0.12088 0
Columns 13 through 16
0 0 0 242.89
0 0 0 112.93
0 0 0 4.5326
0 0 0 0.032869
A line is created for each column so your plot actually has 16 lines, not 4.
The reason they don't appear is because in a log plot, 0 is not defined. However, those line objects are still produced - they are just not shown. As evidence of that, check out the 16 object handles.
h = semilogy(z,PDF_Hp,'LineWidth',4)
% h =
% 16×1 Line array:
%
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
% Line
The reason the "wrong" colors are defined is because you're only asking for the first 4 objects to appear in the legend. But objects 2-3-4 are all non-visible lines.
legend('σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')
Solution
Get rid of the columns with 0s in PDF_Hp
Weitere Antworten (2)
Bjorn Gustavsson
am 15 Sep. 2020
That's peculiar. What happens if you keep the plot-handle from your calls to semilogy and send that to legend?
ph = semilogy(z,PDF_Hp,'LineWidth',4);
xlabel('The Distance Z')
ylabel('The Probability Density Function ')
title('The PDF of Hp Vs The Distance')
legend(ph,'σ^2_I=0.1','σ^2_I=0.2','σ^2_I=0.5','σ^2_I=0.8')
That will at least allow you to modify the line-colors and check how they change in the plot and in the legend.
HTH
Siehe auch
Kategorien
Mehr zu Legend finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!