Radiation pattern plotting.
82 ビュー (過去 30 日間)
古いコメントを表示
Hi, I am having difficulties plotting the following function:
theta = -2:0.01:2;
y = (cos(pi/4*cos(theta*pi))-cos(pi/4))./(sin(theta*pi)*sin(pi/4));
I'd like it to resemble as closely as possible the plot in the attachment.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/158433/image.jpeg)
I'd certainly appreciate any assistance. I understand that patterncustom() is probably to be used, yet I am not quite sure how. NB The pi/4 argument in the expression for y is the result of substituting l=lambda/4 in the original.
2 件のコメント
Hildo
2016 年 11 月 28 日
編集済み: Walter Roberson
2016 年 11 月 29 日
Appear that is a cylindrical coordinates plot ( y as function(theta,r) ). But you are just seeing a slice on the y-axis.
採用された回答
Omkar Savkur
2021 年 6 月 29 日
Hi Yuval, building off David's answer, you can use the polarpattern function to help plot the radiation power. You can interact with the plot and specify the plot parameters all in one line.
% Angle in degrees
theta = (-1:.001:1)*180;
% lambda is arbitrary in this calculation; pick a value
lambda = 1;
% d is entire dipole length, both halves
d = lambda/4;
k = 2*pi/lambda;
kd2 = k*d/2;
y = 20*log10(abs((cos(kd2.*cosd(theta))-cos(kd2))./sind(theta)));
y = y-max(y);
% normalize y to obtain directivity; new max is 0 dB
y(y<-40) = -40;
figure(1)
polarpattern(theta,y,'AngleDirection','cw','AngleAtTop',0,'AngleResolution',30)
2 件のコメント
Gerard Marcial Sopsop
2021 年 7 月 5 日
Hello, may I ask what "k" and "kd2" in the code stands for in the formula of radiation pattern? Thanks.
Chirag
2023 年 1 月 11 日
その他の回答 (5 件)
David Goodmanson
2017 年 11 月 20 日
Hi Yuval,
I made a stylistic change to your code and defined theta at the very start with the factor of pi, rather than waiting to do that in the trig functions. Also the code below has the necessary range of -pi to pi and no more, rather than the duplicate overplotting that happens with -2pi to 2pi.
Since the formula you are using is the linear quantity E rather than intensity ~~E^2, the conversion to dB is 20*log(....) rather than 10*log(....).
theta = (-1:.001:1)*pi;
lambda = 1; % lambda is arbitrary in this calculation; pick a value
d = lambda/4; % d is entire dipole length, both halves
k = 2*pi/lambda;
kd2 = k*d/2;
y = 20*log10(abs((cos(kd2.*cos(theta))-cos(kd2))./sin(theta)));
y = y-max(y); % normalize y to obtain directivity; new max is 0 dB
y(y<-40) = -40;
figure(1)
polarplot(theta,y)
rlim([-40 0])
set(gca,'thetazerolocation','top','thetadir','clockwise')
The entire figure from the book can be obtained by concatenating the y's for for other values of d into a 5x2001 matrix and plotting, or the y's can be plotted sequentially using the 'hold on' command.
The figure is a classic one but must have been done by hand, because if you look at the trace for d = 3*lambda/4, it comes close to passing through the point theta = 150, r = 10 but misses the corresponding one theta = 30, r = 10 by a lot more. Those two spots should be symmetric. The pc of course does a lot better in that regard, but lacks the same aesthetic.
0 件のコメント
Tamir Suliman
2016 年 11 月 29 日
Hello please check my code at
2 件のコメント
Tamir Suliman
2016 年 12 月 4 日
編集済み: Tamir Suliman
2016 年 12 月 4 日
Sir if you look at the answer you could use polar to plot the function
polar(theta,y)
you could use view([90 -90])
to rotate it accordingly the script that I sent has a similar function and similar plots
this is the result of plotting your function
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/186355/image.png)
Dan Klemfuss
2017 年 11 月 18 日
Good Evening. Can you please check you equation? I've plotted several antenna gain patterns before but the resulting plot doesn't quite match the expected output. You should be able to generate the plot using the following if you obtain the right equation:
theta = -pi:0.01:pi;
y = (cos(pi/4*cos(theta*pi))-cos(pi/4))./(sin(theta*pi)*sin(pi/4));
figure('Name','3-dB beamwidth=87°','NumberTitle','off','MenuBar','none');
polarplot(theta, y,'k')
rlim([min(y)-5 max(y)+5]);
ax = gca;
ax.ThetaDir = 'clockwise';
ax.ThetaZeroLocation = 'top';
plotTitle = sprintf('Radiation Pattern');
title(plotTitle)
0 件のコメント
Yuvan Sankar
2022 年 3 月 4 日
theta = (-1:.001:1)*pi; lambda = 1; % lambda is arbitrary in this calculation; pick a value d = lambda/4; % d is entire dipole length, both halves k = 2*pi/lambda; kd2 = k*d/2; y = 20*log10(abs((cos(kd2.*cos(theta))-cos(kd2))./sin(theta))); y = y-max(y); % normalize y to obtain directivity; new max is 0 dB y(y<-40) = -40; figure(1) polarplot(theta,y) rlim([-40 0]) set(gca,'thetazerolocation','top','thetadir
0 件のコメント
Akash Pawar
2023 年 1 月 16 日
% Angle in degrees
theta = (-1:.001:1)*180;
% lambda is arbitrary in this calculation; pick a value
lambda = 1;
% d is entire dipole length, both halves
d = lambda/4;
k = 2*pi/lambda;
kd2 = k*d/2;
y = 20*log10(abs((cos(kd2.*cosd(theta))-cos(kd2))./sind(theta)));
y = y-max(y);
% normalize y to obtain directivity; new max is 0 dB
y(y<-40) = -40;
figure(1)
polarpattern(theta,y,'AngleDirection','cw','AngleAtTop',0,'AngleResolution',30)
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Polar Plots についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!