sky plot for the satellite

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vimal kumar chawda
vimal kumar chawda 2019 年 5 月 18 日
回答済み: SHAO Jingheng 2022 年 6 月 22 日
Hello Schlor
I am having azimuth, zenit and elevation. So how do I plot skyplot for the satellite .
I have code as below are as
polarplot(azi,ele)
It should be only visible only 0° to 90°.
Can anyone help me?
Thank you
  3 件のコメント
vimal kumar chawda
vimal kumar chawda 2019 年 5 月 20 日
Hallop
I have uploaded the variable of all but please consider azi and ele.
Thank you
vimal kumar chawda
vimal kumar chawda 2019 年 5 月 20 日
Here we are seeing plot as a top view but I need 90 at center. That means its skyplot.
So how can I bring 90° at center and other corner of the image is direction like North =N ; e,w,s.
figure
pax = polaraxes;
theta = 0:0.01:2*pi;
rho = sin(2*theta).*cos(2*theta);
polarplot(theta,rho)
pax.ThetaDir = 'clockwise';
pax.FontSize = 12;
Please find the code below?
Thank you

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採用された回答

vimal kumar chawda
vimal kumar chawda 2019 年 5 月 23 日
I got the result please find the below:
you can use this and found out how skyplot are works . In normal case we are getting polarplotresult top view. In this we get from aside. One can plot for different structure. Hope you enjoy.
pax=polaraxes;
pax.ThetaDir = 'clockwise';
fig3=polarplot(Azi,rad2deg(Elevation))
pax.RDir = 'reverse';

その他の回答 (2 件)

Joga Setiawan
Joga Setiawan 2020 年 8 月 1 日
This is an example to get the skyplot of an antenna tracking a LEO satellite, shown in the subplot(223)
t_AZ_EL=[0 41 0.4;
30 43.6 2.4;
60 45.6 3.9;
90 48.1 5.6;
120 51.1 7.4;
150 54.7 9.5;
180 58.5 11.5;
210 62.9 13.4;
240 68 15.4;
270 73.6 16.9;
300 79.7 17.3;
330 86.7 20.3;
360 94.3 21.3;
390 102.1 21.7;
420 110.6 21.6;
450 117.9 21;
480 125.7 19.7;
510 131.9 18.3;
540 137.9 16.5;
570 143.2 14.7;
600 148.3 12.5;
630 151.7 10.9;
660 155.1 9.1;
690 158.4 7.1;
720 161.1 5.4;
750 163.8 3.4;
780 166 1.7;
810 167.3 0.7]
t=t_AZ_EL(:,1);
AZ=t_AZ_EL(:,2);
EL=t_AZ_EL(:,3);
subplot(221)
plot(t,AZ)
grid
xlabel('time (s)')
ylabel('Azimuth (deg)')
subplot(222)
plot(t,EL)
grid
xlabel('time (s)')
ylabel('Elevation (deg)')
subplot(223)
polarplot(AZ*pi/180,EL)
ax = gca;
d = ax.ThetaDir;
ax.ThetaDir = 'clockwise';
ax.ThetaZeroLocation = 'top';
ax.RDir = 'reverse';
rlim([0 90])
rtickangle(20)
thetaticks([0 45 90 135 180 225 270 315])
thetaticklabels({'North','45','East','135','South','225','West','315'})

SHAO Jingheng
SHAO Jingheng 2022 年 6 月 22 日
Thx

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