# Radar System PPI polar display

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Jose Iglesias on 12 Mar 2022
Commented: William Rose on 12 Mar 2022
I am trying to plot a radar Plan Position Indicator (PPI) slant range with the following code. This code is describing a scene with a radar at 2 km altitude scanning the area contained in a box. The x axis ranges from -90 to 0 degrees, and to 0 degrees to 90 degrees.The y axis ranges from 2 to 12km and the x axis is in respect 10 km split into two section of 0 to 5 km on either side. I tried this code and it worked and I was able to plot the B-scan slant range and true ground range. However, when I tried to plot the PPI slant range with a polar display which is supposed to look like a traditional radar screen, it looks like lines contained within the radar screen instead of dots representing point locations. I attached the code and the other script which has the two equations being used. The equations are phi = atan(x/y) and rho = sqrt(x^2 +y^2 +(h-H)^2) -(h-0.2). I set the polarplot in terms of phi and rho but my plot looks like lines instead of dots on the polar display. I also included a pic of the scene and my current polar plot. Any suggestions?
Script 1
% x = [5 5 5 5 4 4 4 4 4 3 3 3 3 3 3 0 0 0 0 0 0 2 5 5 5 5 5 5];
% y = [10 9 3 0 10 9 8 3 0 10 9 8 7 3 0 10 6 5 4 3 0 5 10 6 5 4 3 0];
% h = [1.8 2 2 2 2 1.8 2 2 2 2 2 1.8 2 2 2 2 2 1.8 2 2 2 1.8 2 2 1.8 2 2 2];
% phi = [-26.6 -29.1 -59.1 -90 -21.8 -24.0 -26.6 -53.1 -90 -16.7 -18.4 -20.6 -23.2 -45 -90 0 0 0 0 0 0 21.8 26.6 39.8 45 51.3 59.0 90];
% r = [11.32 10.49 6.16 5.39 10.95 10.01 9.17 5.39 4.47 10.63 9.70 8.73 7.87 4.69 3.61 10.20 6.32 5.32 4.47 3.61 2 5.59 11.36 8.06 7.30 6.71 6.16 5.39];
%% build scenario
box_x=[];
box_x=[box_x -5:0.5:5];
box_x=[box_x (zeros(1,21)+5)];
box_x=[box_x 5:-0.5:-5];
box_x=[box_x (zeros(1,21)-5)];
box_y=[];
box_y=[box_y zeros(1,21)];
box_y=[box_y 0:0.5:10];
box_y=[box_y (zeros(1,21)+10)];
box_y=[box_y 10:-0.50:0];
box_z=zeros(1,84);
%%
%%
%%
ridgebasef_x=[];
ridgebasef_x=[ridgebasef_x 5:-0.25:0];
ridgebasef_x=[ridgebasef_x 0:-0.25:-5];
ridgebasef_y=[];
ridgebasef_y=[ridgebasef_y (zeros(1,21)+4)];
ridgebasef_y=[ridgebasef_y 4:0.25:9];
ridgebasef_z=[];
ridgebasef_z=[ridgebasef_z (zeros(1,42))];
%%
ridgetop_x=[];
ridgetop_x=[ridgetop_x 5:-0.25:0];
ridgetop_x=[ridgetop_x 0:-0.25:-5];
ridgetop_y=[];
ridgetop_y=[ridgetop_y (zeros(1,21)+5)];
ridgetop_y=[ridgetop_y 5:0.25:10];
ridgetop_z=[];
ridgetop_z=[ridgetop_z (zeros(1,42)+0.2)];
ridgebaseb_x=[];
ridgebaseb_x=[ridgebaseb_x 5:-0.25:0];
ridgebaseb_x=[ridgebaseb_x 0:-0.25:-5];
ridgebaseb_y=[];
ridgebaseb_y=[ridgebaseb_y (zeros(1,21)+6)];
ridgebaseb_y=[ridgebaseb_y 6:0.20:10];
ridgebaseb_z=[];
ridgebaseb_z=[ridgebaseb_z (zeros(1,42))];
% done with Scenario
%% Call conversion to PPI_SR
figure(1)
polarplot(box_Az,box_Rng,'k');hold on; %'k' gives black
polarplot(ridgebasef_Az,ridgebasef_Rng,'.-g');%'.-g' gives dashed green line
polarplot(tower_Az,tower_Rng,'r');%'g' gives red line
polarplot(ridgetop_Az,ridgetop_Rng,'.-c');%'.-c' gives dashed cyan line
polarplot(ridgebaseb_Az,ridgebaseb_Rng,'y');%'y' gives yellow line
hold off;
xlim([-90 90])
ylim([2 12]);
grid on
xlabel('AZIMUTH (deg)')
ylabel('SLANT RANGE (km)')
title('BSCAN (SLANT RANGE)');
% figure
% hold on
% plot(box_x,box_y,x,y)
% set(gca,'XTickLabel', [-90 -70 -50 -30 -10 0 10 30 50 70 90])
% set(gca, 'YTickLabel', [0 1 2 3 4 5 6 7 8 9 10])
Script 2
function [outaz outR] = FN_PPI_SR(tx,ty,tz,rx,ry,rz)
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
outR = sqrt((tx-rx).^2+(ty-ry).^2+(tz-rz).^2)-(tz-0.2);
% x^2 y^2 (h-H)^2
outaz = 57.3*atan((tx-rx)./(ty-ry));
end

William Rose on 12 Mar 2022
Edited: William Rose on 12 Mar 2022
[Edited: Fixed a few typos in my text below. Code, figure unchanged.]
Angles should be in radians, not degrees, for polarplot().
I have changed atan() to atan2(), which is the better function to use if you have both x and y data.
I have changed the angle range from +-90 degrees to 0..180 degrees.
I use rlim() and thetalim() rather than xlim() and ylim().
I deleted the xlabel(), ylabel() commands since they do not work with polar plots.
I am showing just the ridgetop, road, and tower for simplicity. You can add the rest.
Your way of defining the target location vectors seems cumbersome to me. See how I have done it differently for the road, ridgetop, and tower. My way is clearer and uses a lot fewer lines of code.
I modfied the radius calculation in your function because I think it is a mistake to subtract tz-0.2 from every slant range.
##### 2 CommentsShow 1 older commentHide 1 older comment
William Rose on 12 Mar 2022
@Jose Iglesias, you're welcome. Good luck with your work.

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