Something strange using polar plot function in MATLAB

Question

I have a simple function of theta and I want to plot this function in dB using the polarplot function in MATLAB. But when I make the graph from -40 to 0, the graph seems to have a strange part around horizontal axis. My MATLAB code (R2016a) is:

%% Define range of plotting angle.
ceta= [10^-9:0.0001:2*pi];
% ceta starts not from pure zero to avoid 0/0 in some cases.

E =  abs( ( cos((cos(ceta))*pi/2) ) ./ ( sin(ceta) ) ); 

power_dB = 10.*log10(E.^2); 
power_dB = power_dB - max(power_dB);
max(power_dB)
polarplot(ceta,power_dB);
rlim([-40 0]);

The obtained figure is this: plot resulting from code

Matt · Accepted Answer · 2018-11-19 23:15:26Z

4

Your values for E are very near to 0 when ceta = 0, pi, or 2pi. This is resulting in very large values when you take the log of E.

You can just remove the points from ceta and E when E is very low. See the code block below.

E =  abs( (  cos((cos(ceta))*pi/2) ) ./ ( sin(ceta) ) ); 
ceta(E<1e-2) = [];
E(E<1e-2) = []; 
power_dB = 10.*log10(E.^2); 
power_dB = power_dB - max(power_dB);
max(power_dB)
polarplot(ceta,power_dB);
rlim([-40 0]);

Gives:

answered Nov 19, 2018 at 23:15

Matt

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4 Comments

user288086 Over a year ago

i have a question, when 10*log10(E.^2) tends to large value i think rlim([ -40 0]) should remove it so why that is not happened?

Matt Over a year ago

rlim is only controlling the figure axis. What's happening is as you plot value larger than -40 (in terms of magnitude) its passing through the origin and continuing to the edges of the plot. Try this, remove the line I suggested and add in power_dB(power_dB < -60) = -60 right before your polar plot. You'll see the lines come back, but stop at 2 circles out from the center. This represents the -60 value once they've passed through the origin.

Matt Over a year ago

Also, looking at this a second time better solution might have been to just use power_dB(power_dB < -40) = -40, rather than my suggested change.

Adriaan Over a year ago

E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta)); would suffice, cutting down on the LISP style usage of brackets.

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