How to make a root locus plot square aspect ratio
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I have tried:
s = tf('s');
sysL = (s^2+1)/(s*(s^2+4));
rlocus(sysL);
v=[-4 4 -4 4];
axis(v);
axis('square');
But I get a message that says: Warning: This plot type does not support this option for the "axis" command.
I am using 2013b
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Accepted Answer
Star Strider
on 7 Oct 2015
Edited: Star Strider
on 7 Oct 2015
It does not support axis('square'). It does support axis('equal'), but that overrides your axis(v) call.
s = tf('s');
sysL = (s^2+1)/(s*(s^2+4));
rlocus(sysL);
v=[-4 4 -4 4]; axis(v);
hpos = get(gcf, 'Position'); % Get 'Position'
set(gcf, 'Position',[hpos([1 2 4]) hpos(4)]) % Set Width = Height
Experiment to get the result you want.
4 Comments
Star Strider
on 8 Nov 2023
Comparing them —
s = tf('s');
sysL = (s^2+1)/(s*(s^2+4));
figure
rlocus(sysL);
figure
rlocus(sysL);
hpos = get(gcf, 'Position'); % Get 'Position'
set(gcf, 'Position',[hpos([1 2 4]) hpos(4)]) % Set Width = Height
axis('equal');
It works as desired in R2023b.
.
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