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Something strange using polar plot function in MATLAB
1
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:
matlab plot polar-coordinates
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
asked Nov 19 '18 at 22:56
user288086user288086
134
add a comment |
1
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:
matlab plot polar-coordinates
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
asked Nov 19 '18 at 22:56
user288086user288086
134
add a comment |
1
1
1
0
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:
matlab plot polar-coordinates
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
asked Nov 19 '18 at 22:56
user288086user288086
134
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:
matlab plot polar-coordinates
matlab plot polar-coordinates
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
asked Nov 19 '18 at 22:56
user288086user288086
134
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
asked Nov 19 '18 at 22:56
user288086user288086
134
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
edited Nov 19 '18 at 23:29
Cris Luengo
19.3k51947
19.3k51947
asked Nov 19 '18 at 22:56
user288086user288086
134
asked Nov 19 '18 at 22:56
user288086user288086
134
asked Nov 19 '18 at 22:56
user288086user288086
134
134
add a comment |
add a comment |
1 Answer
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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 '18 at 23:15
MattMatt
1,159112
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?
– user288086
Nov 19 '18 at 23:31
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
Nov 19 '18 at 23:44
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.
– Matt
Nov 19 '18 at 23:45
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta));would suffice, cutting down on the LISP style usage of brackets.
– Adriaan
Nov 30 '18 at 22:21
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
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 '18 at 23:15
MattMatt
1,159112
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?
– user288086
Nov 19 '18 at 23:31
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
Nov 19 '18 at 23:44
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.
– Matt
Nov 19 '18 at 23:45
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta));would suffice, cutting down on the LISP style usage of brackets.
– Adriaan
Nov 30 '18 at 22:21
add a comment |
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 '18 at 23:15
MattMatt
1,159112
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?
– user288086
Nov 19 '18 at 23:31
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
Nov 19 '18 at 23:44
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.
– Matt
Nov 19 '18 at 23:45
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta));would suffice, cutting down on the LISP style usage of brackets.
– Adriaan
Nov 30 '18 at 22:21
add a comment |
4
4
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 '18 at 23:15
MattMatt
1,159112
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 '18 at 23:15
MattMatt
1,159112
answered Nov 19 '18 at 23:15
MattMatt
1,159112
answered Nov 19 '18 at 23:15
MattMatt
1,159112
answered Nov 19 '18 at 23:15
MattMatt
1,159112
1,159112
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?
– user288086
Nov 19 '18 at 23:31
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
Nov 19 '18 at 23:44
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.
– Matt
Nov 19 '18 at 23:45
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta));would suffice, cutting down on the LISP style usage of brackets.
– Adriaan
Nov 30 '18 at 22:21
add a comment |
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?
– user288086
Nov 19 '18 at 23:31
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
Nov 19 '18 at 23:44
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.
– Matt
Nov 19 '18 at 23:45
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta));would suffice, cutting down on the LISP style usage of brackets.
– Adriaan
Nov 30 '18 at 22:21
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?
– user288086
Nov 19 '18 at 23:31
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?
– user288086
Nov 19 '18 at 23:31
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
Nov 19 '18 at 23:44
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
Nov 19 '18 at 23:44
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.
– Matt
Nov 19 '18 at 23:45
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.
– Matt
Nov 19 '18 at 23:45
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta)); would suffice, cutting down on the LISP style usage of brackets.– Adriaan
Nov 30 '18 at 22:21
E = abs(cos(cos(ceta)*pi/2) ./ sin(ceta)); would suffice, cutting down on the LISP style usage of brackets.– Adriaan
Nov 30 '18 at 22:21
add a comment |
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