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:
plot resulting from code










share|improve this question





























    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:
    plot resulting from code










    share|improve this question



























      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:
      plot resulting from code










      share|improve this 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







      matlab plot polar-coordinates






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 19 '18 at 23:29









      Cris Luengo

      19.3k51947




      19.3k51947










      asked Nov 19 '18 at 22:56









      user288086user288086

      134




      134
























          1 Answer
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          oldest

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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:



          enter image description here






          share|improve this answer
























          • 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











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          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:



          enter image description here






          share|improve this answer
























          • 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
















          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:



          enter image description here






          share|improve this answer
























          • 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














          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:



          enter image description here






          share|improve this answer













          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:



          enter image description here







          share|improve this answer












          share|improve this answer



          share|improve this answer










          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



















          • 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


















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