dB scale (log scale) of a polar plot graph
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this is my function. how can i show it in dB scale? thanks everyone wants to help.
theta = 0:0.01:2*pi;
x = 60; % k*a
result =(((cos(theta)).*(sin((x/2).*sin(theta))/(x/2).*sin(theta))));
polar(theta,result)
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답변 (6개)
Jonathan LeSage
2013년 10월 17일
편집: Jonathan LeSage
2013년 10월 17일
You can simply convert the results directly to the decibels scale and plot the transformed results using the polar function. I suggest you refer to the definition of the decibel first. Additionally, you can check out the mag2db function. Here are some useful links to get you started:
Once you have the results in terms of decibels, you can plot using the polar function as you did before!
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Ibrahim Samy
2017년 10월 24일
i think you better use polardb(theta , result, -20, '-k')
-20 for minimal db u can see
sixwwwwww
2013년 10월 17일
편집: sixwwwwww
2013년 10월 17일
Dear Atarli,
Here is the conversion to decibel:
result =(((cos(theta)).*(sin((x/2).*sin(theta))/(x/2).*sin(theta))));
result_dB = 10 * log(result);
polar(theta,abs(result_dB))
polar(theta, angle(result_dB))
I hope it helps. Good luck!
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sixwwwwww
2013년 10월 18일
You can plot it but in case of negative values of result you should make two plots separately as I did. One plot will be for absolute value and other will be for phase value
Vivek Selvam
2013년 10월 17일
But careful about the gain conversion as given the tool description (polar_dB makes a plot of gain=10*log10(g) versus polar angles phi)
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Vivek Selvam
2013년 10월 18일
log10() is imaginary for negative values and negative for values between 0 and 1. You might want to solve that.
murat alboga
2014년 5월 14일
편집: murat alboga
2014년 5월 14일
Selamun aleykum everyone, hello... I have one questin abouth broadside array and polar pattern as dB. I have project in matlab.That is broadside array pattern.Below code, i achieved array factor as a magnitude, but when i converted array factor to dB graph is changed and corrupted. I'm tried figure out that problem but i didn't.Please can you help me... code:
N=5;
d=1;
k=2*pi;
theta=0:0.01:M+1;
psi=k.*d.*cos(theta);
AF=sinc((N.*psi./2)/pi)./sinc((psi./2)/pi);
figure;
polar(theta,AF);
mnAF=max(AF);
figure;
polar(theta,20*log10(AF));
array factor as magnitude
dB scale:
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Yasir Ahmed
2018년 1월 24일
Yes that happens because the array response in certain directions is very close to zero and on a logarithmic scale that's a big negative value. Polar plot can only handle values zero and above. So one way around this problem is to divide the vector by the minimum value of the vector so that on a log scale the minimum value is zero (20*log10(1)=0). This will work quite well if the range of values in the vector is not that big. For more visit:
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