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)
1 个评论
murat alboga
2014-5-14
移动：Voss
2024-6-30
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:
回答（5 个）
Jonathan LeSage
2013-10-17
编辑：Jonathan LeSage
2013-10-17
0 个投票
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!
2 个评论
atarli
2013-10-18
Ibrahim Samy
2017-10-24
i think you better use polardb(theta , result, -20, '-k')
-20 for minimal db u can see
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!
4 个评论
Azzi Abdelmalek
2013-10-17
What about negative values?
sixwwwwww
2013-10-17
For negative amplitude values we will complex dB which can be used as above. Thanks for correction
atarli
2013-10-18
sixwwwwww
2013-10-18
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
2013-10-17
0 个投票
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)
2 个评论
atarli
2013-10-18
Vivek Selvam
2013-10-18
log10() is imaginary for negative values and negative for values between 0 and 1. You might want to solve that.
Yasir Ahmed
2018-1-24
0 个投票
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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