dB scale (log scale) of a polar plot graph

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)

Answer 1

[Jonathan LeSage]

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:

• http://en.wikipedia.org/wiki/Decibel
• http://www.mathworks.com/help/signal/ref/mag2db.html

Once you have the results in terms of decibels, you can plot using the polar function as you did before!

[atarli]

thanks for all of your helps, i thought to plot it with polar(theta,log10(result)) command. what do you think about it?

[Ibrahim Samy]

i think you better use polardb(theta , result, -20, '-k')

-20 for minimal db u can see

Answer 2

[sixwwwwww]

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!

[Azzi Abdelmalek]

What about negative values?

[sixwwwwww]

For negative amplitude values we will complex dB which can be used as above. Thanks for correction

[atarli]

thanks for all of your helps, i thought to plot it with polar(theta,log10(result)) command. what do you think about it?

[sixwwwwww]

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

Answer 3

[Vivek Selvam]

You can try this: Polar Plot Tool - dB

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)

[atarli]

thanks for all of your helps, i thought to plot it with polar(theta,log10(result)) command. what do you think about it?

[Vivek Selvam]

log10() is imaginary for negative values and negative for values between 0 and 1. You might want to solve that.

Answer 4

[atarli]

thanks for all of your helps, i thought to plot it with polar(theta,log10(result)) command. what do you think about it?

Answer 5

[murat alboga]

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:

Answer 6

[Yasir Ahmed]

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:

http://www.raymaps.com/index.php/fundamentals-of-a-uniform-linear-array-ula/