dB scale (log scale) of a polar plot graph
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atarli on 17 Oct 2013
Answered: Yasir Ahmed on 24 Jan 2018
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
Jonathan LeSage on 17 Oct 2013
Edited: Jonathan LeSage on 17 Oct 2013
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!
Ibrahim Samy on 24 Oct 2017
i think you better use polardb(theta , result, -20, '-k')
-20 for minimal db u can see
sixwwwwww on 17 Oct 2013
Edited: sixwwwwww on 17 Oct 2013
Here is the conversion to decibel:
result_dB = 10 * log(result);
I hope it helps. Good luck!
sixwwwwww on 18 Oct 2013
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 on 17 Oct 2013
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)
Vivek Selvam on 18 Oct 2013
log10() is imaginary for negative values and negative for values between 0 and 1. You might want to solve that.
murat alboga on 14 May 2014
Edited: murat alboga on 14 May 2014
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:
array factor as magnitude
Yasir Ahmed on 24 Jan 2018
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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