How to find damping ratio from halfpower method?
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xout=acc_u_2(:,1);
fs=20;
m = [length(xout), 2^(nextpow2(length(xout)))];
N = 12000;
xdft = fft(xout,N);
xdft = xdft(1:N/2+1);
psdx = (1/(fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:(fs/N):(fs/2);
plot(freq,10*log10(psdx))
xlabel('Frequency [Hz]', 'FontSize', 12)
ylabel('PSD [V$^2$/Hz]', 'FontSize', 12)
grid on
fr1=psdx(2:1201); fr2=psdx(1202:2401); fr3=psdx(2402:3601); fr4=psdx(3602:4801); fr5=psdx(4802:6001);
psd_peaks=[max(10*log10(fr1)) max(10*log10(fr2)) max(10*log10(fr3)) max(10*log10(fr4)) max(10*log10(fr5))];
index = [find(10*log10(psdx)==psd_peaks(1)) find(10*log10(psdx)==psd_peaks(2)) find(10*log10(psdx)==psd_peaks(3)) find(10*log10(psdx)==psd_peaks(4)) find(10*log10(psdx)==psd_peaks(5))];
freq_true = freq(index) %given frequencies from the peaks
The above code gives me the following PSD plot and i have calculated the natural frequencies. I now need to use the half power method to calculate the damping ratio;
from the documentation i used the following code lines
bw = powerbw(psdx,freq_true)
bw = powerbw(10*log10(psdx),freq_true)
However they produced input errors, any help would be appreciated.
3 commentaires
dpb
le 1 Mai 2019
Well, it tells you the BW for whichever peak it worked on (the maximum I presume if you passed the full spectrum) is 0.0025 Hz. Given the sharp nature of the peaks, for -3 dB that doesn't look unreasonable number from your plot.
If you return the additional optional outputs, it'll also tell you the low/high frequencies it calculated and the power contained therein...
Read the documentation carefully to undersrtand the inputs/outputs and read the Algorithm section to understand what it's actually doing behind the scenes...
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