I have a doubt about the unit of amplitude on y axis on linear plot of a sound wave generated in Matlab.
In logarithmic scale it is dB but I am confused about what it should be on the linear scale and what is the maths behind the conversion that would be a plus if someone has any idea bout it.
In this image the value of power at 404.2 Hz in -1.26 dB on log scale and 404.2 on linear scale..What is the maths behind this conversion and what should be the ideal unit on Y axis on linear scale?
I currently use "V" which I think is wrong.
Also I have attached section of my code which is generating this plot.
--------------------------
fSep=1/(N*h);
disp(['FFT frequency separation = ' num2str(fSep) ' Hz'])
[Y,f,YdB]=SimpleLineSpectrum2(y,h,0,fny);
figure
subplot(211)
plot(f,Y,'LineWidth',1.5),grid
title(['Line Spectrum of ' name ],'fontweight','bold','fontsize',10)
ylabel('Linear power(V)')
xlabel('Frequency (Hz)')
axis([fL fR 0 max(Y)])
subplot(212)
plot(f,YdB,'LineWidth',1.5),grid
title('Line Spectrum in dB','fontweight','bold')
ylabel('power (dB)')
xlabel('frequency (Hz)')
axis([fL fR dBmin 0])
pause
--------------------------------------------------------
and this is what "Simplelinespectrum " is doing
% normalize by 2*np
np=length(y); % length of the input vector
%===remove the mean
yavg=mean(y);
y=y-yavg;
%===calculate the fast Fourier transform
Y=fft(y(1:np));
%===generate the frequencies in the frequency grid
nint=fix(np/2);
i=1:nint+1;
f(1:nint+1)=(i-1)/(np*h);
%===take the absolute value and normalize
Ya=2*abs(Y(1:nint+1))/(np); % normalization
% Ya=abs(Y(1:nint+1)); % no normalization
%===generate the frequencies and magnitudes to be
% returned consonant with frLeft and frRight
fL=frLeft;
iL=fix(1+np*h*fL);
fR=frRight;
iR=fix(1+np*h*fR);
retYa=Ya(iL:iR);
retf=f(iL:iR);
Ymax=max(retYa);
YdB=20*log10(retYa/Ymax);
-------------------------------------------------
Dear concern,
dB is used to quantify the ratio of two values in logarithmic scale which conveniently represent very large or small numbers on a same scale.
For converting the ratio of two power values in dB, we use ans(dB) = 10*log10(ratio) and ratio=10(ans(dB)/10).
For converting the ratio of two voltage or current values in dB, we use ans(dB)=20*log10(ratio) and ratio=10(ans(dB)/20).
You can verify the concept of multiplicative factor of 20 and 10 using P = V2/R = I2*R and taking R=1.
For your information, dBm is used to quantify the value of Power (not a ratio) in logarithmic scale which is generally used in wireless communication and other areas.
Hope you find this answer hepful.
For converting the ratio of two power values in dB, we use
ans(dB) = 10*log10(ratio)
ratio=10(ans(dB)/10).
Now if the ratio=power/power and the unit of power is in WATT we convert it to dB scale and if the unit is in miliWatt we get dBm in log scale. Even though there is no unit of the ratio , we still write dBm for power in mW unit.
So if u convert from dB to linear, the unit of power will be in Watts.
Dear concern,
dB is used to quantify the ratio of two values in logarithmic scale which conveniently represent very large or small numbers on a same scale.
For converting the ratio of two power values in dB, we use ans(dB) = 10*log10(ratio) and ratio=10(ans(dB)/10).
For converting the ratio of two voltage or current values in dB, we use ans(dB)=20*log10(ratio) and ratio=10(ans(dB)/20).
You can verify the concept of multiplicative factor of 20 and 10 using P = V2/R = I2*R and taking R=1.
For your information, dBm is used to quantify the value of Power (not a ratio) in logarithmic scale which is generally used in wireless communication and other areas.
Hope you find this answer hepful.
Hi,
it seems that your data is not power but voltage as called "linear power (V)" in the top image.
The derivation of the level in dB would be 20 * log(V)
In addition, the dB values are shifted to get relative values and a maximum of 0 dB by dividing the voltage by the voltage maximum value.
This is done in the last line of your program:
YdB=20*log10(retYa/Ymax);
Jacques
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