means squared error and dB
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For curves in X-Y coordinate system where the Y axis is in decibels (dB), I want to calculate the mean squared error (MSE) between two curves? If the Y values were linear, I could use MSE equation. Can I use MSE when the y values are in dB or should I converts the Y values from dB to linear then compute the MSE and take it to dB?
For instance, Y1 is the y values (in dB) of one curve. Then, Y2 is Y1 values plus or minus a random number from -5 to 5. Therefore, Y2 samples are X dBs above or below those of Y1 where X is in the [-5,5] dB interval.
1) Now that two curves Y1 and Y2 are given in dB, to calculate their MSE, can I use the MSE equation directly or linear conversion is needed (script below)? If the Y values in dB, is there a better way to quantify the average difference than MSE?
Y1=-60:1:60; %y values in dB for curve 1
Y2=Y1+randi([-5,5],size(Y1)) %y values in dB for curve 2
MsedB1 = sum((Y1-Y2).^2)/numel(Y1); %is this MSE correct?
Y1_Lin = 10.^(Y1*0.1); %y values for curve 1 in linear
Y2_Lin = 10.^(Y2*0.1); %y values for curve 2 in linear
MsedB2 = 10*log10(sum((Y1_Lin-Y2_Lin).^2)/numel(Y1_Lin)); %or is this MSE correct?
1 个评论
Mathieu NOE
2020-12-31
hello
the MSE can be expressed in dB, computed as 10*log10(power2/power1) where power1 = sum((Y1_Lin.^2) and power2 = sum((Y2_Lin.^2)
there is no reason to divide by numel(Y1_Lin)
if you want , there is nothing that speaks against defining your own metrics like an averaged dB difference between the two datas
it may give very similar values compare to a "true" MSE in dB and will save a few conversions from dB to lin and vice versa
but your code seems incorrect to me :
MsedB2 = 10*log10(sum((Y1_Lin-Y2_Lin).^2)/numel(Y1_Lin)); %or is this MSE correct? NO !!
采纳的回答
David Goodmanson
2021-1-1
Hi MatG,
As with a lot of things, the answer depends. In your example, which is the first case below, the variation in power is not very extreme, which means that the two methods will not differ by very much. On the other hand, the second case is for a square wave that spends half its time around the value 1, and the other half of the time around the value 0, plus a bit of noise. Here the difference between the two methods is very large.
t = 0:.001:2*pi;
n = length(t);
r = 2*rand(1,n)-1; % random between +-1
% case 1
P = 10.^((5/10)*r);
figure(1)
plot(t,P)
grid on
mean_of_P_in_dB = 10*log10(mean(P))
mean_of_dB = mean(10*log10(P))
% mean_of_P_in_dB = 0.9531
% mean_of_dB = 0.0396
% case 2
y = (1+square(t))/2; % 1 or 0
P = (y +.001*r).^2; % add small amount of noise, 60 dB down
figure(2)
plot(t,P)
grid on
mean_of_P_in_dB = 10*log10(mean(P))
mean_of_dB = mean(10*log10(P))
% mean_of_P_in_dB = -3.0102
% mean_of_dB = -34.3295
In the second case, the average power is about 1/2, so the mean of P converted to dB is -3 dB. That is a reasonable evaluation of the situation. When direct averaging dB, the signal spends half its time at 0 dB and half down in the region of -60 dB, leading to the somewhat absurd conclusion that the average power is -34 dB. Make the noise 10 times smaller than that, and the supposed average power is now -44 dB. This is not reasonable. Most cases are like your example, though, and averaging dB can make sense.
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