# Estimating harmonic loudness

I'm working with data that looks something like this:

sec Hz amp
0.0 2630 0.001
0.1 2340 0.1
0.2 2220 1.0
0.3 2200 0.7
0.4 2200 0.2
0.5 2200 0.01

sec Hz amp
1.0 463 0.001
1.1 534 0.1
1.2 440 0.4
1.3 440 0.7
1.4 440 0.2
1.5 440 0.01


This data is for two separate harmonics. You can see that the frequency drifts a little with the attack of the note and that the amplitude beings to fade out towards the end of each harmonics. I'm not showing all the samples.

Here's what I'm doing to establish the relative loudness of these harmonic.

Harmonic 1 has a maximum amplitude in the frequency domain of 1.0. Harmonic 2 has a maximum amplitude in the frequency domain of 0.7. So there amplitude ratio is equivalent to 3.1 dB.

I now take into account that loudness is dependent on frequency and do so via the following table:

(freq,dbs)
(0, -50)
(20, -43)
(30, -33)
(40, -27)
(50, -23)
(60, -21)
(70, -19)
(80, -17)
(90, -16)
(100, -15)
(200, -8)
(300, -6)
(400, -6)
(600, -6)
(700, -7)
(800, -8)
(1500, -9)
(2000, -7)
(2200, -6)
(2500, -4)
(3000, -1.5)
(3700, 0)
(4000, -0.5)
(5000, -4)
(6000, -9)
(7000, -14)
(9000, -16)
(10000, -15)
(12000, -25)
(15000, -40)
(20000, -55)
(30000, -70)


Harmonic 1 has a frequency of 2200 Hz, which gives a value of -5 dB in the look up table. Harmonic 2 has a frequency of 440 Hz, which gives a value of -6 dB in the look up table. So that's a difference of 1 dB.

The total difference is then 3.1 dB + 1 dB = 4.1 dB. Right?

So a difference of 10 dB is equivalent to a doubling of loudness:

2^(dB/10)=increase in loudness, with
2^(4.1/10) = 1.33


So the first harmonic is approximately 33% louder than the second.

Am I on the right track with regard to estimating the difference in loudness? Any obvious mistakes. Any obvious improvements?

• You should rather use a smooth curve rather than such a badly quantized lookup table. – leftaroundabout Mar 27 '12 at 22:57