Overall Sound Pressure Level between two Frequencies

[roldy]

Homework Statement

The band levels for a noise are 70 dB reference 20$$\mu$$Pa at 31.5 Hz and decrease 3 dB for each octave. Find the overall sound pressure level between 31.5 and 8000 Hz and the A-weighted sound pressure level for this frequency range.

Homework Equations

$$SPL=20\frac{P_{rms}}{P_{ref}}$$

The Attempt at a Solution

I think my misunderstanding of the problem is prohibiting me from calculating the correct answer.

I used excel to do the calculations

The band level noise for each octave are as follows

31.5 Hz: 70 dB
63 Hz: 67 dB
125 Hz: 64 dB
250 Hz: 61 dB
500 Hz: 58 dB
1000 Hz: 55 dB
2000 Hz: 52 dB
4000 Hz: 49 dB
8000 Hz: 46 dB

The reason I did this was because the problem stated that the band level noise decreased 3 dB for each octave.

Then I added in the corrections to the dB values above. The correction values for each octave band level are as follows.

31.5 Hz: -39.4 dB
63 Hz: -26.2 dB
125 Hz: -16.1 dB
250 Hz: -8.6 dB
500 Hz: -3.2 dB
1000 Hz: 0 dB
2000 Hz: 1.2 dB
4000 Hz: 1.0 dB
8000 Hz: -1.1 dB

The final dB values:
31.5 Hz: 30.6 dB
63 Hz: 40.8 dB
125 Hz: 47.9 dB
250 Hz: 52.4 dB
500 Hz: 54.8 dB
1000 Hz: 55 dB
2000 Hz: 53.2 dB
4000 Hz: 50 dB
8000 Hz: 44.9 dB

I then square them
31.5Hz: 936.36
63 Hz: 1664.64
125 Hz: 2294.41
250 Hz: 2745.76
500 Hz: 3003.04
1000 Hz: 3025
2000 Hz: 2830.24
4000 Hz:2500
8000 Hz: 2016.01

Then the average
2335.051111
and then square root
48.32236657

This gives me the rms pressure value for 31.5 to 8000 Hz. I then use the equation
$$SPL=20log\frac{P_{rms}}{P_{ref}}$$ to find the overall sound pressure level. The answer I get is 127.66 dB. The correct answer is 73.0 dB and the A-weighted value is 60.8 dBA. What did I do wrong?

 

[Steve4Physics]

This is an old (12+ years at time of answering) question. Since communications with the OP is unlikely I’m posting a fairly detailed answer.

The OP appears to have ssumed that the overall dB value depends on the average of the bands’ intensities. This is wrong. The overall dB values depends on the sum of the bands’ intensities.

It’s worth noting that intensity is proportional to pressure squared. And squaring (in a later step) can be avoided by remembering $$20log\frac{P_{rms}}{P_{ref}} = 10log\frac{P_{rms}^2}{P_{ref}^2}$$We can first find the unweighted (no A-weighting) dB values as follows:

As noted by the OP, the band levels are
31.5 Hz: 70 dB
63 Hz: 67 dB
125 Hz: 64 dB
250 Hz: 61 dB
500 Hz: 58 dB
1000 Hz: 55 dB
2000 Hz: 52 dB
4000 Hz: 49 dB
8000 Hz: 46 dB

We can calculate the value of $\frac{P_{rms}^2}{P_{ref}^2}$ for each band using $10^{\frac {dB}{10}}$. This gives the following values:
31.5Hz: 10000000
63 Hz: 5011872.33627272
125Hz: 2511886.43150958
250Hz: 1258925.41179417
500Hz: 630957.344480193
1kHz: 316227.766016838
2kHz: 158489.319246111
4kHz:79432.8234724282
8kHz: 39810.7170553497

The total of these is 20007602.1498474. Converting back to dB gives
10log(20007602.1498474) = 73.0dB matching the official answer.

To find the dB(A) values, the procedure is the same, except that we start from the A weighted values correctly given by the OP in Post #1:
31.5 Hz: 30.6 dB
63 Hz: 40.8 dB
125 Hz: 47.9 dB
250 Hz: 52.4 dB
500 Hz: 54.8 dB
1000 Hz: 55 dB
2000 Hz: 53.2 dB
4000 Hz: 50 dB
8000 Hz: 44.9 dB