I came across this problem that i am having a hard time solving. It is a bit technical..

Say you have two speakers outputting the exact same signal, and you place yourself just in between the two speakers. And let's pretend the combined SPL of these speakers is 85 dB. When you turn one of them off, what is the SPL at your position?

When I first tried solving this, I used the equation 85 = 20Log(x / 0.00002), x being the total pressure level. Rearranging and calculating gives approx. x = 0.3556 Pa. Halving x and putting it into the same formula gives an SPL of approx. 78.98 dB, a 6 dB decrease.

My problem is, that when I used an SPL meter between two real speakers, and turned one of them off, the measured decrease was only 3 dB.

This corresponds to the equation 85 = 10Log(10^(x/10) * 2), the formula for adding SPLs from different sound sources, which gives approx. x = 81.99.

I have re-checked the calculations and they are correct, assuming I am using the right formulas. Can anyone explain this?

Grateful for any help.

fagerhult wrote:

I came across this problem that i am having a hard time solving. It is a bit technical.. Say you have two speakers outputting the exact same signal, and you place yourself just in between the two speakers. And let's pretend the combined SPL of these speakers is 85 dB. When you turn one of them off, what is the SPL at your position? When I first tried solving this, I used the equation 85 = 20Log(x / 0.00002), x being the total pressure level. Rearranging and calculating gives approx. x = 0.3556 Pa. Halving x and putting it into the same formula gives an SPL of approx. 78.98 dB, a 6 dB decrease. My problem is, that when I used an SPL meter between two real speakers, and turned one of them off, the measured decrease was only 3 dB. This corresponds to the equation 85 = 10Log(10^(x/10) * 2), the formula for adding SPLs from different sound sources, which gives approx. x = 81.99. I have re-checked the calculations and they are correct, assuming I am using the right formulas. Can anyone explain this? Grateful for any help.

A halving of two acoustic sources does indeed equate to -3dB. However a
halving of two voltages would be -6dB.
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Recording Studio Suntans