Sound Power Level (SPL) of a loudspeaker

[Rorshach]

This is a simple problem from a textbook I am reading, and everything below is written word by word and sign by sign from said textbook. Formulas given in the book just don't give the result authors claim they do:

Homework Statement

An input of 1 W produces a SPL of 115 dB at 1 m. What is the SPL at 6.1 m?

Homework Equations

The Attempt at a Solution

SPL = 115 - 10log(0.22/1) = 115 - 15.7 = 99.3 dBThe assumption made in the 20log(6.1) factor is that the loudspeaker is operating in a free field and that inverse square law is valid in this case. This is a reasonable assumption for a 20-ft distance if the loudspeaker is remote from reflecting surfaces. A loudspeaker is rated at a sound- pressure level of 115 dB on axis at 1 m with 1 W into 8Ω. If the input were decreased from 1 to 0.22 W, what would be the sound pressure level at 1 m distance?

SPL = 115 - 10log(0.22/1)
= 115 - 6.6
= 108.4 dB

Note that 10log is used because two powers are being compared.

 

[TSny]

SPL = 115 - 10log(0.22/1) = 115 - 15.7 = 99.3 dBThe assumption made in the 20log(6.1) factor...

Looks like the book has a typographical error. It appears that the quantity given in green was meant to be 20log(6.1).

 

[Rorshach]

I am sorry, in that case it was my typo. In the book it stands 20log(6.1). It's the second equation that doesn't work out. I made sure it is exactly like in the textbook, so no typo there.

 

[TSny]

Are you saying that the following numbers don't work out?

SPL = 115 - 10log(0.22/1)
= 115 - 6.6
= 108.4 dB

It looks OK to me. Never mind, there is a sign error. The equation SPL = 115 - 10log(0.22/1) should not have a minus sign on the right. It should be positive. That is, it should be

SPL = 115 + 10log(0.22/1) = 115 - 6.6 = 108.4 dB

 

[Rorshach]

that is right. I have a question: When should it be minus sign and when should it be plus? What determines that?

 

[TSny]

that is right. I have a question: When should it be minus sign and when should it be plus? What determines that?

The change in sound level is always 10Log(I_f/I_i) where I_f is the final intensity and I_i is the initial intensity.

For a point source of power P, I = P/(4πr²). So, if you keep the power constant while changing r, you have that the change in sound level is

10Log(I_f/I_i) = 10Log(r_i²/r_f²) = 20Log(r_i/r_f) = -20Log(r_f/r_i)

 

[Rorshach]

but here we are dealing with pressure, and it would have to be SPL = 115 + 10log(0.22/1) for the correct result- how do I correlate power and pressure correctly?

 

[TSny]

If p is the "sound pressure" (i.e., the amplitude of the pressure variation in a sound wave), then it turns out that the intensity of the sound, I, is proportional to p². So, the sound pressure level can be written as

SPL = 10 Log[I/I₀] = 10 Log[p²/p₀²] = 20 Log[p/p₀]

The "0" subscript refers to some standard reference level (usually corresponding to the threshold of hearing). If you are only interested in changes in SPL, then

ΔSPL =10 Log[I_f/I_i] = 20 Log[p_f/p_i].

In your particular problem, you did not have to work with any sound pressure values, p.

 

[Rorshach]

I still don't understand how I'm supposed to get
SPL = 115 + 10log(0.22/1)

In the first equation it was very intuitive, and produced expected result. But second equation just makes no sense:
in the logarithm (just like authors state in the book is 10log, because two powers are being compared) we have a fraction with wattage in nominator (after change, 0.22W) and denominator (before change, 1W, refference value). And for this equation to produce expected result we have to have either - 10log(1/0.22), which makes no sense since refference value is in nominator, or +log10(0.22/1), which I have no idea where it came from.

 

[TSny]

I'm not sure where you are having difficulty.

In terms of power changes, the change in SPL is

ΔSPL = 10 Log[(final power)/(initial power)]

So, in your specific problem, ΔSPL = 10 Log[(.22 W)/(1 W)] = 10 Log[.22/1] = -6.6 decibels

Logarithms obey the property Log[a/b] = - Log[b/a]

So, Log[(final power)/(initial power)] = - Log[(initial power)/(final power)]

So, you can just as well write the change in SPL as

ΔSPL = - 10 Log[(initial power)/(final power)] = - 10 Log [1/(.22)] = -6.6 decibles.

 

[Rorshach]

I mean the general formula for this would be:
SPL = G - ΔSPL
where G is a given value of SPL at a given distance.

and all the signs in the formula are the same, only values of SPL, G and ΔSPL change and can be negative or positive, right?

 

[TSny]

Usually, the symbol Δ indicates a change given by the final value minus the initial value. Thus,

ΔSPL = SPL_f - SPL_i.

Rearranging this gives

SPL_f = SPL_i + ΔSPL.

Note the plus sign on the right hand side.

ΔSPL is related to the change in intensity, I, of the sound according to ΔSPL = 10Log[I_f/I_i].
If I_f > I_i then ΔSPL is positive. If I_f < I_i then ΔSPL is negative.

So, SPL_f = SPL_i + 10Log[I_f/I_i].