# Where does this decibel equation come from, why a factor of 20?

For sound levels, it is said that decibels follow this equation:

$$dB = 20 \log(\frac{p_s}{p_0})$$

Where $$\p_s\$$ is the pressure of the measured sound wave (in Pascals) and $$\p_0\$$ is some fixed reference pressure.

But where does the 20 come from in this particular equation?

I know that the 10 comes from the equation when relating quantities of power:

$$\dB = 10 \log(P / P_0)\$$ for measured power $$\P\$$ and reference power $$\P_0\$$.

So for something like voltage, where $$\P = V^2/R\$$, you have $$\dB= 10 \log((V^2/R) / (V_0^2/R)) = 10 \log((V/V_0)^2) = 20 \log(V/V_0)\$$ so I understand that if you have some squared quantity in there you can pull out the exponent of 2 in the log function and that gets you the multiplier of 2 on the outside, hence the 10 becoming 20 in the case of voltage rather than power.

But in the sound equation I can't derive the origin of the 20 by somehow relating power to pressure. I imagine power is proportional to the square of pressure but I can't figure out where, when it comes to sound waves.

• Thinking this through (and I could be wrong...), well, dB is always a power ratio, just by definition. Volts are only one component of power (volts * amps = power). Similarly, pressure is not power. Pressure * flow is power in a hydraulic system. So it is 20 * (log(pressure) / log(pressure). – mkeith Jan 20 at 4:59
• @mkeithYeah. dB is always a power ratio. So when you convert linear power ratios to dB you just use 10. But since power is proportional to square of amplitude (ampltidues including SPL, voltage, and current) when you use voltage or current ratios to calculate dB, you use 20. Because although you are using voltage or current ratios to calculate dB, you are actually converting it to a power ratio because dB is always a power ratio. – DKNguyen Jan 20 at 5:01
• @DKNguyen I had a question about that as well but didn't know if it was worth a new question or not, but is there an equation that relates amplitude and power so that there is a decibel relationship from amplitude? – user525966 Jan 20 at 5:02
• It's the one you are using with the 20. You basically use the 20 whenever your ratio is some unit whose squared is proportional to the power. Voltage and current are both amplitudes so they apply. With sound, I think pressure is the amplitude (whatever the case, it follows the squared power relation) – DKNguyen Jan 20 at 5:03
• I get that aspect, what I mean is the equation that relates sound power and sound amplitude, i.e. how do we know power is proportional to the square of amplitude? – user525966 Jan 20 at 5:04

Sound power ($$\P\$$) is proportional to the sound intensity ($$\I\$$):

$$P = A\ I$$

where $$\A\$$ is the surface area.

And sound intensity ($$\I\$$) is proportional to the square of sound pressure:

$$I = \frac{p^2}{z_0}$$

where $$\p\$$ is the sound pressure level (SPL) and $$\z_0\$$ is the specific acoustic impedance (think of it as something like a characteristic impedance). Just like in $$\\mathrm{P = V^2/R}\$$.

Putting all together yields us $$dB = 10 \log (\frac{P}{P_0}) = 20 \log (\frac{p}{p_0})$$

This will be too mathematical, so be warned!

We all know that the work ($$\W\$$) is defined by the product of the force ($$\F\$$), and the displacement caused by that force ($$\dx\$$):

$$W = \int \mathbf F \ dx = \int \mathbf F \ \mathbf v\ dt$$

and thus, the power is

$$P = dW/dt = \mathbf F \ \mathbf v$$

So, sound power ($$\P_s\$$) is calculated in the same manner:

$$P_s = \mathbf {F_s \ v_p}$$

where $$\\mathbf {F_s}\$$ is the sound force vector and $$\\mathbf {v_p}\$$ is the particle velocity vector (or, particle flow rate). The particles here are the particles inside the medium the force is applied (e.g. air molecules).

Since we know that the pressure is defined as the applied force per unit area, we obtain

$$\mathbf {F_s} = p \ \mathbf A \\ \therefore P_s = p \ \mathbf A \ \mathbf{v_p} = \mathbf{A\ I}$$

where $$\p\$$ is the sound pressure level (SPL), $$\\mathbf A\$$ is the surface area vector. Note that the term $$\p \ \mathbf{v_p}\$$ is replaced with $$\\mathbf I\$$, which as defined as Sound Intensity (i.e. the sound power applied to a surface area).

When the force $$\\mathbf {F_s}\$$ is applied to an area $$\\mathbf A\$$ in a medium (results in the acoustic pressure, $$\p\$$), the particles in that medium will flow with a rate (flow rate, $$\\mathbf v\$$) but present opposition to that pressure. This opposition is called acoustic impedance, $$\z\$$. Just like the impedance in electricity: The opposition to electron flow caused by the voltage (yes, voltage is a force!).

$$p = \mathbf {z \ v}$$

The final results can be obtained by using the absolute values of the vectors above. Also, remember that the particles' velocities are caused by the sound force itself, so the absolute values (i.e. magnitudes) instead of the vectors can be used.

So,

$$p = z \ v \Rightarrow v = p / z \\ I = p \ v \Rightarrow I = p (p / z) = p^2/z \\ P = A \ I = A \ p^2 / z$$

Finally,

$$\mathrm{ dB = 10 \log (P/P_0) =} 10 \log (\frac{A\ p^2/z} {A\ p_0^2/z_0})$$

Where $$\P_0\$$ is the reference sound power and $$\p_0\$$ is the reference SPL. Now $$\z = z_0\$$ here for the same medium (characteristic sound impedance); just like the R in V²/R (same load).

Putting all together yields

$$\mathrm{ dB = 10 \log (P/P_0)} = 10 \log (\frac{A\ p^2/z} {A\ p_0^2/z_0}) = 20 \log (\frac{p} {p_0})$$

I hope these are enough and satisfactory.

• How do we know power is proportional to intensity, or that intensity is proportional to the square of sound pressure? That's more or less what my question is after in the first place - I know that the 20 has to come from that relationship being a square one, but I don't know where that relationship itself comes from. – user525966 Jan 20 at 5:06
• @user525966 Good question. Try asking on the physics stack exchange: "How do we know sound pressure squared is proportional to the power in the sound wave?" – DKNguyen Jan 20 at 5:08
• @user525966 so you need the full proof of the definitions? Alright then, I'm editing my answer. – Rohat Kılıç Jan 20 at 6:28
• @user525966 see the edit. – Rohat Kılıç Jan 20 at 8:01
• I thought voltage was more of a "electric potential difference" or "electric pressure" and not a force in itself? I'm also still unclear how this results in a squared relationship that would explain the factor of 20. Part of it may be my inexperience with this stuff, I'll need to re-read this slowly. Thank you for the detailed addition to the post, upvoted. – user525966 Jan 20 at 17:53