Understanding decibels

Given two power quantities \(a\) and \(b\), the decibels can be calculated from their ratio using the following formula.

\begin{equation} \label{eq:power-ratio-to-decibels} \mathrm{dB} = 10 \log_{10} \left( \frac{a}{b} \right) \\ \end{equation}

Therefore, given the decibels, the ratio between the two values can also be obtained using a similar formula.

\begin{equation} \label{eq:decibels-to-power-ratio} \frac{a}{b} = 10^{\frac{\mathrm{dB}}{10}} \end{equation}

However, if those values are root-power quantities, a factor of 20 must be used.

\begin{equation} \label{eq:root-power-ratio-to-decibels} \mathrm{dB} = 20 \log_{10} \left( \frac{a}{b} \right) \\ \end{equation}

Again, given the decibels, the ratio can also be obtained using 20 as the factor.

\begin{equation} \label{eq:decibels-to-root-power-ratio} \frac{a}{b} = 10^{\frac{\mathrm{dB}}{20}} \end{equation}

To understand why 20 must be used for certain units, we can inspect the equivalence between the two formulas for watts (power quantity) and volts (root-power quantity). First, it’s important to understand that power is proportional to the squared voltage, assuming the load is constant.

\begin{align*} P = V \times I = V \times \frac{V}{R} = \frac{V^2}{R} \end{align*}

This equivalence can be replaced in the formula of decibels for power quantities (again, assuming both power quantities are under the same load), resulting in the previous formula for root-power quantities.

\begin{align*} \mathrm{dB} &= 10 \log_{10} \left( \frac{P_2}{P_1} \right) \\ &= 10 \log_{10} \left( \frac{V_2^2/R}{V_1^2/R} \right) \\ &= 10 \log_{10} \left( \left( \frac{V_2}{V_1} \right)^2 \right) \\ &= 10 \cdot 2 \log_{10} \left( \frac{V_2}{V_1} \right) \\ &= 20 \log_{10} \left( \frac{V_2}{V_1} \right) \end{align*}

One common misunderstanding about decibels stems from the fact that the input signal is usually not measured in volts, but in “absolute” decibels using a constant reference value; in other words, the input is often expressed in dBV or dBu. Note, however, that the amplification is expressed in dB, since it measures two variables (input and output signals).

Here are the formulas for converting from voltage to different “suffixed decibel” units.

Suffix	Formula	Example
dBV	\[20 \log_{10} \left( \frac{V_{\mathrm{in}}}{1} \right)\]	\[20 \log_{10} \left( \frac{2}{1} \right) = 20 \log_{10}(2) \approx 6.02\]
dBu	\[20 \log_{10} \left( \frac{V_{\mathrm{in}}}{\sqrt{0.6}} \right)\]	\[20 \log_{10} \left( \frac{2}{0.774} \right) \approx 20 \log_{10}(2.581) \approx 8.23\]
dBFS	\[20 \log_{10} \left( \frac{V_{\mathrm{in}}}{V_{\mathrm{max}}} \right)\]	\[20 \log_{10} \left( \frac{2}{4.5} \right) \approx 20 \log_{10}(0.444) \approx -7.04\]

All of these values express the same quantity, an input voltage of 2 volts RMS, but with different units. The key detail is that the input value can’t be expressed in decibels without another reference value, which is usually constant and varies depending on the unit. Also notice how a factor of 20 is used in the formulas, since we are dealing with a root-power quantity, volts, and not a power quantity like watts.

Since the input voltage and the gain of the amplifier are known, the output voltage can be obtained with the formula from the previous section. First, the ratio between the input and output voltages is calculated by replacing the amplifier gain, 5 dB.

\begin{align*} \frac{V_{\mathrm{out}}}{V_{\mathrm{in}}} &= 10^{\frac{\mathrm{dB}}{20}} \\ &= 10^{\frac{5}{20}} \\ &= 10^{0.25} \\ &\approx 1.778 \end{align*}

Then, the input voltage of 2 volts RMS is replaced to obtain the output voltage.

\begin{align*} V_{\mathrm{out}} &\approx 1.778 \times 2 \\ &\approx 3.556 \end{align*}

Notice how the output voltage grows logarithmically, not linearly; if the input of 2 volts RMS was directly multiplied by the gain of 5 dB, the output would become 10 volts RMS.

The following figure plots the resulting sinusoidal wave after being amplified by a gain of 5 dB. Again, its peak and RMS values are highlighted.

From the plot, we can also observe that the RMS value of the output is below the clip limiter, while the peaks are above it and would not be processed by the device.

Similarly to how the input voltage was converted to dBV, the same can be done to the output voltage.

\begin{split} \mathrm{dBV} &= 20 \log_{10} \left( \frac{V_{\mathrm{out}}}{1} \right) \\ &\approx 20 \log_{10} \left( \frac{3.556}{1} \right) \\ &\approx 20 \log_{10}(3.556) \\ &\approx 20 \times 0.551 \\ &\approx 11.02 \end{split}

Just like with any other value expressed in decibels, this 11.02 expresses a ratio between two values, in this case the output voltage and a constant reference value of 1 volt. This new value, although expressed in decibels, is completely independent from the input or the amplification, it simply indicates a ratio between a signal level and a constant.

Also note that, by subtracting the input and output dBV, the gain can be obtained. This can be done with a simple subtraction because the input and output dBV values were calculated using the same logarithmic scale, which is also used to express the gain of the amplifier. In other words, a value expressed in “suffixed” decibels can be added to another value in “plain” decibels to obtain a result back in “suffixed” decibels because all decibels use the same logarithmic formula \eqref{eq:power-ratio-to-decibels}.