A.24  Decibel (dB) and Related Definitions

The decibel (dB) expresses a ratio between two powers $P_1$ and $P_2$:

Hence, dB is a relative, not an absolute quantity. If $T_{\mathrm{\text{dB}}}=0$, then the powers are equal. If $T_{\mathrm{\text{dB}}}$ is positive, $P_1$ is larger than $P_2$ and vice-versa if $T_{\mathrm{\text{dB}}}<0$.

There are other definitions²² related to dB such as dBW, which is the relative power in dB of a signal with respect to 1 watt (W). An important distinction is that, while dB should always be a “relative” measure indicating a ratio of powers, dBW is an “absolute” measure that indicates one specific power value. For example, assuming that $P$ is power in watts, its conversion to $\widehat{P}$ in dBW is given by

For example, 100 W corresponds to 20 dBW.

Similarly, dBm is defined with respect to 1 milliwatt (mW). Assuming that $P$ is power in watts,

For example, if a receiver detects a signal level of $-13$ dBm, it means this signal has a power that is 13 dB smaller than a milliwatt. For example, on a 3G cell phone using five “bars” to indicate signal strength, the 1-bar may correspond to $-113$ dBm and 5-bars to $-100$ dBm.

Another definition derived from dB is dBc, which is related to the presence of a carrier, such as in radio communications. In this case, the reference is the power of the strongest carrier. Therefore, typically, the dBc value of a signal component is negative.

A more complicated definition is dBm0, used for example in audio and telephony. It means the level compared to a milliwatt after the value is adjusted to make a reference (“correct”) value be 0 dBm.

One unfortunate fact is that sometimes dB is erroneously interpreted as the absolute value of a signal power. For example, in sound engineering, the dB sound pressure level or dB SPL is widely used and it represents the ratio between the measured sound pressure level and the reference point. However, the “SPL” suffix and the reference power value are often omitted and dB may be incorrectly understood as an absolute power value in these cases.

In digital signal processing, most of the time the correct unit is unknown and, consequently, there is no interest on specifying the reference value. If the signals are assumed to be in volts and obtained from a resistance of 1 ohm, in many cases the correct unit would be dBW instead of dB. For example, in spectral analysis, it is common to convert a power spectral density $S(f)$ in W/Hz to $10{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_{10}S(f)$, which should be interpreted as dBW/Hz. For example, the periodogram.m function in Matlab/Octave shows graphs in dB/Hz but the informed unit could be dBW/Hz.

Sometimes dB is used to express voltage ratios, not power ratios. For example, assuming purely resistive impedance $R$, the power associated to a voltage $V$ is $P=V^2∕R$. In this case Eq. (A.112) can be written as

Both voltage values in Eq. (A.113) should be measured over the same impedance. As an example of a possible mistake, consider an amplifier that requires 1 V from a 1,000 Ohms source to output 40 V over a 10 Ohms speaker. In this case, it is not strictly correct to use Eq. (A.113) and state that this amplifier has a “gain” of $20{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_{10}(40∕1)\approx 32$ dB. In fact, the power ratio in this case is $10{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_{10}((40^2∕10)∕(1∕1000))\approx 52$ dB.

There are also definitions of absolute values for voltage ratios such as dBmV, which for the cable industry represents decibels relative to 1 millivolt across 75 Ohms (the impedance of a coaxial cable). Hence, 0 dBmV corresponds to $10{\text{log}<!--\; FUNCTION\; APPLICATION\; -->}_{10}({(10^{-3})}^2∕75)\approx -78.75$ dBW or -48.75 dBm.

Some other examples of dB usage:

• To convert from dBW to dBm, simply add 30.
• If a signal is transmitted with power 5 dBm and goes through a channel that attenuates it by 3 dB, it arrives at the receiver with 2 dBm of power.
• A sinusoid with amplitude $A=10$ volts has average power $P=A^2∕2=50$ W, which corresponds to $\approx 16.99$ dBW.
• A DC signal with amplitude $A=10$ volts, has average power 100 watts or, equivalently, 20 dBW.
• Assume one wants to design a (polar) signal that assumes only two amplitude values $-A$ and $A$ with power -9 dBm. The first step is to convert the power -9 dBm to $10^{-9∕10}=0.126$ mW. Assuming $A$ is in volts, this signal has power $A^2$ watts and the amplitude should be $A=\sqrt{0.126\times 10^{-3}}=0.011$ V.