Decibel notation is a convenient way of expressing ratios of quantities that may or may not span many orders of magnitude. It is also used to express the amplitude of various signal parameters such as voltage or current relative to a given reference level.

A power ratio, *P2:P1*, in dB is calculated as,

\Large dB= 10\log {\left(\frac{{ p }_{ 2 } }{{ p }_{ 1 }} \right)}

For example, if we are comparing a 30-watt received power to a 15-watt specification, we could say that the received power exceeded the specification by,

\Large 10\log { \left( \frac { 30 }{ 15 } \right) } =3dB

If the impedance associated with two power levels is constant, then the power is proportional to the voltage (or current) squared. In this case, we can also express voltage (or current) ratios in dB,

\Large 10\log { \left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right) } =10{ \log { \left( \frac { { v }_{ 2 } }{ { v }_{ 1 } } \right) } }^{ 2 }=20\log { \left( \frac { { v }_{ 2 } }{ { v }_{ 1 } } \right) }

Antenna or amplifier gains are usually reported in dB. So are cable or filter losses. An amplifier that receives a 4-watt signal and produces a 100-watt signal has a gain of,

\Large 10\log { \left( \frac { 100 }{ 4 } \right) =14dB }

A cable whose input signal has an amplitude of 3.2 volts and whose output signal has an amplitude of 2.8 volts exhibits a gain of,

\Large 20\log { \left( \frac { 2.8 }{ 3.6 } \right) } =-2.18dB

or a loss of,

\Large 20\log { \left( \frac { 3.6 }{ 2.8 } \right) } =2.18dB

### Expressing Signal Amplitudes in dB

Signal amplitudes can also be expressed in decibels as a ratio of the amplitude to a specified reference. For example, a 100-μvolt signal amplitude can also be expressed as,

\Large 20\log { \left( \frac { 100\mu V }{ 1\mu V } \right) } =40dB\mu V\quad

### Power of using Decibel

So what is the benefit of expressing signal amplitudes in dB? Previously, we compared a 30-watt receiver to a 15-watt specification. And we showed that the receiver was 3 dB over the specification. In this case, if the powers had been expressed in dB(W),

\Large 30W=10\log { \left( \frac { 30W }{ 1W } \right) } =14.7dB(W)

\Large 15W=10\log { \left( \frac { 15W }{ 1W } \right) } =11.7dB(W)

We could have calculated the ratio as,

\Large 14.7dB(W)-11.7dB(W)=3dB

Rather than dividing amplitudes to determine the ratio, we can simply subtract amplitudes expressed in dB(·). Again, as long as the impedance is constant, it won’t matter whether we are working with units of power, voltage or current.

### dBm

One of the most common units expressed in decibels is dB(mW) or dB relative to 1 milliwatt. This is almost always written in the abbreviated form, dBm (i.e. without the “W” and without the parentheses). Many oscilloscopes and spectrum analyzers optionally display voltage amplitudes in dBm. Since dBm is a unit of power, we must know the impedance of the measurement in order to convert dBm to volts. For example, a voltage expressed as 0 dBm on a 50-Ω spectrum analyzer is.

\Large 0dBm=\frac { x }{ 1mW } \quad \rightarrow x=1mW

\Large x=\frac { \left| { V }^{ 2 } \right| }{ 50 \Omega }

\Large V=\sqrt { (1mW)*50\Omega } =0.22V