A logarithmic expression of the ratio of two power levels: \[ 10 \times \log_{10}\left(\frac{P1}{P2}\right) \]

P1	P2	\( \frac{P1}{P2} \)	dB
1	1	1	0
2	1	2	3
10	1	10	10
0.5	1	0.5	-3
7.2	1.6	4.5	6.5
100	1	100	20
1000	1	1000	30

Since a dB is a ratio, it is not an absolute quantity, like a watt, etc.

To make it an absolute quantity, we must specify a reference.

dB suffix

The suffix can turn dB into an absolute quantity dbM.

10 MHz sinewave at 950mVpp into \( 50\Omega \).

\[ Vpp = 950mV \] \[ Vrms = \frac{Vpp}{2 \times \sqrt{2}} = \frac{950mV}{2 \times \sqrt{2}} = 336mV_{rms} \]

Therefore, the power in the \( 50\Omega \) load is:

\[ P = \frac{\left(Vrms\right)^2}{R} = \frac{0.336^2}{50} = 2.256mW \]

To express this in dBm, the reference is 1mW:

\[ dBm = 10\times\log_{10}\left(\frac{2.256mW}{1mW}\right) = 3.53dBm \]

In dBµ, the reference is 1µW:

\[ dBµ = 10\times\log\left(\frac{2.256 \times 10^{-3}}{1 \times 10^{-6}}\right) = 33.53dBµ \]

Voltage ratios

The dB relates to power, so we have to calculate power.

\[ 10\times\log\left[\frac{\frac{V1^2}{R}}{\frac{V2^2}{R}}\right] \]

\[ 10\times\log\left[\frac{V1^2}{R}\times\frac{R}{V2^2}\right] \]

\[ 10\times\log\left[\frac{V1^2}{V2^2}\right] \]

\[ 10\times\log\left[\left(\frac{V1}{V2}\right)^2\right] \]

\[ 2\times10\times\log\left[\frac{V1}{V2}\right] \]

\[ 20\times\log\left[\frac{V1}{V2}\right] \]

Thus, our 950mVpp can be expressed in dBmv

\[ V = \sqrt{P \times R} \]

\[ dBmv = 20\times\log\left(\frac{336mV}{1mV}\right) = 50.5dBmV \]