Decibel (dB)

The bel (B) is a unit of measurement for the ratio of a physical quantity (power, intensity, magnitude, etc.) and a specified or implied reference level in base-10 logarithm. As it is a ratio of two quantities with the same unit, it is dimensionless.

For example, consider a power amplifier with input signal power $P_{in}=100\; mW$ and output signal power $P_{out}=10\;W$, then the power gain of the amplifier is $P_{out}/P_{in}=100$, which can be more concisely expressed in base-10 log scale:

$$L_B=\log_{10} \frac{P_{out}}{P_{in}}=\log_{10} \frac{10}{0.1} =\log_{10} 100 =2\; bel(B)$$ (484)

The unit bel (B) was first used in early 1920's in honor of Alexander Bell (1847 – 1922), a telecommunication pioneer and founder of the Bell System (the Bell Labs).

As bel (B) is often too big a unit (a gain of 100 is only 2 B), a smaller unit of decibel (dB), 1/10 of the unit bel (B), is more widely used instead. Now the power gain above can be expressed as:

$$L_B=\log_{10}\frac{P_{out}}{P_{in}}=\log_{10} 100 =2\;B=20\;dB, \;\;\;\; \;\;\;\;\; L_{dB}=10 \log_{10}\frac{P_{out}}{P_{in}}=20\;dB$$ (485)

Similarly a 1,000 fold power gain is expressed as:

$$L_B=\log_{10}\frac{P_{out}}{P_{in}}=\log_{10} 1000 =3\;B=30\;dB, \;\;\;\; \;\;\;\; L_{dB}=10 \log_{10}\frac{P_{out}}{P_{in}}=30\;dB$$ (486)

Given the input power $P_{in}$ and the power gain in decibel, e.g., $L_{dB}=30\;dB$, the output power can be obtained as:

$$\frac{P_{out}}{P_{in}}=10^{L_{dB}/10}=10^{30/10}=10^3, \;\;\;\; \;\;\;\;P_{out} =10^3\;P_{in}=1,000\;P_{in}$$ (487)

As another example, the sound level is measured in decibel, in terms of the ratio of the sound intensity (power per area, e.g., $W/m^2$) and the threshold of human hearing ( $10^{-12}\; W/m^2$) as the reference. The human hearing has a large range from 0 dB (threshold) to 140 dB (military jet takeoff, $10^{14}$ times the threshold, i.e., $10^2\; W/m^2$). 160 dB sound level will cause instant membrane/eardrum perforation.

In general, power (and energy) is always proportional to the amplitude of certain quantity squared (e.g., $P=V^2/R=I^2 R$, $E=mv^2/2$, $E=kx^2/2$). Therefore a different definition is used for ratios between two amplitudes, for example, the output and input voltages $V_{out}$ and $V_{in}$ of a voltage amplifier, we have:

$$L_{dB}=10 \log_{10} \frac{V^2_{out}}{V^2_{in}} =20 \log_{10} \frac{V_{out}}{V_{in}}\;dB$$ (488)

If the input to a voltage amplifier is 10 mV and the output voltage is 1 V, then the voltage gain in terms of decibel is:

$$20 \log_{10} \frac{V_{out}}{V_{in}}=20 \log_{10} \frac{1,000}{10}=40\; dB$$ (489)

If the output voltage is 10 V, then

$$20 \log_{10} \frac{V_{out}}{V_{in}}=20 \log_{10} \frac{10,000}{10}=60\; dB$$ (490)

We see that the difference of one order of magnitude in the gain corresponds to 20 dB.

Given the input voltage $V_{in}$ and the voltage gain in decibel, e.g., $L_{dB}=60\;dB$, the output voltage can be obtained as:

$$\frac{V_{out}}{V_{in}}=10^{L_{dB}/20}=10^{60/20}=10^3, \;\;\;\; \;\;\;\;\; V_{out}=10^3 V_{in}=1,000 V_{in}$$ (491)

A related issue is the half power point. Recall that for a second order system, when $\zeta$ is small (e.g., $\zeta<0.2$), the magnitude $\vert H(j\omega)\vert$ of the frequency response function has a peak at $\omega=\omega_p\approx \omega_n$. The bandwidth of the peak is defined as the difference between two cut-off frequencies $\omega_1$ and $\omega_2$ ( $\omega_1<\omega_n < \omega_2$) at which

$$\vert H(j\omega_1)\vert^2=\vert H(j\omega_2)\vert^2=\frac{1}{2} \vert H(j\omega_p)\vert^2\;\;\; \;\;\;\; \vert H(j\omega_{1,2}) \vert=0.707\; \vert H(j\omega_p) \vert$$ (492)

The ratio between the half-power point and the peak in decibel is

$$20 \log_{10} \left( \frac{ \vert H(j\omega_{1,2})\vert}{\vert H(j\omega_p) \vert} \right) =20 \log_{10} 0.707=-3.01\;dB \approx -3\;dB$$ (493)