Decibels with Power Gain

Two levels of power can be compared using a unit of measure called the bel, which is defined by the following equation:
$$\mathrm{B}=\log _{10} \frac{P_{2}}{P_{1}}$$
However, to provide a unit of measure of less magnitude, a decibel is defined, where
$$1 \text { bel }=10 \text { decibels }(\mathrm{dB})$$
The result is the following important equation, which compares power levels $ P_{2} $ and $ P_{1} $ in decibels:
$$\bbox[10px,border:1px solid grey]{\mathrm{dB}=10 \log _{10} \frac{P_{2}}{P_{1}} \text { decibels }(\mathrm{dB}) } \tag{1}$$
If the power levels are equal $ \left(P_{2}=P_{1}\right) $, there is no change in power level, and $ \mathrm{dB}=0 $. If there is an increase in power level $ \left(P_{2}>P_{1}\right) $, the resulting decibel level is positive. If there is a decrease in power level $ \left(P_{2} < P_{1} \right) $, the resulting decibel level will be negative. For the special case of $ P_{2}=2 P_{1} $, the gain in decibels is
$$\mathrm{dB}=10 \log _{10} \frac{P_{2}}{P_{1}}=10 \log _{10} 2=\mathbf{3} \mathbf{d B}$$
Therefore, for a speaker system, a 3-dB increase in output would require that the power level be doubled. In the audio industry, it is a generally accepted rule that an increase in sound level is accomplished with 3-dB increments in the output level. In other words, a 1-dB increase is barely detectable, and a 2-dB increase just discernible.
A 3-dB increase normally results in a readily detectable increase in sound level. An additional increase in the sound level is normally accomplished by simply increasing the output level another $ 3 \mathrm{~dB} $. If an 8-W system were in use, a $ 3-\mathrm{dB} $ increase would require a 16-W output, whereas an additional increase of $ 3 \mathrm{~dB} $ (a total of $ 6 \mathrm{~dB} $ ) would require a $ 32-\mathrm{W} $ system, as demonstrated by the calculations below:
$$\mathrm{dB}=10 \log _{10} \frac{P_{2}}{P_{1}}=10 \log _{10} \frac{16}{8}=10 \log _{10} 2=3 \mathbf{d B}$$
$$\mathrm{dB}=10 \log _{10} \frac{P_{2}}{P_{1}}=10 \log _{10} \frac{32}{8}=10 \log _{10} 4=6 \mathbf{d B}$$
For $ P_{2}=10 P_{1} $
$$\mathrm{dB}=10 \log _{10} \frac{P_{2}}{P_{1}}=10 \log _{10} 10=10(1)=10 \mathbf{d B}$$
resulting in the unique situation where the power gain has the same magnitude as the decibel level. For some applications, a reference level is established to permit a comparison of decibel levels from one situation to another. For communication systems a commonly applied reference level is
$$P_{\text {ref }}=1 \mathrm{~mW} \quad \text { (across a } 600-\Omega \text { load) }$$
Equation (1) is then typically written as
$$\mathrm{dB}_{m}=\left.10 \log _{10} \frac{P}{1 \mathrm{~mW}}\right|_{600 \Omega}$$
Note the subscript $ m $ to denote that the decibel level is determined with a reference level of $ 1 \mathrm{~mW} $.In particular, for $ P=40 \mathrm{~mW} $,
$$\mathrm{dB}_{m}=10 \log _{10} \frac{40 \mathrm{~mW}}{1 \mathrm{~mW}}=10 \log _{10} 40=10(1.6)=16 \mathbf{d B}_{m}$$
whereas for $ P=4 \mathrm{~W} $,
$$\mathrm{dB}_{m}=10 \log _{10} \frac{4000 \mathrm{~mW}}{1 \mathrm{~mW}}=10 \log _{10} 4000=10(3.6)=36 \mathbf{d B}_{m}$$
Even though the power level has increased by a factor of $ 4000 \mathrm{~mW} / $ $ 40 \mathrm{~mW}=100 $, the $ \mathrm{dB}_{m} $ increase is limited to $ 20 \mathrm{~dB}_{m} $. In time, the significance of $ \mathrm{dB}_{m} $ levels of $ 16 \mathrm{~dB}_{m} $ and $ 36 \mathrm{~dB}_{m} $ will generate an immediate appreciation regarding the power levels involved. An increase of $ 20 \mathrm{~dB}_{m} $ will also be associated with a significant gain in power levels.

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