# Decibels with Power Gain   Whatsapp  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.

## Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250