The definition of voltage gain is \$V_{\text{out}}/V_{\text{in}}\$.
However, I read some articles about the gain in decibels, and I have a confusion now.
Here is an article about it: https://en.wikipedia.org/wiki/Gain#Voltage_gain
Here, I understand the definition of Power gain in decibels, which is $$ \text{Gain} = 10 \log \left( {P_{\text{out}} \over P_{\text{in}}} \right)\text{ dB} $$
However, I can't understand why Voltage gain in decibels is $$ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} $$
If \$ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} \$ is derived from $$ 10 \log {\left( {V_{\text{out}}^2 \over R_{\text{out}}}\right) \over \left({V_{\text{in}}^2 \over R_{\text{in}}}\right) }\text{ dB} $$ then this is the power gain, not the voltage gain, isn't it? However, the Wikipedia says it is a formula for the Voltage gain in decibels. I thought the voltage gain in decibels would be \$ 10 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} \$. Actually, the example section in that linked page uses voltage gain \$ V_{\text{out}} \over V_{\text{in}} \$.
Why did \$V^2/R\$ suddenly come out from the voltage gain in decibels?
Answer
As you say, the decibel is a unit of power ratio.
\$ G\ [\mathrm{dB}] = 10 \log_{10}\left(\dfrac{P_1}{P_2}\right)\$.
When the input and output impedances are equal and then we can express the gain in terms of voltage as
\$ G\ [\mathrm{dB}] = 20 \log_{10}\left(\dfrac{V_1}{V_2}\right) \$
I wouldn't call this the "voltage gain in decibels." I'd rather say it's the decibel gain, calculated from the voltage gain.
Sometimes, you will see a voltage gain expressed in decibel according to this formula even when the input and output impedances are different. There is no technical justification for this --- it's simply a shorthand practice that's become common through usage.
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