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To convert decibels (dB) to a linear scale, use the inverse logarithmic formula:
For power ratios: \[ \text{linear ratio} = 10^{\frac{\text{dB}}{10}} \]
For amplitude ratios such as voltage or current: \[ \text{linear ratio} = 10^{\frac{\text{dB}}{20}} \]
The decibel is a logarithmic way of expressing a ratio. It does not directly represent an absolute quantity unless a reference is specified.
If the original dB value describes a power ratio, the definition is:
\[ \text{dB} = 10 \log_{10}\left(\frac{P_2}{P_1}\right) \]
Solving for the linear ratio:
\[ \frac{P_2}{P_1} = 10^{\frac{\text{dB}}{10}} \]
Examples:
\(3\ \text{dB}\): \[ 10^{3/10} \approx 1.995 \approx 2 \] So, +3 dB is about 2× power.
\(10\ \text{dB}\): \[ 10^{10/10} = 10 \] So, +10 dB is 10× power.
\(-10\ \text{dB}\): \[ 10^{-10/10} = 0.1 \] So, −10 dB is 0.1× power.
If the dB value describes an amplitude ratio such as voltage or current, the definition is:
\[ \text{dB} = 20 \log_{10}\left(\frac{A_2}{A_1}\right) \]
Therefore:
\[ \frac{A_2}{A_1} = 10^{\frac{\text{dB}}{20}} \]
Examples:
\(6\ \text{dB}\): \[ 10^{6/20} \approx 1.995 \approx 2 \] So, +6 dB is about 2× voltage/current amplitude.
\(20\ \text{dB}\): \[ 10^{20/20} = 10 \] So, +20 dB is 10× amplitude.
\(-20\ \text{dB}\): \[ 10^{-20/20} = 0.1 \] So, −20 dB is 0.1× amplitude.
Because power is proportional to the square of amplitude:
\[ P \propto A^2 \]
For example, in a resistive circuit:
\[ P = \frac{V^2}{R} \]
Substituting amplitude into the power definition gives:
\[ 10\log{10}(A^2) = 20\log{10}(A) \]
That is why amplitude-based quantities use 20 instead of 10.
| dB | Power ratio | Amplitude ratio |
|---|---|---|
| -20 dB | 0.01 | 0.1 |
| -10 dB | 0.1 | 0.316 |
| -6 dB | 0.251 | 0.501 |
| -3 dB | 0.501 | 0.708 |
| 0 dB | 1 | 1 |
| +3 dB | 1.995 | 1.413 |
| +6 dB | 3.981 | 1.995 |
| +10 dB | 10 | 3.162 |
| +20 dB | 100 | 10 |
These are useful for quick estimates in RF, audio, control, and instrumentation work.
Ask: What does the dB value describe?
If it came from watts, milliwatts, power gain, power loss: use \[ 10^{\frac{\text{dB}}{10}} \]
If it came from voltage gain, current gain, sound pressure, field strength: use \[ 10^{\frac{\text{dB}}{20}} \]
For voltage or current, the 20-log rule assumes the comparison is made under the same impedance conditions. If impedance changes, voltage ratio and power ratio are no longer interchangeable.
If the value is something like dBm, dBW, dBV, dBu, then it is not just a ratio; it is referenced to a specific quantity.
Examples:
dBm: referenced to 1 mW \[ P(\text{mW}) = 10^{\frac{\text{dBm}}{10}} \]
dBW: referenced to 1 W \[ P(\text{W}) = 10^{\frac{\text{dBW}}{10}} \]
dBV: referenced to 1 V \[ V = 10^{\frac{\text{dBV}}{20}} \text{ V} \]
dBu: referenced to 0.775 V \[ V = 0.775 \cdot 10^{\frac{\text{dBu}}{20}} \text{ V} \]
Example:
To convert dB to linear:
Power: \[ \text{linear} = 10^{\frac{\text{dB}}{10}} \]
Amplitude: \[ \text{linear} = 10^{\frac{\text{dB}}{20}} \]
The only real decision is whether the dB value refers to power or amplitude. Once that is clear, the conversion is straightforward.
If you want, I can also give you: