dBm and dBW

Decibels themselves are unitless — they describe ratios. To express an absolute power level in a dB-like scale, we reference to a fixed power.

• dBW: power referenced to $1\text{W}$: $P(\text{dBW})=10{\log }_{10}\left(\frac{P}{1\text{W}}\right)$. So $1\text{W}=0\text{dBW}$, $10\text{W}=10\text{dBW}$, $100\text{W}=20\text{dBW}$, $1\text{mW}=-30\text{dBW}$.
• dBm: power referenced to $1\text{mW}$: $P(\text{dBm})=10{\log }_{10}\left(\frac{P}{1\text{mW}}\right)$. So $1\text{mW}=0\text{dBm}$, $1\text{W}=30\text{dBm}$, $100\text{W}=50\text{dBm}$.

Conversion

Since $1\text{W}=1000\text{mW}=10^3\text{mW}$, the conversion between dBW and dBm is

$P(\text{dBm})=P(\text{dBW})+30.$

To convert from dBW to dBm, add 30. To convert from dBm to dBW, subtract 30. This is just $10{\log }_{10}(1000)=30\text{dB}$.

A worked example

$P_1=100\text{W}$:

• $P_1(\text{dBW})=10{\log }_{10}(100)=20\text{dBW}$.
• $P_1(\text{dBm})=10{\log }_{10}(100,000)=50\text{dBm}$. Or via conversion: $20+30=50\text{dBm}$.

$P_2=1\text{mW}$:

• $P_2(\text{dBW})=10{\log }_{10}(0.001)=-30\text{dBW}$.
• $P_2(\text{dBm})=10{\log }_{10}(1)=0\text{dBm}$. Or via conversion: $-30+30=0\text{dBm}$.

Gain to go from $P_2$ to $P_1$:

• Linear: $G=100/0.001=100,000$.
• $G(\text{dB})=10{\log }_{10}(100,000)=50\text{dB}$.

Check: $P_1(\text{dBW})-P_2(\text{dBW})=20-(-30)=50\text{dB}$. And $P_1(\text{dBm})-P_2(\text{dBm})=50-0=50\text{dB}$. Same gain, either way.

The pattern: a gain in dB is the difference of two power levels in either dBW or dBm. The result is the same.

Where these show up

• dBm is the standard in RF and microwave engineering. Receiver sensitivities are quoted in dBm, transmitter outputs in dBm, channel powers in dBm. $0\text{dBm}=1\text{mW}$ is a convenient reference because it’s near typical mobile-phone receiver levels.
• dBW is more common in satellite communications and high-power applications where the levels are tens of watts to kilowatts.

What you cannot do

You cannot add or subtract dBm/dBW values to combine two physical powers at a junction:

$P_3=P_1+P_2\text{ (linear)}⟹P_3(\text{dBm})\ne P_1(\text{dBm})+P_2(\text{dBm}).$

Adding dBm values corresponds to multiplying powers, not adding them. To find the power at a combiner output in dBm, convert each input to linear (watts), add, then convert back. See the decibel rules for details.