Conversion between DB, DBm, and DBw

Decibel (decibel,db) is a ratio function that operates on dimensionless parameters:

DB=˙10LOG10 (x), where x is unitless D B =˙10 l o G Ten (x), where x is unitless dB \dot = Ten log_{10} (x), \quad \text{ Where $x $ is unitless}

For example, the amplifier gain is a dimensionless value, which is the ratio of output power to input power:

Poutpin=gtherefore Gain in db =10log10g=˙g (DB) P o u t p i n = g therefore Gain in DB = ten l o g ten g =˙g (d B) \frac {P_{out}} {P_{in}} = g \ \text{therefore Gain in DB $= 10log_{10}g \dot = g (DB) $}

We may have seen the following statement:

Output power of 5 dBw

Input power is + dBm

The power here is not a dimensionless value, why can you say so? Careful observation of the power of the expression, you can find the unit is dBm or dBw.

Assuming we have the power p, and now we have p in addition to 1 watts, the result is a value of P relative to the 1.0 watt power, which is a dimensionless value.

For example, if P = 2500 MW, then p/1w = 2.5. This means that the power P is 2.5 times times larger than 1 watts.

Since p/1w is a dimensionless value, we can use that value in decibels. There are the following definitions:

P (dBw) =˙10log10 (P1W) p (d B w) =˙10 l o g (P 1 W) p (dBw) \dot = Ten log_{10} (\frac{p}{1w})

Thus, we can easily derive the definition of dBm:

P (dbm) =˙10log10 (P1MW) p (d B m) =˙10 l o G ten (P 1 m W) p (dbm) \dot = Ten log_{10} (\FRAC{P}{1MW})

For example, if P = 100 watts, then P (dBw) = Dbw,p (dbm) = 0 dbm, if p = 1mW, then P (dBw) = -30 dbw,p (dbm) = dbm.