# DBm and dB

1. Feb 18, 2010

### kd001

How do I convert a value in dBm to dB?

2. Feb 18, 2010

### mathman

from wikipedia:

dBm (sometimes dBmW) is an abbreviation for the power ratio in decibels (dB) of the measured power referenced to one milliwatt (mW). It is used in radio, microwave and fiber optic networks as a convenient measure of absolute power because of its capability to express both very large and very small values in a short form. Compare dBW, which is referenced to one watt (1000 mW).

Since it is referenced to the watt, it is an absolute unit, used when measuring absolute power. By comparison, the decibel (dB) is a dimensionless unit, used for quantifying the ratio between two values, such as signal-to-noise ratio.

3. Feb 18, 2010

### elect_eng

You really can't convert one to the other because one number represents power, and the other is a relative value.

A dBm value is the dB value referenced to 1 milliWatt.

Value[dBm]=10*log(P/1mW) where P is measured in mW.

A dB value represents a ratio.

Value[dB]=10*log(P1/P2) where P1 and P2 are in the same units.

If you have 10 dBm of power, you can say this is 10 dB relative to 1 milliWatt. You could also say you have -20 dB referenced to 1 Watt. So, you can see the issue of trying to establish a general conversion method. It's best to just understand the meanings and use them properly in a given context.

4. Feb 18, 2010

### kd001

My problem is that I've got some signal strength data in dBm. I've also got an equation which relates signal strenght in dB to electron density. I've got to use the signal strengh data to calculate the electron density. If I use values in dBm instead of db in the equation how would that affect the result?

Thanks

5. Feb 18, 2010

### f95toli

I am guessing your equation relates GAIN (you can't measure signal strenght in dB since it is a relative unit) to density; so in order to use that equation you need to know how many dBm that went into the system and how many dBm that came out; the difference is the gain.

6. Feb 18, 2010

### kd001

f95toli, that's very helpful but can you clarify a bit more. I know the difference between the dbm that went into the system and the dBm that came out (i.e signal loss in dBm). How could I then use that in an equation?

EDIT: I've attached the equation. L is signal loss in dB. I know signal loss in dBm and all the other variables except for Ne (the electron density) which I need to work out. I don't need the absolute value of Ne, just the change in Ne that would result in the given change in signal strength. I know how to intergrate the equation.

#### Attached Files:

• ###### formula.jpg
File size:
3 KB
Views:
275
Last edited: Feb 18, 2010
7. Feb 18, 2010

### f95toli

Well, if you e.g. have 20 dBm going into a system, and 0 dBm coming out the gain was -20 dB (i.e. the loss was 20 dB).

It might be helpful to convince yourself that this is correct by converting the dBm to mW and the gain/loss to "*times". In my example above we have 20 dBm=100 mW going into the system and 0dBm=1mW coming out; that means that the gain was 1/100=-20dB (i.e. the signal coming out was 20dB=100 times lower than the signal going in)

The fact that you can simply add/substract to get gain/loss in dB is one reason why dBm is