# Convert dBm to dBW

Discussion in 'Homework Help' started by BruceBly, Jan 15, 2013.

1. ### BruceBly Thread Starter New Member

Jul 26, 2012
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0
I am trying to follow an example formula in my text on how to convert dBm to dBw. The formula given is

+10dBm=10log P^2/0.001

log^-1(1)=P^2/0.001 => 10 = P^2/0.001

P^2=0.001 W

dBW=10log 0.001W/1W=-20 dBw

I believe 10 dBm to be a bad example. I don't fully understand how to do this formula. I have found two ways to put this in my calculator and get the correct answer but if I substitute 10 dBm for 20 dBm I don't get the correct answer, therefore I am doing it wrong somewhere. How did the text get log^-1 out of log when it moved it to the other side of the equation? Can someone explain to me how to do this conversion.

2. ### mlog Member

Feb 11, 2012
276
36
Unless I'm missing something, dbm is referencing 1 milliwatt and dbw is referencing 1 watt. Since the ratio is 1000, shouldn't the conversion involve adding or subtracting 30 db? In other words, 10 times log 1000 = 10 times 3 = 30.

3. ### crutschow Expert

Mar 14, 2008
20,492
5,801
Yes, the OP's math is off.

dBW = 10 log (0.001/1) = -30 dBw, not -20 dBw.

4. ### BruceBly Thread Starter New Member

Jul 26, 2012
25
0
I believe the math is correct. I have used several conversion apps online and it gives me -20dBW as the answer. I just don't understand how to calculate this on my own without a online converter.

5. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
The key is in understand what the measurements mean and careful tracking of units helps out quite a bit.

A Bel is defined as the base-10 logarithm of the ratio of two powers.

x = log(P/Po) Bel

Note that, unlike the typical situation, we have to explicitly provide the unit of Bel here because the definition is that there is 1 Bel per log(P/Po). It's a bit unusual, but it is completely consistent with how we use units in more typical situations.

Just like a deciliter is one-tenth of a liter, a deciBel (dB) is one-tenth of a Bel. So we have:

x = log(P/Po) Bel * (10dB/Bel) = 10 log(P/Po) dB

dBm is simply defined with the constraint that Po=1mW and dBw (or dBW) is defined with Po=1W. So, recapping, we have:

x = 10 log(P/Po) dB

x = 10 log(P/1mW) dBm

x = 10 log(P/1W) dBw

Hence we can work the problem piecewise:

Find P assocaited with 10dBm

10dBm = 10 log(P/1mW) dBm

10 = 10 log(P/1mW)

1 = log(P/1mW)

10 = P/1mW

P = 10*1mW = 10mW

Now, what is this in dBw?

x = 10 log(P/1W) dBw

x = 10 log(10mW/1W) dBw

x = 10 log(1/100mW) dBw

x = 10 * (-2) dBw

x = -20dBw

Q.E.D.

Notice how tracking the units keeps everything nice and neat and tidy and makes it really hard to make a mistake.

Now, let's figure out how, in general, to convert from dBm to dBw.

Given x (a value expressed in dBm)

x = 10 log(P/1mW) dBm

(x/10dBm) = log(P/1mW)

10^(x/10dBm) = P/1mW

P = 1mW*10^(x/10dBm)

Now plug this into the definition to get y (so as not to reuse x) in dBw

y = 10 log(P/1W) dBw

y = 10 log(1mW*10^(x/10dBm)/1W) dBw

y = 10 log( [1mW/1W] * [10^(x/10dBm)] ) dBw

y = 10 {log(1mW/1W) + log(10^(x/10dBm) } dBw

y = 10 {log(1/1000) + log(10^(x/10dBm) } dBw

y = 10 {-3 + x/10dBm } dBw

y = ( x/dBm - 30 ) dBw

or

y = x dBw/dBm - 30dBw

Again, see how tracking the units keeps everything nice and tidy?

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6. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,951
1,758
0 dBm is 1 milliwatt or 0.001 as shown above.

0 dBW is 1 watt or 1000 times 1 mW.

10*log10(0.001) = -30 dBW

7. ### BruceBly Thread Starter New Member

Jul 26, 2012
25
0
Find P assocaited with 10dBm

10dBm = 10 log(P/1mW) dBm

10 = 10 log(P/1mW)

1 = log(P/1mW)

10 = P/1mW

P = 10*1mW = 10mW

I am unclear as to why when you move the log over to the other side the one turns into a 10. What calculation are you using to do this?

8. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
1 = log(P/1mW)

10^1 = 10^log(P/1mW)

10^1 = 10

10^log(x) = x, so 10^log(P/1mW) = P/1mW

Hence

1 = log(P/1mW)

goes to

10 = P/1mW

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9. ### BruceBly Thread Starter New Member

Jul 26, 2012
25
0
Thanks, I believe I understand the function of log now. Here is another problem I believe is correct. I believe using the 10dBm as the example problem in the book was a poor example.

38dBm = 10 log(P/1mW) dBm

38= 10 log(P/1mW)

3.8= log(P/1mW)

309.57= P/1mW

P = 6309.57*1mW = 6.30957W

x = 10 log(P/1W) dBW

x = 10 log(6.30957W/1W) dBW

x = 10 log(6.30957W/1W) dBW

x = 10 * (6.30957W) dBW

x = 8dBW

10. ### WBahn Moderator

Mar 31, 2012
23,384
7,096
The final result is correct, but it doesn't follow from the line directly above it . I think you just omitted the log from that line. You also have a units error in that same line.

x = 10 log(6.30957W/1W) dBW

The W's cancel out, which is good because you cannot take the log of any quantity that had dimensions.

x = 10 log(6.30957) dBW

x = 8dBW

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