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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    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
    13,000
    3,229
    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
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    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
    17,726
    4,788
    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
    3,373
    1,157
    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
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    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
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    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
    17,726
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    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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