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dBu, dBV, dBm

Discussion in 'Recording' started by roirat, Dec 2, 2004.

  1. roirat

    roirat Guest

    I was trying recently to figure out why 2Vrms is 6.02 dBV instead of 3.01 dBv, and I learned that it is of course because the power in constant-load circuit is proportional to the square of the voltage. So,

    dBV = 20 * log (Vmeas / 1V)

    Here's my question: Does the same thing apply to dBm? If 0dBm is 1 milliwatt into a 600 ohm load, then is 2 milliwatts into the same load 3.01 dBm or 6.02 dBm?

    Here's a good explanation of dB in general (tho it doesn't quite get to my question):


  2. roirat

    roirat Guest

    I'm pretty sure i've answered my own question.

    dBm = 10 * log (Pmeas / 1mW)

    Not that I think I'll ever actually use this stuff, but I get curious...

  3. jcnoernberg

    jcnoernberg Guest

    post your q here if you still need help...


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