On Sat, 17 Jan 2004 14:02:26 -0600, John Fields
wrote:
On Sat, 17 Jan 2004 17:59:46 +0000, Don Pearce
wrote:
On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:
Yes. Perhaps yet another way to skin the same poor cat...
Since
dB(V) = 20 log (V/Vref),
No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.
---
Hmmm...
Perhaps the confusion arises from the notation.
I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---
Then, since 6dB represents about a 2:1 change in voltage and
P = EČ/R,
a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.
Now, since
dB(W) = 10 log (P1/Pref)
Again here, this is an equation giving dB, NOT dB(W)
---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---
and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.
So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.
dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.
---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---
John, your total confusion over dB, dB(V) and dB(W) here surprises me
more than somewhat.
---
Hopefully my explanation of the notation I use clears it up for you. ;-)
I think the moral here is to stick to the notation that everybody else
understands.
d
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