Going back to fundamentals --
Given a = duty cycle = average power
and define "^2" = "squared"
Then P[ave] = a P[ref]
P = V^2/R
V[ave]^2/R = aV[ref]^2/R, the Rs cancel leaving
V[ave]^2 = aV[ref]^2
10 log(V[ave]^2) = 10 log (aV[ref]^2), which is equivalent to
20 log (V[ave]) = 20 log (aV[ref])
= 20 log (a) + 20 log (V[ref])
In the last equation one sees the duty cycle isolated as "20 log (a)" when
referring to power in terms of voltage.
Best regards,
Don Umbdenstock
Sensormatic
> ----------
> From: Doug McKean[SMTP:[email protected]]
> Reply To: Doug McKean
> Sent: Friday, October 19, 2001 2:41 PM
> To: [email protected]
> Subject: Re: Is This Right?
>
>
> >
> > More to the proof discussion launched by the duty cycle question,
> given
> >
> > > dB = 10 log (P1/P2)
> > >
> > > Let "a" be the duty cycle ratio, with 0<a<1, so that P1 = aP2.
> > >
> > > Then dB = 10 log (aP2/P2) = 10 log (a). Eq.
> > > (1)
> > >
> > If 10 log (P1/P2) = 10 log (V1^2/V2^2) = 10 log (V1/V2)^2 = 20 log
> (V1/V2),
> >
> > Then does it follow that,
> >
> > dB = 10 log (aV2^2/V2^2) = 10 log (aV2/V2)^2 = 20 log (a) ? Eq. (2)
> >
> >
> > If this is true, then
> >
> > duty cycle "a" = 10 log (a) from Eq. (1) and
> >
> > = 20 log (a) from Eq. (2)
> >
> > What am I missing?
>
> The original intention of the calculations.
>
> The first relation, dB = 10 log (aP2/P2) = 10 log (a)
> is a "power" relation.
>
> The second relation, dB = 10 log (aV2^2/V2^2) = 20 log (a)
> is a "voltage relation. "
>
> Equating the two is invalid since you're trying to equate
> two different concepts. Doesn't mean anything.
>
> - Doug McKean
>
>
>
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