09-13-11: Almarg
Paperw8, we had discussed the definition of db for electrical signals in another thread a while back, and as I indicated then, with all due respect you are simply wrong. Please do some further research, and I think you will see that:
db = 10log(P1/P2) = 20log(V1/V2)
where P1/P2 is the ratio of two powers, and V1/V2 is the ratio of two voltages. 6db is 6db, regardless of whether it is derived on the basis of power (where it represents a factor of 4) or voltage (where it represents a factor of 2).
you were wrong when we discussed this matter earlier and you are still wrong. one of the problems is that people cite equations that they saw somewhere by rote without understanding what the equations really mean. look at the equation that you cited for power_dB:
10log(P1/P2) = 20log(V1/V2)
the left hand side is a ratio of *power* levels while the right hand side is a ratio of *voltage* levels. power and voltage are 2 different things. the right hand side expresses how power levels change in response to changes in voltage levels. however, the equation that you cite is only correct when you are referring to a configuration where a voltage is being delivered to a load; there are other circumstances in which the equation that you cite is *not* correct (i explained that earlier so i won't go through that explanation again).
so to give some context as to how the equation that you cite above really works, take a given voltage, v1, delivered to a load, z. the current, i1, through z in that case is:
i1=v1/z
the power delivered to z, p1, in that case is:
p1=v1*i1=v1*v1/z=v1^2/z
now deliver a voltage, v2, to z be described as:
v2=(1/2)*v1
now the current, i2, through z is:
i2=v2/z=(1/2)*v1/z
the power delivered to z in this case, p2, is:
p2=v2*i2=[(1/2)*v1]*[(1/2)*v1/z]
p2=(1/4)*v1^2/z
since:
p1=v1^2/z
p2 can be expressed as:
p2=(1/4)*p1
so what this all means is when the voltage delivered to a load is reduced to one-half it's original value, the power delivered to that load is reduced to one-quarter it's original value. so to use the equation for dB we have:
v_dB=10*log(v2/v1)=10*log(1/2)=10*(-0.3)=-3dB
p_dB=10*log(p2/p1)=10*log(1/4)=10*(-0.6)=-6dB
that's how this stuff really works: there is no magic where somehow -3dB corresponds to a reduction to half its original value when talking about voltage but magically -6dB corresponds to a reduction to half its original value when talking about power.