Wadia S7i direct to amp

the digital preamplifier is operating on voltage (and not power), so 3dB is a halving of voltage level.

wrong! every 6dB (i.e. every 12 steps) is halving of the voltage amplitude. You correctly wrote that the digital amp works in the voltage domain & not the power domain but you did not understand that concept fully.
Did you do a sanity check to see if every 6 counts starting from 100 down to 0 would yield exactly 50dB of amplitude control before you wrote you prev post?

We have 100 steps with each step giving us 0.5dB => 50dB of voltage control.
What is 50dB in linear/numeric? 316.22.
So, we can attenuate the digital music signal by a factor of 316.22 using the DSP.
setting 100 = full scale signal amplitude
setting 88 = 1/2 full scale
setting 76 = 1/4 FS
setting 64 = 1/8 FS
setting 52 = 1/16 FS
setting 40 = 1/32 FS
setting 28 = 1/64 FS
setting 16 = 1/128 FS
setting 04 = 1/256 FS
setting 00 = 1/307.175 FS

& 20log10(1/307.175) = -49.75dB, which I'll round off to -50dB, which is exactly the amplitude control range of the digital preamp.

12-04-10: Bombaywalla
wrong! every 6dB (i.e. every 12 steps) is halving of the voltage amplitude. You correctly wrote that the digital amp works in the voltage domain & not the power domain but you did not understand that concept fully.

actually, you are incorrect. the basic decibel measurement is computed as follows:

10*log10(q1/q2)

where q1 and q2 are quantities that are being measured
(ref: http://en.wikipedia.org/wiki/Decibel).
[wikipedia is not a peer-reviewed reference so you always have to view statements there with some degree of caution, so feel free to challenge this statement if you have a better reference - i just don't feel like digging through my own books since the wikipedia equation matches my own knowledge on the subject]

as it turns out, power is proportion to voltage**2. so when the quantities being measured are power levels, and since power is a function of voltage, the equation for power can also be represented as a ratio of voltages:

10*log10((vo**2/r)/(vi**2/r))

[i am assuming equal resistance values for simplicity]

however, because of the way that logarithms work, you can also express this equation:

20*log10(vo/vi)

this is the equation that you presented - i'm just trying to explain to you how the equation was derived. so, a 3dB reduction in voltage corresponds to a 6dB reduction in power.

you will notice that the wadia white paper refers to *volume* output. volume output is a power domain concept. so the wadia white paper on digital volume control indicates that you start to lose significant (i.e. non-interpolation bits) when the digital attenuation causes a 36dB reduction in *volume* output.

i took a look at the wadia manual, and i think that there is an error in the manual. the manual states that there is a 0.5dB change in *volume* for each step in the digital volume control. from my calculations, and the wadia white paper on digital volume control, it appears that there is actually a 1dB change in volume for each step in the digital volume control.

so, a 3dB reduction in voltage corresponds to a 6dB reduction in power.

wrong! a 3dB reduction in POWER corresponds to a 6dB reduction in VOLTAGE.

i took a look at the wadia manual, and i think that there is an error in the manual.

I had a good laugh at this one!!! :-)

it appears that there is actually a 1dB change in volume for each step in the digital volume control.

Oh, is it??
so, now we have 100 steps & 100dB of volume control??
You better call up Wadia & tell them that they actually have 100dB of volume control & not 50dB.....

Paperw8: a 3dB reduction in voltage corresponds to a 6dB reduction in power.

Bombaywalla: wrong! a 3dB reduction in POWER corresponds to a 6dB reduction in VOLTAGE.

I haven't taken the time to read through most of this thread, but I'll just say with respect to the quoted comments that neither is correct.

A 3db reduction in power corresponds to a 3db reduction in voltage, for a given impedance. A 6db reduction in power corresponds to a 6db reduction in voltage, for a given impedance.

To see that, consider that a 3db reduction in voltage corresponds to the voltage being reduced by a factor of 0.707. Since, as I think we all agree, P = Esquared/R, that 3db reduction in voltage results in power being reduced by a factor of 0.707 x 0.707 = 0.5, which is also 3db.

Regards,
-- Al

12-05-10: Bombaywalla
wrong! a 3dB reduction in POWER corresponds to a 6dB reduction in VOLTAGE.

unlike much of the subjective discussion in this forum, this is a factual matter; your *opinion* on the matter is irrelevent to the actual *facts*.