Wadia S7i direct to amp

that's why you don't start losing significant bits (from the 16-bit input stream recovered from the cd) until you hit a digital output level of about 65 (out of 100). each step in the digital output level knocks off about 0.5dB, so by the time you get to 65, you are down about 18dB, which effectively reduces the signal by a factor of about 64; i.e. the signal is about 64 times weaker in comparison to the unattenuated digital signal....

yes, Paperw8, thanks for taking the time to explain this to me but I already understand the concepts of DSP.
I agree that (100-65)*0.5 = 17.5dB, which you are rounding up to 18dB. I seem to have 1 issue in your calculations - how did you arrive at 18dB being an attenuation factor of 64??
it's a voltage attenuation of 18dB i.e. 20log10(x)=18dB. So, what should x be to get 18dB?

12-04-10: Bombaywalla
yes, Paperw8, thanks for taking the time to explain this to me but I already understand the concepts of DSP.
I agree that (100-65)*0.5 = 17.5dB, which you are rounding up to 18dB. I seem to have 1 issue in your calculations - how did you arrive at 18dB being an attenuation factor of 64??
it's a voltage attenuation of 18dB i.e. 20log10(x)=18dB. So, what should x be to get 18dB?

since each increment in the digital output level corresponds to 0.5dB, each increment of 6 corresponds to 3dB. the digital preamplifier is operating on voltage (and not power), so 3dB is a halving of voltage level. in the binary domain, a halving corresponds to 1-bit. so each 3dB knocks off 1-bit so when you knock off 6-bits you have reduced the binary value by 2**6=64.

as far as actual amplitude, bits 17-21/22 are fraction bits but i am just speaking generally, so if you just consider the 21/22-bit binary word, each 3dB would correspond to a right shift of the binary word as you knock off the least significant bit. so, for example, the binary value 100 corresponds to a base-10 value of 4. however, if you right shift the binary value, binary-100 becomes binary-10, which is a base-10 value of 2. if you do a right shift on binary-10, you get binary-1, which is a base-10 value of 1.

anyway, that's the idea...

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.....