Wadia S7i direct to amp

12-03-10: Bombaywalla
Paperw8, I'm pretty sure that the Wadia resolution is 24 bits (& not 21 bits) after it upsamples.

i'll say this again: don't take my word for it, READ THE WADIA WHITEPAPERS and draw your own conclusions. wadia has several whitepapers posted on their website that anyone can download and read.

i'll say this again: don't take my word for it, READ THE WADIA WHITEPAPERS and draw your own conclusions. wadia has several whitepapers posted on their website that anyone can download and read.

OK, I see that I am right & you are also right - after upsampling (marketing verbage used by Wadia is "resolution enhancement") the word length is 24 bits (which is what I wrote) & then, after processing by DigiMaster algo, they chuck the least 2 significant bits & reduce the rez to 21/22 bits (which is what you wrote).

On a side note - I think all of us would be better of if you dropped your high-handedness altogether. You might have a good reason to attain your attitude but realize that you are on a PUBLIC forum where all sorts come to visit & discuss. If you feel compelled to share your knowledge & uplift the "ignorant" masses here do it with an attitude of a school teacher - you'll be appreciated much more. Just my suggestion to you being a long-term member here.

12-03-10: Bombaywalla
OK, I see that I am right & you are also right - after upsampling (marketing verbage used by Wadia is "resolution enhancement") the word length is 24 bits (which is what I wrote) & then, after processing by DigiMaster algo, they chuck the least 2 significant bits & reduce the rez to 21/22 bits (which is what you wrote).

as you saw in the white papers, the wadia has a 22-bit DAC, so the wadia digital preamplifier operates on the 21/22-bit signal. 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. this, in turn, effectively knocks off the 6 least significant bits from the 21/22-bit signal. thus, by the time you are at 65 on the digital output level, you only have the most significant 15/16 bits remaining in the original 21/22-bit signal and you are at the point where you are starting to lose significant information (since the original signal from the cd was a 16-bit signal).

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...