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# Basic DAC Question

C

#### [email protected]

Jan 1, 1970
0
I have three questions about theoretical THD+N distortion while
creating
sine waves with DACs.

First question: Assuming a perfect DAC with perfect analog
components,
obviously a DAC with fewer bits will create a sinewave with larger
steps,
and thus will have a higher percentage of THD+N, but how do I
calculate
the exact percentage?

Second question: I know about the Nyquist limit and it seems to me
that
as the Nyquist limit is approached the sinewave will have bigger
steps
no matter how many bits it has and thus will have a higher percentage
of THD+N, but how do I calculate the exact percentage?

Third question; In the real world I wouldn't have perfect analog
components; in fact I would purposely introduce a lowpass filter
at the output of the DAC to attenuate the switching noise. How
much would that change the answers to the questions above?

BACKGROUND:

We need to replace an old system that generates 20 Hz to 20 kHz sine
waves with a 12 bit DAC that puts out a 4096-step sine wave -- the
same number of steps whether it is putting out 20 Hz or 20 kHz.
A variable oscillator changes the clock rate of a counter that
gets the values from an EPROM lookup table.

We were discussing replacing the above with a modern DAC -- either
16 bits at 44.1 ksps or 24 bits at 96 ksps. The objection was
raised that at 20Khz we are putting out 4096 x 20,000 sps, or
81.92 Msps. I am guessing that 96 ksps with a added filter at
the DAC output is good enough. The final power stage starts
slew-rate limiting at 30-40 kHz with large signals and the small-
signal response is 3dB down at 50Khz and way down in the mud at
100KHz. I just don't see how it needs over 80 megasamples per
second to keep the THD+N reasonably low. Am I right?

J

#### John Larkin

Jan 1, 1970
0
I have three questions about theoretical THD+N distortion while
creating
sine waves with DACs.

First question: Assuming a perfect DAC with perfect analog
components,
obviously a DAC with fewer bits will create a sinewave with larger
steps,
and thus will have a higher percentage of THD+N, but how do I
calculate
the exact percentage?

Second question: I know about the Nyquist limit and it seems to me
that
as the Nyquist limit is approached the sinewave will have bigger
steps
no matter how many bits it has and thus will have a higher percentage
of THD+N, but how do I calculate the exact percentage?

Third question; In the real world I wouldn't have perfect analog
components; in fact I would purposely introduce a lowpass filter
at the output of the DAC to attenuate the switching noise. How
much would that change the answers to the questions above?

BACKGROUND:

We need to replace an old system that generates 20 Hz to 20 kHz sine
waves with a 12 bit DAC that puts out a 4096-step sine wave -- the
same number of steps whether it is putting out 20 Hz or 20 kHz.
A variable oscillator changes the clock rate of a counter that
gets the values from an EPROM lookup table.

We were discussing replacing the above with a modern DAC -- either
16 bits at 44.1 ksps or 24 bits at 96 ksps. The objection was
raised that at 20Khz we are putting out 4096 x 20,000 sps, or
81.92 Msps. I am guessing that 96 ksps with a added filter at
the DAC output is good enough. The final power stage starts
slew-rate limiting at 30-40 kHz with large signals and the small-
signal response is 3dB down at 50Khz and way down in the mud at
100KHz. I just don't see how it needs over 80 megasamples per
second to keep the THD+N reasonably low. Am I right?

If you just want a sine wave, use an Analog Devices DDS chip. For
under $10, the whole job is done for you, DAC and all. The effective sample rate will be so high, 20 MHz maybe, that the most primitive output filter will do. John J #### Jim Granville Jan 1, 1970 0 I have three questions about theoretical THD+N distortion while creating sine waves with DACs. First question: Assuming a perfect DAC with perfect analog components, obviously a DAC with fewer bits will create a sinewave with larger steps, and thus will have a higher percentage of THD+N, but how do I calculate the exact percentage? Second question: I know about the Nyquist limit and it seems to me that as the Nyquist limit is approached the sinewave will have bigger steps no matter how many bits it has and thus will have a higher percentage of THD+N, but how do I calculate the exact percentage? Third question; In the real world I wouldn't have perfect analog components; in fact I would purposely introduce a lowpass filter at the output of the DAC to attenuate the switching noise. How much would that change the answers to the questions above? BACKGROUND: We need to replace an old system that generates 20 Hz to 20 kHz sine waves with a 12 bit DAC that puts out a 4096-step sine wave -- the same number of steps whether it is putting out 20 Hz or 20 kHz. A variable oscillator changes the clock rate of a counter that gets the values from an EPROM lookup table. We were discussing replacing the above with a modern DAC -- either 16 bits at 44.1 ksps or 24 bits at 96 ksps. The objection was raised that at 20Khz we are putting out 4096 x 20,000 sps, or 81.92 Msps. I am guessing that 96 ksps with a added filter at the DAC output is good enough. The final power stage starts slew-rate limiting at 30-40 kHz with large signals and the small- signal response is 3dB down at 50Khz and way down in the mud at 100KHz. I just don't see how it needs over 80 megasamples per second to keep the THD+N reasonably low. Am I right? What is this being used for ?- ie what specs do you need to meet ? You are right that 20KHz is easier than 20Hz, because the upper frequency can have post filtering, to push any harmonics down. Having said that, going from 82Msps to 96Ksps is a drop of nearly 1000:1, which is a large system change. Will it matter that you now have just over 4 samples per full cycle at 20KHz ? -jg J #### Jim Granville Jan 1, 1970 0 John said: If you just want a sine wave, use an Analog Devices DDS chip. For under$10, the whole job is done for you, DAC and all. The effective
sample rate will be so high, 20 MHz maybe, that the most primitive
output filter will do.

John

Good idea. They also have an ARM variant ADuC7128), with DDS included,
so you might be able to take that, and clone part of the present
system interface, into a one chip retro-replacement.

-jg

P

#### Phil Allison

Jan 1, 1970
0
[email protected]>
I have three questions about theoretical THD+N distortion while
creating sine waves with DACs.

First question: Assuming a perfect DAC with perfect analog
components,
obviously a DAC with fewer bits will create a sinewave with larger
steps,
and thus will have a higher percentage of THD+N, but how do I
calculate
the exact percentage?

** It depends on the nature of the digital signal the DAC is converting and
the bandwidth of the post filter.

At the expense of a small increase in noise, THD at the output can be almost
eliminated by "dithering " the incoming data stream.

Second question: I know about the Nyquist limit and it seems to me
that
as the Nyquist limit is approached the sinewave will have bigger
steps
no matter how many bits it has and thus will have a higher percentage
of THD+N, but how do I calculate the exact percentage?

** Wrong - s/n and linearity are not affected. Sampling theory shows that
just over two precise samples per cycle *fully* characterises a sine wave. A
sine wave only has three parameters to capture - amplitude, frequency and
phase.

No others.

You need to read tests results from CD players to get the idea in your head
of how nearly perfect the process is.

Third question; In the real world I wouldn't have perfect analog
components; in fact I would purposely introduce a lowpass filter
at the output of the DAC to attenuate the switching noise.

** A "reconstruction filter" is essential to any D to A.

We need to replace an old system that generates 20 Hz to 20 kHz sine
waves with a 12 bit DAC that puts out a 4096-step sine wave -- the
same number of steps whether it is putting out 20 Hz or 20 kHz.

** Converting at over 80 MHz is quite a feat.

Care to name that DAC ?

A variable oscillator changes the clock rate of a counter that
gets the values from an EPROM lookup table.

We were discussing replacing the above with a modern DAC -- either
16 bits at 44.1 ksps or 24 bits at 96 ksps. The objection was
raised that at 20Khz we are putting out 4096 x 20,000 sps, or
81.92 Msps. I am guessing that 96 ksps with a added filter at
the DAC output is good enough. The final power stage starts
slew-rate limiting at 30-40 kHz with large signals and the small-
signal response is 3dB down at 50Khz and way down in the mud at
100KHz. I just don't see how it needs over 80 megasamples per
second to keep the THD+N reasonably low. Am I right?

** No - you are totally wrong !!

The sampling rate of an audio DAC is fixed.

The term is " samples per second " - not " samples per cycle "

Go read Nyquist and Shannon again, cos you have missed to most important
bit.

Pun intended.

....... Phil

J

#### John Larkin

Jan 1, 1970
0
Your numbers don't hold together. 20 kHz sine, with a 4096 step
DAC, requires 80 Mhz operation of the DAC, or roughly 10 Ns. Look
at the basics first.

It's not necessary to hit every step in a sine lookup table, which is
why you can get a high-resolution 20 KHz sine wave at a mere 44 KHz
(CD) sample rate. At higher frequencies, you can start making
many-address hops in the table without penalty. The filter fixes it
all up.

The advantage of higher sample rates (say, 96 KHz instead of 44) is
that the filter need not be so good, and the zero-order-hold (sinc)
rolloff is mostly eliminated.

John

J

#### John Devereux

Jan 1, 1970
0
Jim Granville said:
Good idea. They also have an ARM variant ADuC7128), with DDS included,
so you might be able to take that, and clone part of the present
system interface, into a one chip retro-replacement.

I have been curious about that chip since it was released - do you
happen to know what it was actually designed for?

J

#### Jim Granville

Jan 1, 1970
0
John said:
I have been curious about that chip since it was released - do you
happen to know what it was actually designed for?

The press release says this " For smart sensing applications, the
ADuC7128 features a 32-bit on-chip DDS that operates at 21 MHz." -

there are some sensore, like LVDS, that like sine drive.
But there are some things they seems to have overlooked, and
you get the impression this was a simple cut and paste job.

For example, quadrature sine drive could have been useful,
allowing Sin/Cosine meters to be drivem using the Sine ROM.
They also have no digital phase path, from the DDS.

It seems you must read the DDS via the ADC, and do the
same with a response channel, and then use SW to
calculate the Phase.

-jg

G

#### Guy Macon

Jan 1, 1970
0
Jim said:
Will it matter that you now have just over 4 samples per full
cycle at 20KHz ?

Phil said:
s/n and linearity are not affected. Sampling theory shows that
just over two precise samples per cycle *fully* characterises a
sine wave. A sine wave only has three parameters to capture -
amplitude, frequency and phase. No others.

John said:
It's not necessary to hit every step in a sine lookup table, which is
why you can get a high-resolution 20 KHz sine wave at a mere 44 KHz
(CD) sample rate. At higher frequencies, you can start making
many-address hops in the table without penalty. The filter fixes it
all up.

With the specified large signal slew rate limit of 30 kHz and small
signal rolloff at 50Khz, most of the frequency cmponents that make
a 20 kHz 4 samples per cycle stepped waveform different from a pure
sine wave are too high for this system to reproduce. Add a filter
at the DAC output (which you have to do anyway) and it gets even
better. Factor in the fact that THD+N is specified over some
frequency range and 4 samples per cycle looks even better. If the
THD+N meter was perfect and measured only up to, say, 30kHz, a
harmonic at 40 kHz would not be measured. Real THD+N meters will
let a bit of the out of band harmonics in, but less and less as
the harmonics get higher.

That being said, I have to admit that I also don't know how to
calculate the theoretical THD+N in percent starting from the
number of bits and the sample rate. I have seen lookup tables
for number of bits vs. THD, but not for the sample rate to
signal frequency ratio vs. THD. There must be a formula for
calculating those numbers, but I can't find it.

C

#### [email protected]

Jan 1, 1970
0
I just got in a sample board of the old design and looked at it with
a
scope. I was told that it had a 12 bit DAC that puts out a 4096-step
sine wave at 20Hz to 20kHz. On the board I saw an Analog Devices
DAC312
which isn't fast enough.

I set the output to 20Hz, and measured the DAC clock at 82kHz. That
seemed right (20*4096=81,920.) As I raised the output frequency the
DAC clock went up, but at 626 Hz and 2.6MHz the DAC clock suddenly
dropped in half -- but the output frequency didn't. This happened
again at 1251Hz, 2501Hz. 5001 Hz. and 10001 Hz. At the top end I
saw little stair steps in the output. About 128 of them in each
cycle.

It looks to me like the board is going to a different part of the
EPROM for a sinewave with fewer steps each time it drops the DAC
clock in half. So the 4096 steps are only at low frequencies.
There are only 256 steps above 5kHz and 128 steps above 10kHz.

Here is how I think they are doing it:

20*4096 = 81,920
625*4096 = 2,560,000
1,250*2048 = 2,560,000
2,500*1024 = 2,560,000
5,000*512 = 2,560,000
10,000*256 = 2,560,000
20,000*128 = 2,560,000

They also sent me a manual, which is a good thing because they also
misinformed me when they said it always puts out sine waves. There
is also a setting labeled "PEAK-RMS 1.246" that looks like a sinewave
with some hard clipping. It looks nice and clean at 20Hz but at
20kHz
not so nice at the output. I need to find out how high they need
to go with it. No official word but one of the technicians claims
that he has only seen the PEAK-RMS 1.246 setting used at 50, 60,
and 400 Hz. Ain't discovering customer requirements grand? I wish
they would just tell me what they are trying to accomplish instead
of giving me specs that are wrong.

C

#### CBFalconer

Jan 1, 1970
0
I just got in a sample board of the old design and looked at it
with a scope. I was told that it had a 12 bit DAC that puts out
a 4096-step sine wave at 20Hz to 20kHz. On the board I saw an
Analog Devices DAC312 Datasheet
which isn't fast enough.

If you want serious discussion read the following sig and the URL.

--
If you want to post a followup via groups.google.com, ensure
you quote enough for the article to make sense. Google is only
an interface to Usenet; it's not Usenet itself. Don't assume
your readers can, or ever will, see any previous articles.
More details at: <http://cfaj.freeshell.org/google/>

J

#### Jim Granville

Jan 1, 1970
0
That being said, I have to admit that I also don't know how to
calculate the theoretical THD+N in percent starting from the
number of bits and the sample rate. I have seen lookup tables
for number of bits vs. THD, but not for the sample rate to
signal frequency ratio vs. THD. There must be a formula for
calculating those numbers, but I can't find it.

Some (most?) spice engines allow table entry, and will do fourier plots,
so you could enter the sine LUT into a table and
run the fourier ?
I recall ~1yr ago, bumping into a table limit in B2Spice,
and got them to fix it for this type of use, but no, I have
not done this specific table usage.

-jg

S

#### Steve at fivetrees

Jan 1, 1970
0
CBFalconer said:
If you want serious discussion read the following sig and the URL.

--
If you want to post a followup via groups.google.com, ensure
you quote enough for the article to make sense. Google is only
an interface to Usenet; it's not Usenet itself. Don't assume
your readers can, or ever will, see any previous articles.
More details at: <http://cfaj.freeshell.org/google/>

CB: do us all a favour and get a newsreader that threads. Then we can look
forward to your signal-to-noise ratio improving significantly.

Steve
http://www.fivetrees.com

J

#### [email protected]

Jan 1, 1970
0
I just got in a sample board of the old design and looked at it with
a
scope. I was told that it had a 12 bit DAC that puts out a 4096-step
sine wave at 20Hz to 20kHz. On the board I saw an Analog Devices
DAC312
which isn't fast enough.

I set the output to 20Hz, and measured the DAC clock at 82kHz. That
seemed right (20*4096=81,920.) As I raised the output frequency the
DAC clock went up, but at 626 Hz and 2.6MHz the DAC clock suddenly
dropped in half -- but the output frequency didn't. This happened
again at 1251Hz, 2501Hz. 5001 Hz. and 10001 Hz. At the top end I
saw little stair steps in the output. About 128 of them in each
cycle.

It looks to me like the board is going to a different part of the
EPROM for a sinewave with fewer steps each time it drops the DAC
clock in half. So the 4096 steps are only at low frequencies.
There are only 256 steps above 5kHz and 128 steps above 10kHz.

Here is how I think they are doing it:

20*4096 = 81,920
625*4096 = 2,560,000
1,250*2048 = 2,560,000
2,500*1024 = 2,560,000
5,000*512 = 2,560,000
10,000*256 = 2,560,000
20,000*128 = 2,560,000

They also sent me a manual, which is a good thing because they also
misinformed me when they said it always puts out sine waves. There
is also a setting labeled "PEAK-RMS 1.246" that looks like a sinewave
with some hard clipping. It looks nice and clean at 20Hz but at
20kHz
not so nice at the output. I need to find out how high they need
to go with it. No official word but one of the technicians claims
that he has only seen the PEAK-RMS 1.246 setting used at 50, 60,
and 400 Hz. Ain't discovering customer requirements grand? I wish
they would just tell me what they are trying to accomplish instead
of giving me specs that are wrong.

There seems no(?) easy way to estimate the THD based on DAC bits and
the number of points per cycle and a final filter form.
At 20Hz there's a massive number of points building up a each precise
single cycle and at 16bits, distortion will naturally be about 0.02%.
The filter though will only be stripping off harmonics beyond the
1000th and at a rate dependant on it's rolloff and shape. Up at 20kHz,
the reduced points per cycle may be giving an intrinsic 3% THD but the
filter will now be biting very hard on the 2nd harmonic and all above.
Final distortion entirely dependant on the particular filter used. Pro
rata for all intermediate frequencies. It smells like some kind of
balancing act is going on that is sufficient to always give a low
distortion figure.

Maybe do the new design just the same way as the existing but using a
modern 16 bitter. Maybe speed it up as well. The binary dividing
method has similarity to the workings of a DDS chip but has a far
superior output waveform as there is no distortion added due to
(unfilterable) non harmonic spurs and jitter.

R

#### Rich Grise, Plainclothes Hippie

Jan 1, 1970
0
CB: do us all a favour and get a newsreader that threads. Then we can look
forward to your signal-to-noise ratio improving significantly.

It wouldn't do any harm for C.N to include a little context for us
latecomers.

Cheers!
Rich

A

#### Apostrophe Police

Jan 1, 1970
0
1000th and at a rate dependant on it's rolloff and shape. Up at 20kHz,
^^^^

Please take the time to learn that there is no apostrophe in the
possessive its.

It damages your credibility.

Regards,

J

#### John Larkin

Jan 1, 1970
0
That being said, I have to admit that I also don't know how to
calculate the theoretical THD+N in percent starting from the
number of bits and the sample rate.

The usual expression is that

s/n = 6.02N - 1.249 dB

and is independent of sample rate if the usual sampling rules are
followed, and assuming an ideal dac. This assumes that the signal has
a gaussian distribution and averages 1/4 of ADC full scale. Whatever
the definition of "signal", the improvement in s/n remains 6.02 dB per

John

P

#### Phil Allison

Jan 1, 1970
0
"John Larkin"
Guy Macon

The usual expression is that

s/n = 6.02N - 1.249 dB

and is independent of sample rate if the usual sampling rules are
followed, and assuming an ideal dac. This assumes that the signal has
a gaussian distribution and averages 1/4 of ADC full scale. Whatever
the definition of "signal", the improvement in s/n remains 6.02 dB per

** For the full scale sine wave case (as in the OP's question) the formula
is

Quantisation noise for an ideal DAC

= 6.02N + 1.76 dB.

= about 3 dB better.

........ Phil

P

#### Paul Keinanen

Jan 1, 1970
0
There seems no(?) easy way to estimate the THD based on DAC bits and
the number of points per cycle and a final filter form.

Why do you want to know the _total_ _harmonic_ distortion for a
sampled audio? system ?.

There is always the classical formula for SNR in dB = 1.76 dB +6.02n,
in which n is the number of bits.

Some of the noise components are outside the required audio passband,
especially when some form of noise shaping is used and thus filtered
out.

In a sampled system, you will only get strong _harmonic_ components,
when the produced waveform is a subharmonic of the sampling frequency,
at other generated frequencies, the same noise power is distributed
among a very large number of frequencies, creating a noise floor.

Look at the spectrum for a DDS system, there are usually a noise
floor, but at some frequencies, the noise power is concentrating on a
few discrete spurs, while the frequencies in between are very quiet.

Paul

P

#### Phil Allison

Jan 1, 1970
0
"Paul Keinanen"
Why do you want to know the _total_ _harmonic_ distortion for a
sampled audio? system ?.

** Funny how so many folk are interested in the linearity of an audio system
and wanna know the THD figure - maybe they know more than you.

Good old THD testing is the simplest measure of linearity and when done
across the whole audio band is very informative.

Sometimes two high level, high frequency tones are used for DAC and ADC
tests (where the difference signal is noted) to avoid the post filter
enhancing the THD figure.

........ Phil

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