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Discussion Groups | Comp.DSP | Purpose of interpolation in DACs?

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Purpose of interpolation in DACs? - Ben Jackson - 2006-05-18 18:10:00

I'd like to test my understanding of the usefulness of interpolation
in a DAC:

If you feed a signal to a DAC at Fc < Fclk/2 then there is a primary
image at Fc and (possibly unwanted) aliasing at Fclk +/-Fc and at
harmoics of Fclk +/-Fc.

Now, if you upsample (zero-stuff) the input to a higher Fclk, then the
DAC runs faster, and the DAC-produced images are now at n*Fclk+/-Fc.
However, the zero-stuffed data itself now has the images that would have
appeared in the original DAC output, so the result is the same (?).

So the next step is interpolation -- upsample + digital filter.  The
digital filter is cheaper and can have sharper cutoffs than an analog
filter.  Of course it can't filter beyond n*Fclk/2, but at least now the
new, higher (and fewer, in some sense) n*Fclk+/-Fc images are all that's
left in the output.  This is nice because the analog output filter can
be lower order and have a wider transition band and still have the same
alias suppression.

Am I missing anything about DAC interpolation?  I know there are some
other more exotic uses (eg filtering to use a higher frequency alias
directly to avoid an upconversion step).

-- 
Ben Jackson
<b...@ben.com>
http://www.ben.com/


Re: Purpose of interpolation in DACs? - Bhaskar Thiagarajan - 2006-05-18 19:19:00

"Ben Jackson" <b...@ben.com> wrote in message
news:s...@saturn.home.ben.com...
> I'd like to test my understanding of the usefulness of interpolation
> in a DAC:
>
> If you feed a signal to a DAC at Fc < Fclk/2 then there is a primary
> image at Fc and (possibly unwanted) aliasing at Fclk +/-Fc and at
> harmoics of Fclk +/-Fc.
>
> Now, if you upsample (zero-stuff) the input to a higher Fclk, then the
> DAC runs faster, and the DAC-produced images are now at n*Fclk+/-Fc.
> However, the zero-stuffed data itself now has the images that would have
> appeared in the original DAC output, so the result is the same (?).
>
> So the next step is interpolation -- upsample + digital filter.  The
> digital filter is cheaper and can have sharper cutoffs than an analog
> filter.  Of course it can't filter beyond n*Fclk/2, but at least now the
> new, higher (and fewer, in some sense) n*Fclk+/-Fc images are all that's
> left in the output.  This is nice because the analog output filter can
> be lower order and have a wider transition band and still have the same
> alias suppression.
>
> Am I missing anything about DAC interpolation?  I know there are some
> other more exotic uses (eg filtering to use a higher frequency alias
> directly to avoid an upconversion step).
>
> -- 
> Ben Jackson
> <b...@ben.com>
> http://www.ben.com/

You can use the digital filter to pre-compensate for the sinx/x roll-off you
get from the DAC.

Cheers
Bhaskar


Re: Purpose of interpolation in DACs? - Randy Yates - 2006-05-19 03:02:00

Hi Ben,

I think your analysis so far is correct.

There is another reason for upsampling a signal to a DAC: to increase
the effective resolution of the DAC. In order for this to work, your
digital signal resolution must be wider than the DAC resolution. For
example, if your digital signal was 16 bits and your DAC was only 12
bits, you could use this technique.

The key to understanding this is realizing that the total quantization
noise power (or, in this case, "re"quantization noise, since we're
converting a digital signal to a lower-resolution digital signal)
depends ONLY on the amount of quantization and NOT on the sample
rate. For example, if I drop the resolution by one bit, I get q^2/12
total noise power [1] spread out over the entire digital bandwidth
from 0 to Fs/2, where q is the quantization step-size (ultimately
determined by your DAC reference voltage and resolution).

So let's say, e.g., that you have a digital signal that's 16 bits wide
and originally at 32 kHz sampling rate. If you upsample as you
described, keeping the digital signal at 16 bits, to 64 kHz,
and then *at the 64 kHz rate* requantize the signal to 12 bits, then
you'd introduce (4*q)^2 / 12 quantization noise power into your 16-bit
signal. But that noise power would be spread from 0 to 32 kHz, whereas
your bandwidth of interest (from the original "baseband" sample rate)
is only 0 to 16 kHz. So the total quantization noise power in your
bandwidth of interest is now 1/2 of what it would have been had
you not upsampled.

It's easy to see that every doubling of the sample rate gives you a 3
dB decrease in quantization noise, and so it takes an oversampling
ratio of 4 to get a 6 dB reduction, or 1 bit of extra resolution. In
general, oversampling by M gives you log_4(M) bits of resolution.

--Randy



[1] This makes the standard assumption that the input signal is
sufficiently complex that the quantization noise is white and
uniformly distributed.