0
$\begingroup$

How does an equalizer prevent a sample and hold effect?

Does a reconstruction filter only smooth out the output of an DAC (+S&H) or does it have another function?

$\endgroup$

2 Answers 2

1
$\begingroup$

Equalizers do not prevent the sample and hold effect, though if you are talking about compensation filters, see the edit below.

There are two types of reconstruction filters (also known as low-pass filters): analog and digital. An analog low-pass filter can smooth out the output of a DAC. A digital low-pass filter cannot, since it comes before the DAC. What it can do is reduce the effect of the sample and hold by increasing the sample rate (assuming it is an interpolation filter) and then getting rid of the aliases via the low-pass filtering. The effect of the sample and hold is reduced due to the increased sample rate.

EDIT: As Robert points out, the frequency droop that happens as a result of the zero-order hold can be partially compensated for with digital filters whose frequency domain gain is the inverse of the zero-order hold effect. While this will not eliminate the sample and hold effect, it will mitigate it.

$\endgroup$
3
  • $\begingroup$ Jim, i cannot agree with you. if the zero-order-hold effect of the DAC is known in advance (like, say, it's a "conventional" DAC, not a sigma-delta DAC), you can, if you're willing to put in a little delay, compensate in the frequency domain for the droop (which is -3.92 dB at Nyquist) of the ZOH. at least to some extent. but that compensation would be bad if the same samples went out to a DAC that operated at a high internal sampling frequency (like a $\Sigma\Delta$ DAC). $\endgroup$ Jan 8, 2014 at 17:22
  • $\begingroup$ @robertbristow-johnson Good point. Edited the answer. $\endgroup$
    – Jim Clay
    Jan 8, 2014 at 19:16
  • $\begingroup$ cool. i wish i could edit my comment now (to make it compatible with the current answer). one other issue (not a disagreement, but to sorta be clear): some folks (even profs, like who i had for "Digital Control Systems" at NU) incorrectly ascribe to the S/H before the ADC input, this Zero-Order Hold effect. it doesn't matter how the ADC gets the numbers, just that it gets the correct numbers. the ZOH effect is due to the DAC holding the voltage constant (instead of outputting a dirac-like pulse) between sample instants of time. $\endgroup$ Jan 8, 2014 at 19:52
0
$\begingroup$

Let me first distinguish mathematical exactness and practical engineering: From a pure mathematical perspective, the Sample and Hold process in frequency is has a Sine(x)/x response. Sine(x)/x is transcendental, so we cannot exactly compensate for it with digital filter (equalizer) implementations, either with FIR or IIR structures. From a practical engineering perspective, and within a defined spectral range of interest, an equalizer can sufficiently compensate for the DAC (Digital to Analog Converter) reconstruction such that the distortion due to the sine(x)/x "pass-band droop" is reduced to below a level of concern (This distortion comes from the zero order hold reconstruction technique commonly implemented in D/A Converters).

This is done by simply providing the inverse Sinc response digitally to the waveform within the passband of interest; we can design the filter up to the complexity needed that will for practical purposes push the residual distortion below our requirements on total distortion.

The question then is how complicated will the filter need to be to sufficiently match an inverse Sinc response?

Below is one example of an amazingly simple two multiplier (three coefficient linear-phase FIR filter) solution I have used for compensating the same Sine(x)/(x) droop in CIC filters which could work equally well for a digital filter prior to DAC reconstruction:

CIC Compensation

How this works is explained simply by considering that this filter contains three samples as its impulse response: a large impulse in the center with magnitude $1+\alpha$, and two smaller impulses leading and trailing with magnitude $-\alpha/2$.

impulse response for compensator

What the three impulses in time represent in frequency: It may be more intuitive for many to use the reciprocal property of the Fourier Transform and consider instead what those same three impulses in frequency would be in time: A dominant impulse at $f=0$ represents a DC offset. The two smaller and identical impulses on each side represent a cosine of small magnitude. Thus we get a small cosine with a DC offset. The exact same result is what happens for those three impulses in time- we get a cosine ripple in frequency with an offset to otherwise pass all signals with that ripple as depicted for each of the components of the resulting Fourier Transform below:

Fourier Transform

This works since for small frequency offsets, the $Sin(\omega)/\omega \approx (1+\alpha)- \alpha\cos(\omega)$! This results in a reasonable compensation where we can adjust $\alpha$ to minimize mean squared error from a flat response, with the effectiveness of the result based on our overall occupied bandwidth (the wider the percentage of the sampled bandwidth that we use, the more deviation from ideal will result with just the three coefficient solution used). An example compensation is shown in the plot below.

CIC Compensation FIR

$\endgroup$
2
  • $\begingroup$ "providing the inverse Sinc response digitally" or with an analog circuit; I was first introduced to this practice in the 1990's for reconstructing digital video into analog with less unpleasant pixellation. It can be done with a single under-damped LC low-pass stage, fairly economically even at typical-for-1990 video pixel rates. $\endgroup$
    – TimWescott
    Nov 22, 2021 at 16:03
  • $\begingroup$ @TimWescott right! This would be the preferred approach for a static solution as it can be built right into the reconstruction filter already needed with no added cost or complexity- while the digital would provide reprogrammability for a more flexible solution with equally minimum additional cost or complexity since the three additional taps can be convolved with any filtering already done (the usual digital vs analog trade depending on what you are doing) $\endgroup$ Nov 22, 2021 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.