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I am currently reading the Fringe Contrast section of this document. This section says the following:

Use a high-pass filter (HPF) and an amplifier (G) to maximize the signal amplitude for analog-to-digital conversion (ADC). enter image description here

How does using a high-pass filter and an amplifier maximize the signal amplitude for analog-to-digital conversion? I'm a novice to electronics, so I'd appreciate a more "basic" explanation.

The Pointer
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    1) remove DC offset and 2) expand the input signal range to match the ADC's input range. –  Aug 07 '21 at 14:51

1 Answers1

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It appears that in this application you are only interested in the alternating signal and not in the DC component. Let's give an example:

  • DC offset = 300 mV.
  • AC signal = 100 mVpk-pk.
  • ADC input = -3 to +3 V.

If you feed the raw signal into the ADC you will be using \$ \frac {0.100}{3 -(-3)} = 1.6\% \$ of the full-scale resolution of the ADC. Accuracy won't be as good as it could be.

To fix this:

  • Remove the DC component with a high-pass filter. This might be as easy as a capacitor in series with the signal. The output of the HPF will be the 100 mVpk-pk AC signal oscillating around 0 V.
  • Let's say we want to amplify this to a max of ±2.5 V to leave a little headroom at each side. That means a gain of \$ \frac {2.5 - (-2.5)}{0.1} = 50 \$ for the G block.

This will give the ADC a signal of ±\$\frac 5 6 = 83\%\$ full scale.

Transistor
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  • Thanks for the answer. I have some questions: 1. So we use the high-pass filter to cut off frequencies (specifically, the DC frequency) higher than the AC frequency? 2. Comparing \$ \frac {0.100}{3 -(-3)} = 1.6\% \$ and \$ \frac {2.5 - (-2.5)}{0.1} = 50 \$, why were the ADC input voltage and the AC signal swapped between the numerator/denominator between these two calculations? – The Pointer Aug 07 '21 at 02:19
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    1. You would choose the "cut-off" frequency of the HPF to be lower than the AC frequency of interest so that it does not attenuate it. The mini-graph in the HPF box hints at this. You want to choose the corner frequency lower than the range of interest. DC has a frequency of 0 so it will be stopped by any HPF. 2. The first is just showing you the % of full scale. The second is a gain calculation. Read it again - I think it should make sense. – Transistor Aug 07 '21 at 02:25