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I am currently working with switched capacitor circuits used as sample and hold stages, gain stages, and integrators.

It is obvious that switching capacitors to simulate resistors involves sampling of voltages. That also means that there is a sampling interval 𝞃(tau) at which samples are collected.

Imagine we have a gain stage in which the input is sampled at phase1 and the output is provided at phase2 and is 0V at phase1. A full clock cycle consists of phase1 and phase2 and a duration of 𝞃.

In the following simulation 𝞃 equals 2 microseconds (sampling frequency=500kHz) and the gain of the amplifier is g=1.

simulation

One can read off the diagram that the phase delay is always in the range of [𝞃/2,3𝞃/2] in the time domain neglecting that the output is 0V at phase2. If we transform this phase delay to the frequency domain we get a linear relation between the phasedelay and the input frequency as shown below:

enter image description here

Now I am wondering about the magnitude error.

My thoughts:

I am sure there must be some kind of magnitude error through sampling, but I am not able to prove it. Looking at the z-transform of the problem above, which can be described as the following transfer function

H(z)= gain* z^(-1) with z^(-1)=e^(-jω*𝞃/2)

no magnitude error results, but I zam sure there is some, maybe in the form of sinx/x.

Can someone please explain if there is some sort of magnitude error or correct me if I made a mistake in my calculations?

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  • \$\begingroup\$ What makes you sure? What's your theory? Have you measured an error? \$\endgroup\$ Commented Sep 15, 2021 at 7:31
  • \$\begingroup\$ Imagine feeding a sine voltage with the rms value of 1V into a zero order sample-hold circuit with a sampling frequency just above the double of the sine frequency. You will get an output voltage consisting of rectangles with different heights. You calculate the rms value of this voltage by adding the area of these rectangles in a specific interval. You will see that the rms voltage is lower and decreases the closer the input frequency is to the nyquist frequency of sampling \$\endgroup\$ Commented Sep 15, 2021 at 11:26
  • \$\begingroup\$ No, the RMS reading will be the same (providing you are above the nyquist frequency). The fundamental frequency of that sampled sinewave however will fall rapidly as you approach nyquist. \$\endgroup\$ Commented Sep 15, 2021 at 11:38
  • \$\begingroup\$ See this to confirm \$\endgroup\$ Commented Sep 15, 2021 at 11:39
  • \$\begingroup\$ Even undersampling the sinewave will work \$\endgroup\$ Commented Sep 15, 2021 at 11:41

1 Answer 1

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I think this is what you might be interested in: -

enter image description here

It basically shows that a barely-sufficiently-sampled sinewave has the same sampled RMS value as the original sinewave. The energy at the fundamental frequency of the sinewave is however somewhat lower because of the sampling artefacts. The fundamental is declining in amplitude as per this formula: -

enter image description here

Both pictures taken from links to answers listed in comments under the question.

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