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I am trying to decipher some notes on a Time-interleaved Analog-to-Digital Converter with 4 sub-ADCs (the author is unreachable at the moment), but there are few things obscure to me so I was hoping to get some hints from the community. Quotations " " will be taken verbatim from the document.

The author starts off well enough with the system description: enter image description here So we have an incoming continuous signal $x(t)$ that gets discretized and then dispatched in to the different sub-ADCs according to $$x_n^{(k)} = x_{4n + k}, \text{ for } k=1, \dots , 4. $$The sampling frequency of $x_m$ is $f_s$. The output of each single sub-ADC is $$y_n^{(k)} = x_n^{(k)} + a_0^{(k)} + z_n^{(k)} + \zeta_n^{(k)}, \quad k=1,\dots,4$$ where $a_0^{(k)}$ is "a constant DC offset", $z_n^{(k)}$ is "a time-varying DC offset, modelled as autoregressive AR(1) process" and $\zeta_n^{(k)}$ is some noise. $x_n^{(k)}$ "is assumed filtered so one can have signals in several Nyquist zones in an individual sub-ADC". The document does not seem take into account offset, gain or timing mismatches; it is more focused on deriving a noise model.

There is then a picture that looks exactly like this: enter image description here The caption says "Complex-vector FFT representation of the DC offsets in the different sub-ADCs."

This is what I do not understand and I would like help with. What am I looking at? More precisely, why are these arrows rotated? I do not think that the time-shift between the $x_n^{(k)}$ would justify this type of representation. And why the DC-offsets would show up at integer multiples of the sampling frequency?

Any input is appreciated!

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  • $\begingroup$ Please edit your question to add a proper citation to the work in question. If there is a copy of it available online and not behind a paywall, please link to that as well as citing the work. $\endgroup$
    – TimWescott
    Apr 16, 2023 at 15:21
  • $\begingroup$ @TimWescott I cannot really do that. It is an internal report draft that I am not allowed to share in its entirety. If this violates the regulations, please close my thread. $\endgroup$
    – gangrene
    Apr 16, 2023 at 15:33
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    $\begingroup$ Be sure to ask about the effect of ADC timing jitter on noise. It'll (A) have an effect, and (B) it'll be multiplicative, not additive, so it'll make the analysis more -- uh -- fun. $\endgroup$
    – TimWescott
    Apr 16, 2023 at 16:03

1 Answer 1

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I'm going to start by adding some notation. The author leaves us with $y^{(k)}_n = x_{4n + k} + \cdots$, which leaves us juggling two indexes.

If I'm reading the diagram correctly, the index $m$ is the "actual" sampling, and defines $k$ and $n$ as $m = 4n + k$, where $k = 1 + m \bmod 4$ and $n = \left \lfloor \frac m 4 \right \rfloor$. So you get an approximation of $x_m$ as:

$$\hat x_m = y^{(k)}_n = x_m + w_m \tag 1$$

where $w_m$ collects all the noise terms (DC, slow-varying "DC" and white noise) into one.

Let the spectrum of $x_m = X(\omega)$. Then, using the same notation, $\hat X = X + W$.

Now note that the component of $w_m$ that depends on the $a_0^{(k)}$ repeats with a period of four samples in $m$: $w_m = a_0^{1 + m \bmod 4} + \cdots$. If you look at the spectrum of $a_0^{1 + m \bmod 4}$, it will have four distinct harmonics: one at $\omega = \frac{2 \pi}{T_s} p$ ($p \in \mathbb I$), one at $\omega = \left(2 \pi + \frac \pi 2 \right)\frac 1 {T_s} n$, one at $\omega = \left(2 \pi + \pi \right)\frac 1 {T_s} n$, and one at $\omega = \left(2 \pi - \frac \pi 2 \right)\frac 1 {T_s} n$.

These are the bunches of arrows you see laid out on the frequency line. It's showing you how the DC offsets of the individual ADCs are contributing to tones in the output.

The arrow direction is expressing the sum as a complex number for each frequency point, and flows from the nature of spectral analysis. At $\omega = 0$ the DC offsets just add up. At $f_s/4$, the offsets add up in quadrature, i.e. $a_0^1 + j a_0^2 - a_0^3 - j a_0^4$. At $f_s/2$ the odds oppose the evens, $a_0^1 - a_0^2 + a_0^3 - a_0^4$, then at $3f_s/4$ they add in quadrature, but in the other direction.

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  • $\begingroup$ All clear. Thank you very much! $\endgroup$
    – gangrene
    Apr 16, 2023 at 16:38

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