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I'm trying to understand an analysis of a sampled continuous time system in the Laplace domain. The source analysis is here (PDF page 6, slide marked 11); I'll explain further below. Suppose I have a system which takes the difference between its current input and its input sampled from half a clock cycle ago, so that:

$$ V_o(kT)=V_f(kT)-V_f(kT-\tfrac{T}{2}) \tag{1} $$

Here $V_o(t)$ and $V_f(t)$ are continuous-time functions, $k$ is a nonnegative integer, and $T$ is a positive time interval.

I think that the system function is likely properly examined in terms of a Z-transform, but I want to understand how it could be analyzed via Laplace transform. The source I cited above claims that you can simply treat $kT$ as a continuous time variable and end up with (taking the Laplace transform of the RHS above)

$$ V_f(s)(1-e^{-sT/2}) \tag{2} $$

A (maybe small) detail that confuses me here is that this suggests that the original CT signal had a shift of the form $V_f(t-T/2)$, when actually if you do a substitution $t'=kT$, we have $V_f(t'-\frac{1}{2k}$), which doesn't look right. Is there a way to show that the RHS of equation 1 has the Laplace transform of equation 2, or is there an approximation being made here?

EDIT: I've made a timing diagram to try to make the situation more clear. This is a filled-in version of slide 10 from my source above, but without the gain block A which I don't think makes a difference here. In this diagram I tried to make $V_f(t)$ a very slow-varying signal (slow relative to input sinusoid $V_{in}$). $V_1$ is the output of the sample and hold block.

At time $t=kT-T/2$, the first phase ends and the sample/hold block holds the value of $V_f(t=kT-T/2)$. At the start of phase 2, $V_{in}$ is connected to the system and the output becomes $V_o(t)=V_{in}(t)+V_f(t)-V_f(kT-T/2)$. Phase 2 ends at $t=kT$, and then the output gets held at $V_o(kT)=V_{in}(kT)+V_f(kT)-V_f(kT-T/2)$.

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There are two ways of analyzing the given system. First, we could simply ignore the sampling process and treat the system as a continuous-time system. This is possible if the input signals are sufficiently band-limited, and if the sampling rate satisfies the Nyquist criterion. In this case we simply have the difference between the original noise signal $v_f(t)$ and a shifted version $v_f(t-T/2)$, which in the Laplace domain corresponds to

$$\mathcal{L}\big\{v_f(t)-v_f(t-T/2)\big\}=V_f(s)\left(1-e^{-sT/2}\right)$$

The second way to analyze the system is in the discrete-time domain. Note however that in this case we need a sampling interval of $T/2$ because we need both, the sampled signal $v_f(kT)$ and the sampled version of the shifted signal $v_f(kT-T/2)$. This is not possible if the sampling interval is $T$.

Modeling sampling mathematically by multiplication with an impulse train, and using a sampling interval of $T'=T/2$ results in the following representation of a sampled signal:

$$v_f(t)\cdot\sum_{n=-\infty}^{\infty}\delta(t-nT')=\sum_{n=-\infty}^{\infty}v_f(nT')\delta(t-nT')\tag{1}$$

The Laplace transform of $(1)$ is

\begin{align*} \mathcal{L}\left\{\sum_{n=-\infty}^{\infty}v_f(nT')\delta(t-nT')\right\} &=\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}v_f(nT')\delta(t-nT')e^{-st}dt\\ &=\sum_{n=-\infty}^{\infty}v_f(nT')e^{-snT'}\tag{2} \end{align*}

The expression on the right-hand side of $(1)$ is called the discrete Laplace transform, and it's just the same as the $\mathcal{Z}$-transform with $z=e^{sT'}$. Hence, a delay of one sample corresponds to a multiplication with $z^{-1}=e^{-sT'}=e^{-sT/2}$.

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  • $\begingroup$ I think there are two things happening: one is that the system is being sampled with period $T$, the other is that the output voltage is related to $V_f$ by a copy of $V_f$ delayed by half a sample period, T/2. $\endgroup$
    – Peter K.
    Commented Jun 13, 2023 at 16:17
  • $\begingroup$ @PeterK. Hmm, but how to delay (exactly) by half a sample period if the sample period is $T$? I'm not entirely clear about how the system works ... $\endgroup$
    – Matt L.
    Commented Jun 13, 2023 at 17:00
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    $\begingroup$ The delay may happen in the continuous-time part of the system before any sampling is possible. Agreed having it exactly half a sample and synchronous with sampling is odd, but it's possible. $\endgroup$
    – Peter K.
    Commented Jun 13, 2023 at 17:02
  • $\begingroup$ Thanks a lot for the response, and apologies for lack of clarity. I think @PeterK. is right about what's happening in the system and I updated the question with a timing diagram to try to make the situation clearer. It might also help to mention that a goal of the system is to remove $V_f$ from the output, where $V_f$ is either a DC offset or a very low frequency noise component. So the output samples are being held every $T$ seconds where they contain the value of $V_{in}$ at that time, minus the value of $V_f$ from $T/2$ seconds ago, and this is all in the continuous time domain. $\endgroup$
    – Halleff
    Commented Jun 13, 2023 at 17:56
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    $\begingroup$ @knzy: But note that an impulse train shifted by $T/2$ (with sampling interval $T$) is not enough for representing the difference $V_f(kT)-V_f(kT-T/2)$. That's what I meant by saying that for an exact representation in discrete time you would need a sampling interval of $T/2$, otherwise you don't get both $V_f(kT)$ and $V_f(kT-T/2)$. $\endgroup$
    – Matt L.
    Commented Jun 14, 2023 at 8:55

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