Suppose I have transfer functions for two continuous causal linear-time invariant (LTI) systems: $F_1(s)$ and $F_2(s)$.

Let $D\left\{\cdot\right\}$ denote the function that maps a transfer function from the continuous-time domain to the discrete-time domain (as another transfer function via the $z$-transform) using ZOH discretization.

What is the relationship between: $$D\left\{F_1(s) F_2(s)\right\} \quad \longleftrightarrow \quad D\left\{F_1(s)\right\}D\left\{F_2(s)\right\}$$

  • In other words, what is the difference between discretizing the convolution of two transfer functions versus convolving their discretization?
  • Some brief experimentation has shown they are "almost" equal, but why?
  • Can we be precise about this difference?

As an aside,

  • Could someone point me to a reference on other properties of discretization, e.g., linearity $D\left\{a_1F_1(s) + a_2F_2(s)\right\} = a_1D\left\{F_1(s)\right\} + a_2D\left\{F_2(s)\right\}$?
  • $\begingroup$ Discretization, as defined in your ref link, is a nonlinear process due to the existance of amplitude quantization, at the least. So your last line is not exact in the mathematical sense. However the error is controllable and could be made arbitrarily small by using fine enough quantization. On the other hand, the model conversion phase of dicretization seems a bit more complex to claim much about its linearity. $\endgroup$
    – Fat32
    Aug 6, 2016 at 20:43
  • $\begingroup$ I've numerically tested the linearity claim using scipy's cont2discrete with method='zoh' on several combinations of LTI systems, and it actually seems to hold exactly in every case that I try. The discretization may be nonlinear in terms of the state-space matrices, but so is the mapping from state-space to t.f. ($F(s) = C(sI - A)^{-1}B + D$) -- and so the two nonlinearities may 'cancel'. However I can't find a proof of this claim anywhere online. It may just follow from the exactness of the discretization, and the fact these systems are linear. $\endgroup$ Aug 6, 2016 at 21:58
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    $\begingroup$ @Fat32, let's leave amplitude quantization out of the problem. it's a different issue. $\endgroup$ Aug 7, 2016 at 18:32
  • $\begingroup$ @robertbristow-johnson the ref OP points includes a discussion of quantization during the conversion process that's probably how I came to the conclusion of the nonlinearity of the conversion process, of his last line. I admit that its presence is a lesser issue. But if there is quantization in any process, then its nonlinearty is guaranteed isn't it? That's what I though OP was asking? However, ignoring all the quantization effects, most probably the OP is seeking an answer towards the model conversion aspect of the process, which is answered by Matt I can see. $\endgroup$
    – Fat32
    Aug 8, 2016 at 12:59

1 Answer 1


Assuming you refer to the ZOH-discretization as shown in this figure from the mathworks site

enter image description here

the relationship between the continuous-time signals and the discrete-time signals can be derived as follows. The signal $u(t)$ is given by


where $g(t)$ is a rectangular impulse response (constant in the interval $t\in [0,T]$, zero anywhere else) representing the ZOH, and $T$ is the sampling period. Let $h(t)$ be the impulse response corresponding to the transfer function $H(s)$. Furthermore, let $f(t)$ be the convolution of $g(t)$ and $h(t)$:

$$f(t)=(g\star h)(t)\tag{2}$$

Then the continuous-time output signal $y(t)$ is given by

$$\begin{align}y(t)&=(u\star h)(t)\\ &=\int_{-\infty}^{\infty}u(\tau)h(t-\tau)d\tau\\&=\sum_ku[k]\int_{-\infty}^{\infty}g(\tau-kT)h(t-\tau)d\tau\\&=\sum_ku[k]f(t-kT)\tag{3}\end{align}$$

where I've used Equations $(1)$ and $(2)$. From $(3)$ the discrete-time output signal is easily obtained as


with $f[n]=f(nT)$. So $y[n]$ is simply the discrete-time convolution of the input signal $u[n]$ and the total impulse response $f[n]$.

From this result it immediately follows that linearity must be satisfied, simply because convolution is a linear operation:

$$f(t)=\alpha_1(g\star h_1)(t)+\alpha_2(g\star h_2)(t)=(g\star (\alpha_1h_1+\alpha_2h_2))(t)\tag{5}$$

However, the (ZOH-)discretization of the concatenation of two transfer functions is not equivalent to the concatenation of the two discretized transfer functions. Referring to the figure above, the first case is equivalent to replacing $H(s)$ by the concatenation of $H_1(s)$ and $H_2(s)$. The other case involves concatenating two complete systems as shown in the figure, the first with $H(s)=H_1(s)$, the second with $H(s)=H_2(s)$. The difference between the two cases is that in the first case you only have one ZOH, whereas in the second one you get two ZOHs. The equivalent discrete-time impulse responses are

$$f[n]=(g\star h_1\star h_2)(nT)\tag{6}$$


$$f[n]=(g\star h_1)\star (g\star h_2)(nT)\tag{7}$$

which are generally not identical. The first case in Eq. $(6)$ is a ZOH discretization of the total transfer function $H(s)=H_1(s)H_2(s)$, whereas the second case in Eq. $(7)$ corresponds to a first-order hold (FOH) discretization of the total transfer function.

  • $\begingroup$ There is a typo in equation $(2)$, should be $(g \star h)(t)$ like you've correctly stated in the line above. $\endgroup$
    – Gilles
    Aug 7, 2016 at 10:18
  • $\begingroup$ If I understand this right, after equation $(4)$ we conclude that the discrete impulse response is identical to the sampled continuous impulse response? I'm surprised the wiki article I linked does not mention this basic fact, as it's what will make ZOH discretization of an LTI system interpretable in the frequency domain. $\endgroup$ Aug 7, 2016 at 14:40
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    $\begingroup$ @AaronVoelker: It's not the sampled continuous-time impulse response corresponding to $H(s)$, but the sampled combined impulse response of the ZOH and the continuous-time system, as described by Eq. $(2)$. $\endgroup$
    – Matt L.
    Aug 7, 2016 at 19:51
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    $\begingroup$ @AaronVoelker: As for the wikipedia article, they only talk about state-space models, and they somehow seem to disregard the "signals and systems" viewpoint that I discussed in my answer. $\endgroup$
    – Matt L.
    Aug 7, 2016 at 20:05
  • $\begingroup$ Thanks for the enlightening post and replies. I think I'm missing something about why $(7)$ is FOH. Is it because $(g \ast g)$ is the (causal) FOH impulse response and then because convolution is associative? I can see this kind of intuitively, but not formally obvious why $(g \ast g)$ should be the FOH impulse, and so is there something that makes this connection clearer? $\endgroup$ Aug 9, 2016 at 2:57

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