the function c2d allows to convert a continuous laplace transfer function to a discrete z-transform transfer function. The base method is the Zero Order Holder. In example:

   Pd=c2d(P,0.1); % 0.1 is the sampling time


$$ \frac{0.25918178 }{z - 0.74081822068} $$

$0.74081822068 = e^{-0.1 \ \cdot \ 3}$ as expected.

If we had a delay to the tf such as

$$ e^{-2 s} \ \frac{3}{3+s} $$

the conversion will result obviously in:

$$ z^{-20} \ \frac{0.25918178 }{z - 0.74081822068}$$

But if we introduce a delay not multiple of $0.1$ like this:

$$ e^{-1.95 \ s} \ \frac{3}{3+s} $$

it will produce:

$$ z^{-20} \ \frac{0.1393 z + 0.1199}{z - 0.7408} $$

As stated here, for "ZOH Method for Systems with Time Delays":

  1. Decomposes the delay $\tau$ as $\tau=kT+\rho$ with $0\le\rho<\tau$
  2. Absorbs the fractional delay $\rho$ into $H(s)$.

I ask how it is done the second step (if it is known), so how are calculated the coefficient of the numerator

$$ 0.1393 z + 0.1199 $$

and why that algorithm adds a zero to the transfer function and increases the order of the numerator to compensate the fractional delay. Actually, the algorithm seems provide good results:

enter image description here

Thank you.

  • $\begingroup$ Can I show you how to do $\LaTeX$ markup for your math? And then you'll use that in the future? $\endgroup$ Commented Nov 10, 2023 at 16:22
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    $\begingroup$ @robertbristow-johnson of course, thank you very much, sorry, it is my first question! $\endgroup$ Commented Nov 10, 2023 at 16:27
  • $\begingroup$ Hay, just FYI, you had the $z$ is the numerator in your original posting. Removing it from the numerator is just like multiplying by $z^{-1}$, which adds a delay of one sample to the transfer function. Does that enter into your accounting of total delay? $\endgroup$ Commented Nov 10, 2023 at 18:47
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    $\begingroup$ Sorry @robertbristow-johnson I do not understand your point. I need to find $$ {(1-z^{-1}) \mathcal{Z}{ \left\\{ \mathcal{L^{-1}} \left\\{ \frac {1}{s} \left(\frac{3}{3+s} \right) \right\\}\right\\}}} $$ and this result in $$ \frac{0.25918178}{z−0.74081822068} $$ . No z at numerator. $\endgroup$ Commented Nov 14, 2023 at 10:47
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    $\begingroup$ Is this the expression you meant to say? $$ (1-z^{-1}) \mathcal{Z} \left\{ \mathscr{L^{-1}} \left\{ \frac{1}{s} \left( \frac{3}{3+s} \right) \right\} \right\} $$ $\endgroup$ Commented Nov 15, 2023 at 22:08

1 Answer 1


The zero-order hold (ZOH) discretization is the same as the step-invariant method, which means that the step response of the discrete-time system equals the continuous-time step response at the sample instants. That's what is shown in the figure in your question.

The calculation of the discrete-time transfer function is performed as follows: factoring out a delay of $20$ samples, we're left with the following continuous-time transfer function:

$$H(s)=\frac{3e^{\tau s}}{s+3},\qquad \tau=0.05\tag{1}$$

The Laplace transform of the step response is

$$G(s)=\frac{H(s)}{s}=\frac{3e^{\tau s}}{s(s+3)}=e^{\tau s}\left(\frac{1}{s}-\frac{1}{s+3}\right)\tag{2}$$

The corresponding step response is


For the ZOH (step-invariant) discretization we require the step response of the discrete-time system to be

\begin{align*} g_d[n] = g(nT) &= \left(1-e^{-3(nT+\tau)}\right)u(nT+\tau)\\ &= \left(1-e^{-3\tau}\left(e^{-3T}\right)^n\right)u[n]\tag{4} \end{align*}

where $T$ is the sampling interval. The discrete-time unit step sequence $u[n]$ equals $u(nT+\tau)$ because $|\tau|<T$.

The $\mathcal{Z}$-transform of $(4)$ is


The transfer function of the discrete-time system is

\begin{align*} H_d(z)=(1-z^{-1})G_d(z) &= 1-e^{-3\tau}\frac{1-z^{-1}}{1-e^{-3T}z^{-1}}\\ &= \frac{1-e^{-3\tau}+\left(e^{-3\tau}-e^{-3T}\right)z^{-1}}{1-e^{-3T}z^{-1}}\tag{6} \end{align*}

With $\tau=0.05$ and $T=0.1$ you obtain exactly the numbers returned by Matlab.

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    $\begingroup$ If found myself Friday this same solution, I would write it, but you come earlier. Happy that I was right! Thank you. $\endgroup$ Commented Nov 13, 2023 at 16:21

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