ZOH transfer function and pulse transfer function

Transfer function of ZOH

A zero order hold keeps each sample constant for one sampling period: $h(kT+\tau) = x(kT)$, $0 \le \tau < T$.

Apply a unit impulse at $t=0$. The ZOH holds the value 1 for one sampling period and then drops to zero, so its impulse response is a rectangular pulse:

$$g_{h0}(t) = u(t) - u(t-T)$$

 g(t)
  1 |------+
    |      |
  0 +------+-------> t
    0      T

Taking the Laplace transform and using the shift theorem $\mathcal{L}[u(t-T)] = e^{-Ts}/s$:

$$\begin{aligned} G_{h0}(s) &= \frac{1}{s} - \frac{e^{-Ts}}{s} \\ &= \frac{1-e^{-Ts}}{s} \end{aligned}$$

It can also be found from a general sampled input $x^*(t) = \sum x(kT)\delta(t-kT)$. The output is

$$h(t) = \sum_{k=0}^{\infty} x(kT)\,[u(t-kT) - u(t-(k+1)T)]$$ $$\begin{aligned} H(s) &= \sum_{k=0}^{\infty} x(kT)\,\frac{e^{-kTs} - e^{-(k+1)Ts}}{s} \\ &= \frac{1-e^{-Ts}}{s}\sum_{k=0}^{\infty} x(kT)e^{-kTs} = \frac{1-e^{-Ts}}{s}\,X^*(s) \end{aligned}$$

so $H(s)/X^*(s) = \frac{1-e^{-Ts}}{s}$.

Pulse transfer function

Since $e^{-Ts} = z^{-1}$:

$$\begin{aligned} G_{h0}(z) &= \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s}\right] \\ &= \frac{z-1}{z}\cdot\frac{z}{z-1} = 1 \end{aligned}$$

So the ZOH alone passes sample values unchanged ($h(kT) = x(kT)$). Its effect appears when it is followed by a plant $G_p(s)$:

$$G(z) = \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}G_p(s)\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{G_p(s)}{s}\right]$$

Example: $G_p(s) = \frac{1}{s+1}$, $T = 1$ s.

$$\begin{aligned} \mathcal{Z}\left[\frac{1}{s(s+1)}\right] &= \frac{z(1-e^{-T})}{(z-1)(z-e^{-T})} \\ G(z) &= \frac{z-1}{z}\cdot\frac{z(1-e^{-T})}{(z-1)(z-e^{-T})} = \frac{1-e^{-T}}{z-e^{-T}} \\ &= \frac{0.6321}{z-0.3679} \end{aligned}$$

Answer: $G_{h0}(s) = \frac{1-e^{-Ts}}{s}$; pulse transfer function of ZOH alone $=1$; ZOH with plant: $(1-z^{-1})\mathcal{Z}[G_p(s)/s]$.