Transfer function of zero order hold

A zero order hold (ZOH) holds each sample value constant until the next sample arrives. Its output is a staircase signal. Its transfer function is found from its impulse response.

Impulse response

Apply a unit impulse $\delta(t)$ (sample of strength 1 at $t=0$). The ZOH holds this value for one sampling period $T$ and then drops to zero:

$$g_{h0}(t) = \begin{cases} 1, & 0 \le t < T \\ 0, & \text{otherwise} \end{cases}$$

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

This rectangular pulse is the difference of two unit steps:

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

Laplace transform

Using $\mathcal{L}[u(t)] = \frac{1}{s}$ and the time-shift theorem $\mathcal{L}[f(t-T)u(t-T)] = e^{-Ts}F(s)$:

$$\begin{aligned} G_{h0}(s) &= \mathcal{L}[u(t)] - \mathcal{L}[u(t-T)] \\ &= \frac{1}{s} - \frac{e^{-Ts}}{s} \\ &= \frac{1-e^{-Ts}}{s} \end{aligned}$$

which is the required result.

General input

For an impulse-sampled input $x^*(t) = \sum x(kT)\delta(t-kT)$, the output is $h(t) = \sum_k x(kT)[u(t-kT) - u(t-(k+1)T)]$, whose Laplace transform is $X^*(s)\,\frac{1-e^{-Ts}}{s}$, confirming the same transfer function.

Frequency response (for reference)

Putting $s = j\omega$: $G_{h0}(j\omega) = T\,\frac{\sin(\omega T/2)}{\omega T/2}\,e^{-j\omega T/2}$, so the ZOH is a low-pass filter with a phase lag of $\omega T/2$ (a delay of half a sampling period).