Transfer function of first order hold

A first order hold (FOH) reconstructs the signal between samples by a straight line extrapolated from the last two samples:

$$h(kT+\tau) = x(kT) + \frac{x(kT) - x((k-1)T)}{T}\,\tau, \qquad 0 \le \tau < T$$

Impulse response

Apply a unit impulse at $t=0$, so $x(0)=1$ and all other samples are zero.

• For $0 \le t < T$ ($k=0$): $h = 1 + \frac{1-0}{T}t = 1 + \frac{t}{T}$ (rises from 1 to 2).
• For $T \le t < 2T$ ($k=1$, $x(T)=0$, $x(0)=1$): $h = 0 + \frac{0-1}{T}(t-T) = -\frac{t-T}{T}$ (from 0 to -1).
• For $t \ge 2T$: $h = 0$.

 g(t)
  2 |     /|
    |   /  |
  1 | /    |
  0 +------+------+----> t
    0      T \    | 2T
 -1 |         \---|

Writing it with steps and ramps

$$\begin{aligned} g_{h1}(t) &= u(t) + \frac{t}{T}u(t) - 2u(t-T) - \frac{2(t-T)}{T}u(t-T) \\ &\quad + u(t-2T) + \frac{t-2T}{T}u(t-2T) \end{aligned}$$

Check: for $0\le t<T$ it gives $1+t/T$; for $T\le t<2T$ it gives $1+\frac{t}{T}-2-\frac{2(t-T)}{T} = -\frac{t-T}{T}$; for $t\ge 2T$ it gives 0.

Laplace transform

Using $\mathcal{L}[u(t)]=\frac1s$, $\mathcal{L}[t\,u(t)]=\frac{1}{s^2}$ and the shift theorem:

$$\begin{aligned} G_{h1}(s) &= \frac1s + \frac{1}{Ts^2} - \frac{2e^{-Ts}}{s} - \frac{2e^{-Ts}}{Ts^2} + \frac{e^{-2Ts}}{s} + \frac{e^{-2Ts}}{Ts^2} \\ &= \left(\frac1s + \frac{1}{Ts^2}\right)\left(1 - 2e^{-Ts} + e^{-2Ts}\right) \\ &= \frac{Ts+1}{Ts^2}\,(1-e^{-Ts})^2 \\ &= \frac{Ts+1}{T}\left(\frac{1-e^{-Ts}}{s}\right)^2 \end{aligned}$$

which is the required transfer function of the first order hold.

Remarks

• The FOH follows ramps better than the ZOH, but it has more phase lag at higher frequencies and needs storage of the previous sample.
• Since $G_{h1}(s) = \frac{Ts+1}{T}\,G_{h0}^2(s)$, it can be seen as two ZOH blocks in cascade with a lead term $(Ts+1)/T$.