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ZOH and FOH defined; ZOH transfer function

Short answer

ZOH holds each sample constant for the full sampling period, producing a staircase output. FOH extrapolates a straight line using the two most recent samples. ZOH impulse response is a unit pulse of duration T; its transfer function is G_h0(s) = (1-e^(-Ts))/s.

What do you mean by Zero Order Hold (ZOH) and First Order Hold? Determine the transfer function for ZOH.

Zero order hold and first order hold

A hold circuit reconstructs a continuous signal from samples by extrapolating between sampling instants with a polynomial of order nn.

  • Zero order hold (ZOH): holds the latest sample constant until the next one: h(kT+τ)=x(kT)h(kT+\tau) = x(kT), 0≤τ<T0 \le \tau < T. Output is a staircase. It needs only the present sample.
  • First order hold (FOH): extrapolates a straight line using the slope of the last two samples: h(kT+τ)=x(kT)+x(kT)−x((k−1)T)T τh(kT+\tau) = x(kT) + \dfrac{x(kT)-x((k-1)T)}{T}\,\tau. Transfer function Gh1(s)=Ts+1T(1−e−Tss)2G_{h1}(s) = \frac{Ts+1}{T}\left(\frac{1-e^{-Ts}}{s}\right)^2.
 samples:  .   .   .   .
 ZOH:     |‾‾‾|‾‾‾|‾‾‾|‾‾‾
 FOH:      /   /   /   /
ZOHFOH
One sample usedTwo samples used
Staircase outputRamp segments
Simple, low costNeeds memory of past sample
Lag ωT/2\omega T/2More lag at high frequency

Transfer function of ZOH

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

gh0(t)=u(t)−u(t−T)g_{h0}(t) = u(t) - u(t-T)
 g(t)
  1 |------+
    |      |
  0 +------+-------> t
    0      T

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

Gh0(s)=1s−e−Tss=1−e−Tss\begin{aligned} G_{h0}(s) &= \frac{1}{s} - \frac{e^{-Ts}}{s} \\ &= \frac{1-e^{-Ts}}{s} \end{aligned}

Frequency response

With s=jωs = j\omega:

Gh0(jω)=T sin⁡(ωT/2)ωT/2 e−jωT/2G_{h0}(j\omega) = T\,\frac{\sin(\omega T/2)}{\omega T/2}\,e^{-j\omega T/2}

so the ZOH acts as a low-pass filter with gain TT at low frequency, zero gain at multiples of ωs\omega_s, and a phase lag of ωT/2\omega T/2 (half-period delay).

✓ Answer checked by Claude Opus 5.5 (high effort)

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