--- title: "ZOH transfer function and pulse transfer function" subject: "Digital Control System" chapter: 1 key: "semester_6/digital_control_system:zoh-pulse-transfer-function" source: https://www.bctengineer.com/q/semester_6/digital_control_system/zoh-pulse-transfer-function site: BCT Engineering --- # ZOH transfer function and pulse transfer function Digital Control System · Chapter 1: Introduction to Discrete Time Control System - Asked in: 2075 Baisakh, 8 marks - Video: [2.1 Zero Order Hold (ZOH)](https://www.youtube.com/watch?v=uo2pQn4NCs4) — ACE-Lab - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/zoh-pulse-transfer-function - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-1 ## Question Derive the transfer function of ZOH and also find the pulse transfer function. ## Short answer ZOH impulse response is u(t)-u(t-T), so G_h0(s) = (1-e^(-Ts))/s. The pulse transfer function of the ZOH alone is G_h0(z) = 1. For a plant G_p(s) following ZOH, the discrete equivalent is G(z) = (1-z^(-1)) Z[G_p(s)/s]. ## Answer #### 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: $$ g_{h0}(t) = u(t) - u(t-T) $$ ```text 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: $(1-z^{-1})\mathcal{Z}[G_p(s)/s]$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. G_h0(s) is: A. (1 - e^(-Ts))/s B. 1/(s+T) C. (1 - e^(Ts))/s^2 D. e^(-Ts)/s Answer: A. From impulse response u(t) - u(t-T). (Key point: ZOH transfer function) 2. The pulse transfer function of the ZOH alone is: A. z/(z-1) B. 1 C. 1 - z^-1 D. T Answer: B. (1 - z^-1) Z[1/s] = (z-1)/z x z/(z-1) = 1. (Key point: Pulse TF of ZOH alone) 3. The ZOH + plant pulse TF is: A. Z[Gp(s)] B. (1 - z^-1) Z[Gp(s)] C. (1 - z^-1) Z[Gp(s)/s] D. Z[Gp(s)/s] Answer: C. The 1/s is attached to the plant before z-transform. (Key point: ZOH with plant) 4. For Gp = 1/(s+1) and T = 1 s, G(z) is: A. 0.3679/(z - 0.6321) B. 0.6321/(z + 0.3679) C. 1/(z - 0.3679) D. 0.6321/(z - 0.3679) Answer: D. G(z) = (1 - e^-T)/(z - e^-T). 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