---
title: "Transfer function of first order hold"
subject: "Digital Control System"
chapter: 1
key: "semester_6/digital_control_system:foh-transfer-function"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/foh-transfer-function
site: BCT Engineering
---

# Transfer function of first order hold

Digital Control System · Chapter 1: Introduction to Discrete Time Control System

- Asked in: 2082 Chaitra, 8 marks; 2076 Bhadra, 8 marks; 2068 Jestha, 8 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/foh-transfer-function
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-1

## Question

Show that the transfer function of first order hold circuit is given by Gₕ₁(s)=((Ts+1)/T)·((1-e⁻ᵀˢ)/s)².

## Short answer

The impulse response of a first-order hold is a triangle: 1 + t/T for 0 ≤ t < T and −(t−T)/T for T ≤ t < 2T. Expressing this with step and ramp functions and taking the Laplace transform yields G_h1(s) = ((Ts+1)/T)·((1−e^(−Ts))/s)^2.

## Answer

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 $t \ge 2T$: $h = 0$.

```text
 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 $(Ts+1)/T$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does a first order hold reconstruct the signal between samples?
   A. Holds the last sample constant
   B. Extrapolates a straight line from the last two samples
   C. Sets the output to zero
   D. Fits a parabola through three samples
   Answer: B. The FOH extends a line with slope (x(kT) - x((k-1)T))/T. (Key point: FOH principle)

2. For a unit impulse at t = 0, what is the FOH output for 0 <= t < T?
   A. 1
   B. 1 + t/T, rising from 1 to 2
   C. -t/T
   D. 0
   Answer: B. With x(0)=1 and the previous sample 0, h = 1 + (1-0)t/T. (Key point: FOH impulse response)

3. For T <= t < 2T, what is the FOH impulse response?
   A. -(t-T)/T, falling from 0 to -1
   B. 1 + t/T
   C. Constant 2
   D. Zero
   Answer: A. With x(T)=0 and x(0)=1, h = (0-1)(t-T)/T, and it is zero for t >= 2T. (Key point: Impulse response beyond T)

4. The result G_h1(s) = ((Ts+1)/T)((1-e^(-Ts))/s)^2 can be interpreted as what?
   A. One ZOH in series with an integrator
   B. Two ZOH blocks in cascade with a lead term (Ts+1)/T
   C. A ZOH in parallel with a delay
   D. A pure delay of T
   Answer: B. Since G_h1 = ((Ts+1)/T) G_h0^2, it is two cascaded ZOHs with a lead term. (Key point: Final result and structure)

## Close variants set in other exams

- [Transfer function of zero order hold](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Azoh-transfer-function) — Digital Control System, chapter 1, asked 7×

## Related in Digital Control System

- [Pulse transfer function with first order hold](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Afoh-system-pulse-tf) — Digital Control System, chapter 3
- [Data hold circuit and ZOH derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adata-hold-zoh-derivation) — Digital Control System, chapter 1
- [ZOH and FOH defined; ZOH transfer function](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Azoh-foh-definition) — Digital Control System, chapter 1
- [Series programming of a third-order transfer function](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aseries-programming-z-plus-1) — Digital Control System, chapter 4, asked 2×
- [Ladder programming of a third-order transfer function](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Aladder-z3-4z2) — Digital Control System, chapter 4, asked 2×
- [Short note: Sample and hold circuit](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ashort-note-sample-hold) — Digital Control System, chapter 1

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