---
title: "Pulse TF with PID controller, plant 1/(s+1)"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:pid-pulse-tf-first-order"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pid-pulse-tf-first-order
site: BCT Engineering
---

# Pulse TF with PID controller, plant 1/(s+1)

Digital Control System · Chapter 3: Analysis of Discrete Time Control System

- Asked in: 2073 Bhadra, 8 marks; 2072 Magh, 8 marks (asked 2 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/pid-pulse-tf-first-order
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Obtain the pulse transfer function of the system shown below. [Figure: unity-feedback loop: R(s) → summing junction → E(s) → sampler (T = 1) → PID controller (kₚ=1, kD=0.2, kI=0.2) → ZOH → 1/(s+1) → C(s); ZOH = transfer function of zero order hold circuit]

## Short answer

Compute G_D(z) = (1.4z^2 − 1.4z + 0.2) / (z(z−1)) and G(z) = 0.6321 / (z−0.3679). Then C(z)/R(z) = G_DG / (1 + G_DG) = (0.8850z^2 − 0.8850z + 0.1264) / (z^3 − 0.4829z^2 − 0.5171z + 0.1264).

## Answer

There is one sampler in front of the controller, so the closed-loop pulse transfer function is $C(z)/R(z) = G_D(z)G(z)/[1+G_D(z)G(z)]$, where $G(z)$ is the z-transform of ZOH and plant together.

#### Step 1: Digital PID controller
**Digital PID controller.** Using the positional PID form (Ogata), with $K_P = 1$, $K_I = 0.2$, $K_D = 0.2$:

$$
\begin{aligned}
G_D(z) &= K_P + \frac{K_I}{1-z^{-1}} + K_D(1-z^{-1}) \\
&= \frac{(K_P+K_I+K_D) - (K_P+2K_D)z^{-1} + K_D z^{-2}}{1-z^{-1}} \\
&= \frac{1.4 - 1.4z^{-1} + 0.2z^{-2}}{1-z^{-1}} = \frac{1.4z^2 - 1.4z + 0.2}{z(z-1)}
\end{aligned}
$$

Here $K_P = K - K_I/2$, $K_I = KT/T_i$ and $K_D = KT_d/T$ are the usual digital PID gains. Trapezoidal integration and a backward difference for the derivative give this form.

#### Step 2: ZOH + plant $1/(s+1)$
$$
\begin{aligned}
G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+1)}\right] = (1-z^{-1})\left[\frac{z}{z-1} - \frac{z}{z-e^{-T}}\right] \\
&= \frac{1-e^{-T}}{z-e^{-T}} = \frac{0.6321}{z-0.3679} \quad (T = 1)
\end{aligned}
$$

#### Step 3: Open-loop pulse transfer function
$$
\begin{aligned}
G_D(z)G(z) &= \frac{0.6321(1.4z^2-1.4z+0.2)}{z(z-1)(z-0.3679)} \\
&= \frac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 1.3679z^2 + 0.3679z}
\end{aligned}
$$

#### Step 4: Closed-loop pulse transfer function
$$
\begin{aligned}
\frac{C(z)}{R(z)} &= \frac{0.8850z^2 - 0.8850z + 0.1264}{(z^3 - 1.3679z^2 + 0.3679z) + (0.8850z^2 - 0.8850z + 0.1264)} \\
&= \frac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 0.4829z^2 - 0.5171z + 0.1264}
\end{aligned}
$$

In powers of $z^{-1}$:

$$
\frac{C(z)}{R(z)} = \frac{0.8850z^{-1} - 0.8850z^{-2} + 0.1264z^{-3}}{1 - 0.4829z^{-1} - 0.5171z^{-2} + 0.1264z^{-3}}
$$

**Answer:** $\dfrac{C(z)}{R(z)} = \dfrac{0.8850z^2 - 0.8850z + 0.1264}{z^3 - 0.4829z^2 - 0.5171z + 0.1264}$

**Check:** At $z = 1$ both numerator and denominator equal $0.1264$. The DC gain is 1, so there is zero steady-state error to a step input because of the integral term.

## Check yourself

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

1. Which expression gives C(z)/R(z) here?
   A. G_D(z)G(z)
   B. G_D(z)G(z) / [1 + G_D(z)G(z)]
   C. G(z) / [1 + G_D(z)]
   D. 1 + G_D(z)G(z)
   Answer: B. With a single sampler before the controller, the loop is a standard unity-feedback pulse transfer function. (Key point: Closed-loop form)

2. What is G(z) for the ZOH and 1/(s+1) with T = 1?
   A. 0.3679 / (z - 0.6321)
   B. 0.6321 / (z - 0.3679)
   C. z / (z - 0.3679)
   D. 0.6321z / (z - 1)
   Answer: B. G(z) = (1-z^-1) Z[1/(s(s+1))] = (1 - e^-T)/(z - e^-T) = 0.6321/(z - 0.3679). (Key point: ZOH plus plant 1/(s+1))

3. What is the open-loop pulse transfer function G_D(z)G(z)?
   A. 0.6321(1.4z^2 - 1.4z + 0.2) / [z(z-1)(z-0.3679)]
   B. 0.6321 / [z(z-1)(z-0.3679)]
   C. 1.4(z-0.3679) / [0.6321 z(z-1)]
   D. 0.6321(1.4z^2 - 1.4z + 0.2) / [(z-1)(z-0.3679)]
   Answer: A. The PID has denominator z(z-1) and the plant has denominator (z-0.3679), so the product is as shown. (Key point: Open-loop product)

4. What is the closed-loop denominator, and what does it show at z = 1?
   A. z^3 - 0.4829z^2 - 0.5171z + 0.1264; numerator and denominator are both 0.1264, so DC gain is 1
   B. z^3 - 1.3679z^2 + 0.3679z; DC gain is 0
   C. z^3 - 0.4829z^2 - 0.5171z + 0.1264; DC gain is 0.5
   D. z^3 + 0.4829z^2 + 0.5171z + 0.1264; DC gain is 1
   Answer: A. Adding numerator and denominator gives that polynomial. Equal values at z = 1 give unity DC gain, because the integral action removes steady-state step error. (Key point: Final result and DC gain)

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