---
title: "Pulse TF with digital PID, plant 1/(s(s+2))"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:pid-pulse-tf-plant-s2"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pid-pulse-tf-plant-s2
site: BCT Engineering
---

# Pulse TF with digital PID, plant 1/(s(s+2))

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

- Asked in: 2082 Kartik, 8 marks
- 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-plant-s2
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

For the control system with digital PID controller as shown in the figure, obtain the pulse transfer function. Assume that the sampling period T is 1 second. Consider Kₚ=1, KI=0.2 and Kd=0.2 of PID controller. [Figure: unity-feedback loop: r(t) → summing junction → e(t) → sampler (T = 1) → e(kT) → digital controller → ZOH → 1/s(s+2) → C(t)]

## Short answer

Digital PID with Kp=1, Ki=0.2, Kd=0.2 gives G_D(z)=(1.4z²-1.4z+0.2)/(z(z-1)). ZOH plus plant 1/s(s+2) with T=1 yields G(z)=(0.2838z+0.1485)/((z-1)(z-0.1353)). Closed-loop: C/R = (0.3974z³-0.1895z²-0.1511z+0.0297)/(z⁴-1.7380z³+1.0812z²-0.2865z+0.0297).

## Answer

The digital controller is a PID controller. It is followed by a ZOH and the plant, with one sampler on the error signal, so:

$$
\frac{C(z)}{R(z)} = \frac{G_D(z)G(z)}{1+G_D(z)G(z)}
$$

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

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

#### Step 2: ZOH + plant $1/[s(s+2)]$
$$
\begin{aligned}
G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right] \\
\frac{1}{s^2(s+2)} &= \frac{1/2}{s^2} - \frac{1/4}{s} + \frac{1/4}{s+2} \\
G(z) &= (1-z^{-1})\left[\frac{0.5Tz}{(z-1)^2} - \frac{0.25z}{z-1} + \frac{0.25z}{z-e^{-2T}}\right] \\
&= \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})}
\end{aligned}
$$

With $T = 1$ and $e^{-2} = 0.1353$:

$$
G(z) = \frac{(1+0.1353)z + (1-0.4060)}{4(z-1)(z-0.1353)} = \frac{0.2838z + 0.1485}{(z-1)(z-0.1353)}
$$

#### Step 3: Open-loop pulse transfer function
$$
\begin{aligned}
G_D(z)G(z) &= \frac{(1.4z^2 - 1.4z + 0.2)(0.2838z + 0.1485)}{z(z-1)^2(z-0.1353)} \\
&= \frac{0.3974z^3 - 0.1895z^2 - 0.1511z + 0.0297}{z^4 - 2.1353z^3 + 1.2707z^2 - 0.1353z}
\end{aligned}
$$

#### Step 4: Closed-loop pulse transfer function
$$
\frac{C(z)}{R(z)} = \frac{0.3974z^3 - 0.1895z^2 - 0.1511z + 0.0297}{z^4 - 1.7380z^3 + 1.0812z^2 - 0.2865z + 0.0297}
$$

or

$$
\frac{C(z)}{R(z)} = \frac{0.3974z^{-1} - 0.1895z^{-2} - 0.1511z^{-3} + 0.0297z^{-4}}{1 - 1.7380z^{-1} + 1.0812z^{-2} - 0.2865z^{-3} + 0.0297z^{-4}}
$$

**Answer:** The closed-loop pulse transfer function is as above, with characteristic equation $z^4 - 1.7380z^3 + 1.0812z^2 - 0.2865z + 0.0297 = 0$.

**Check:** At $z = 1$ the numerator is $0.0865$ and the denominator is $0.0864$, so the DC gain is about 1, as expected with integral action.

## Check yourself

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

1. With a single sampler on the error, what is C(z)/R(z)?
   A. G_D(z)G(z)
   B. G_D(z)G(z) / [1 + G_D(z)G(z)]
   C. G_D(z) / [1 + G(z)]
   D. G(z) / [1 + G_D(z)]
   Answer: B. The controller, ZOH and plant form one open-loop pulse transfer function inside a unity-feedback loop. (Key point: Closed-loop structure)

2. How is 1/[s^2 (s+2)] expanded?
   A. 1/(2s^2) - 1/(4s) + 1/(4(s+2))
   B. 1/(2s^2) + 1/(4s) - 1/(4(s+2))
   C. 1/(4s^2) - 1/(2s) + 1/(2(s+2))
   D. 1/s^2 - 1/s + 1/(s+2)
   Answer: A. These are the correct partial fractions. They give the ramp, step and exponential z-transforms used in G(z). (Key point: Partial fractions for plant)

3. What is G(z) at T = 1 for the ZOH and 1/[s(s+2)]?
   A. (0.2838z + 0.1485) / [(z-1)(z-0.1353)]
   B. 0.6321 / (z - 0.3679)
   C. (0.3679z + 0.2642) / [(z-1)(z-0.3679)]
   D. (0.1353z + 0.2838) / [(z-1)(z-0.1485)]
   Answer: A. The pole is at e^-2T = 0.1353 and the numerator is (1.1353z + 0.594)/4 = 0.2838z + 0.1485. (Key point: ZOH plus plant result)

4. Which is the closed-loop characteristic equation?
   A. z^4 - 1.7380z^3 + 1.0812z^2 - 0.2865z + 0.0297 = 0
   B. z^4 - 2.1353z^3 + 1.2707z^2 - 0.1353z = 0
   C. 0.3974z^3 - 0.1895z^2 - 0.1511z + 0.0297 = 0
   D. z^3 - 1.7380z^2 + 1.0812z - 0.2865 = 0
   Answer: A. It is the open-loop denominator plus the open-loop numerator. At z = 1 the DC gain is about 1 because of integral action. (Key point: Characteristic equation)

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