---
title: "Closed-loop TF with digital PID, plant 1/(s(s+1))"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:pid-closed-loop-plant-s1"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/pid-closed-loop-plant-s1
site: BCT Engineering
---

# Closed-loop TF with digital PID, plant 1/(s(s+1))

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

- Asked in: 2082 Chaitra (new course), 4 marks; 2074 Bhadra, 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-closed-loop-plant-s1
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Obtain the closed loop transfer function of the system shown in figure below. Assume proportional gain Kₚ=1, integral gain Kᵢ=0.2, derivative gain Kd=0.2. [Take T = 1 sec] [Figure: unity-feedback loop: R(z) → summing junction → E(z) → sampler (T = 1 s) → digital PID controller → ZOH → 1/s(s+1) → C(z)]

## Short answer

Compute G_D(z) = (1.4z^2 − 1.4z + 0.2) / (z(z−1)) and G(z) = (0.3679z + 0.2642) / ((z−1)(z−0.3679)). Then C(z)/R(z) = G_DG / (1 + G_DG) = (0.5151z^3 − 0.1451z^2 − 0.2964z + 0.0528) / (z^4 − 1.8528z^3 + 1.5907z^2 − 0.6642z + 0.0528).

## Answer

The controller and the ZOH + plant are in cascade with only one sampler in front, so $C(z)/R(z) = G_D(z)G(z)/[1+G_D(z)G(z)]$.

#### Step 1: Pulse transfer function of 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: Pulse transfer function of ZOH + plant
$$
\begin{aligned}
G(z) &= \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}\cdot\frac{1}{s(s+1)}\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] \\
\frac{1}{s^2(s+1)} &= \frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1} \\
G(z) &= (1-z^{-1})\left[\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}\right]
\end{aligned}
$$

With $T = 1$ s and $e^{-1} = 0.3679$:

$$
G(z) = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)}
$$

#### Step 3: Open-loop pulse transfer function
$$
\begin{aligned}
G_D(z)G(z) &= \frac{(1.4z^2-1.4z+0.2)(0.3679z+0.2642)}{z(z-1)^2(z-0.3679)} \\
&= \frac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 2.3679z^3 + 1.7358z^2 - 0.3679z}
\end{aligned}
$$

#### Step 4: Closed-loop pulse transfer function
Adding numerator and denominator gives the characteristic polynomial:

$$
\frac{C(z)}{R(z)} = \frac{G_DG}{1+G_DG} = \frac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 1.8528z^3 + 1.5907z^2 - 0.6642z + 0.0528}
$$

or, in powers of $z^{-1}$:

$$
\frac{C(z)}{R(z)} = \frac{0.5151z^{-1} - 0.1451z^{-2} - 0.2964z^{-3} + 0.0528z^{-4}}{1 - 1.8528z^{-1} + 1.5907z^{-2} - 0.6642z^{-3} + 0.0528z^{-4}}
$$

**Answer:** $\dfrac{C(z)}{R(z)} = \dfrac{0.5151z^3 - 0.1451z^2 - 0.2964z + 0.0528}{z^4 - 1.8528z^3 + 1.5907z^2 - 0.6642z + 0.0528}$

**Check:** At $z = 1$ the numerator is $0.1264$ and the denominator is $0.1265$, so the ratio is about 1. Because the controller has integral action, the steady-state error for a step input is zero, as expected.

## Check yourself

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

1. With one sampler on the error signal and the controller, ZOH and plant in cascade, 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. 1 / [1 + G_D(z)G(z)]
   D. G_D(z) / [1 + G(z)]
   Answer: B. The standard unity-feedback result is the open-loop pulse transfer function divided by one plus itself. (Key point: Closed-loop formula)

2. What is G_D(z) for Kp = 1, Ki = 0.2, Kd = 0.2?
   A. (1.4z^2 - 1.4z + 0.2) / [z(z-1)]
   B. (1.2z^2 - 1.2z + 0.2) / [z(z-1)]
   C. (1.4z^2 - 1.2z + 0.2) / [z(z-1)]
   D. (1.4z - 0.2) / (z-1)
   Answer: A. The numerator coefficients are (Kp+Ki+Kd) = 1.4, -(Kp+2Kd) = -1.4 and Kd = 0.2, over 1 - z^-1. (Key point: Digital PID controller)

3. For 1/[s(s+1)] with a ZOH and T = 1 s, what is G(z)?
   A. (0.3679z + 0.2642) / [(z-1)(z-0.3679)]
   B. 0.6321 / (z-0.3679)
   C. (0.6321z + 0.2642) / [(z-1)(z-0.6321)]
   D. z / [(z-1)(z-0.3679)]
   Answer: A. G(z) = (1 - z^-1) Z[1/(s^2(s+1))], which with e^-1 = 0.3679 gives that result. (Key point: ZOH plus plant)

4. How is the closed-loop denominator formed, and what confirms the result?
   A. Subtract the numerator from the open-loop denominator; DC gain is 0
   B. Add the open-loop numerator to the open-loop denominator; at z = 1 the ratio is about 1 because of integral action
   C. Multiply numerator and denominator; DC gain is infinite
   D. Use only the open-loop denominator; DC gain is 0.5
   Answer: B. 1 + GDG = (D + N)/D, so the closed-loop denominator is D + N. The integrator gives a DC gain of 1 (zero step error). (Key point: Closed-loop denominator and check)

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