---
title: "Controller with integral action, plant 10/((s+1)(s+5))"
subject: "Digital Control System"
chapter: 4
key: "semester_6/digital_control_system:integral-control-design-10-over"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/integral-control-design-10-over
site: BCT Engineering
---

# Controller with integral action, plant 10/((s+1)(s+5))

Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System

- Asked in: 2078 Chaitra, 12 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/integral-control-design-10-over
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4

## Question

Design a suitable digital controller that includes an integral control action. The design specifications are that the damping ratio of the closed loop poles be 0.5 and there will be at least eight samples per cycle of the damped sinusoidal oscillations. The sampling time is assumed to be 0.2 sec. [Figure: unity-feedback loop: R(z) → summing junction → sampler (T) → digital controller → (1-e⁻ᵀˢ)/s → 10/((s+1)(s+5)) → C(z)]

## Short answer

Digital controller: GD(z)=2.204(z-0.8187)(z-0.3679)/[(z-1)(z-0.2009)] with integral action. Static velocity error constant Kv=3.16 s^-1.

## Answer

#### Step 1: Plant with ZOH ($T = 0.2$ s)
$$
\begin{aligned}
\frac{10}{s(s+1)(s+5)} &= \frac{2}{s} - \frac{2.5}{s+1} + \frac{0.5}{s+5} \\
G(z) &= (1-z^{-1})\left[\frac{2z}{z-1} - \frac{2.5z}{z-e^{-0.2}} + \frac{0.5z}{z-e^{-1}}\right] \\
&= \frac{0.1371z + 0.0921}{(z-0.8187)(z-0.3679)} = \frac{0.1371(z + 0.6714)}{(z-0.8187)(z-0.3679)}
\end{aligned}
$$

#### Step 2: Desired poles
With $\zeta = 0.5$ and 8 samples per cycle:

$$
|z| = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac18\right) = 0.6354, \quad \angle z = 45^\circ, \quad z_1 = 0.4493 + j0.4493
$$

#### Step 3: Try PI alone
Integral action requires a controller pole at $z = 1$. With a PI controller $K(z-0.8187)/(z-1)$ cancelling the slow plant pole, the angle at $z_1$ is:

$$
\underbrace{+21.85^\circ}_{\text{zero } -0.6714} \underbrace{-140.79^\circ}_{\text{pole } 1} \underbrace{-79.73^\circ}_{\text{pole } 0.3679} = -198.67^\circ
$$

This is not $-180^\circ$. There is $18.67^\circ$ too much lag, so PI alone cannot place the poles. A lead part is added, which makes the controller PID-like.

#### Step 4: Controller with integral + lead action
$$
G_D(z) = K\frac{(z - 0.8187)(z - 0.3679)}{(z-1)(z-\beta)}
$$

Both plant poles are cancelled. The angle condition at $z_1$ is:

$$
21.85^\circ - 140.79^\circ - \angle(z_1 - \beta) = -180^\circ \;\Rightarrow\; \angle(z_1-\beta) = 61.06^\circ
$$

$$
\beta = 0.4493 - \frac{0.4493}{\tan 61.06^\circ} = 0.2009
$$

#### Step 5: Gain
$$
K = \frac{|z_1-1|\,|z_1-0.2009|}{0.1371\,|z_1+0.6714|} = \frac{0.7107 \times 0.5134}{0.1371 \times 1.2074} = 2.204
$$

$$
G_D(z) = 2.204\,\frac{(z-0.8187)(z-0.3679)}{(z-1)(z-0.2009)}
$$

#### Step 6: Check
$$
G_DG = \frac{0.3022(z + 0.6714)}{(z-1)(z-0.2009)}
$$

The characteristic equation is $z^2 - 0.8986z + 0.4038 = 0$, with roots $z = 0.4493 \pm j0.4493$. ✔ That is $\zeta = 0.5$ with exactly 8 samples per cycle.

$$
K_v = \frac{1}{T}\cdot\frac{0.3022 \times 1.6714}{1 - 0.2009} = 3.16\ \text{s}^{-1}
$$

The pole at $z = 1$ gives zero steady-state error for a step input.

**Answer:** $G_D(z) = 2.204\dfrac{(z-0.8187)(z-0.3679)}{(z-1)(z-0.2009)}$, with $K_v = 3.16\ \text{s}^{-1}$.

## Check yourself

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

1. What gives the controller integral action in the discrete domain?
   A. A zero at z = 1
   B. A pole at z = 0
   C. A gain greater than 1
   D. A pole at z = 1
   Answer: D. A pole at z = 1 corresponds to the integrator, which gives zero steady-state error to a step. (Key point: Integral action)

2. Why can a PI controller cancelling only the slow pole not place the poles at z1?
   A. The angle is -198.67 degrees, not -180, so 18.67 degrees too much lag remains
   B. The magnitude is too large
   C. PI controllers have no zero
   D. The plant has no zero
   Answer: A. The PI pole at 1 adds lag; the total angle is -198.67 degrees, so a lead part must be added. (Key point: Why PI alone fails)

3. The final controller is K(z-0.8187)(z-0.3679)/((z-1)(z-beta)). What do the two zeros do?
   A. Cancel the pole at 1
   B. Cancel both plant poles
   C. Add integral action
   D. Remove the delay
   Answer: B. The zeros equal the two plant poles (0.8187 and 0.3679) so they are cancelled. (Key point: Controller form)

4. With angle(z1 - beta) = 61.06 degrees, beta is:
   A. 0.4493
   B. 0.8187
   C. 0.2009
   D. -0.2009
   Answer: C. beta = 0.4493 - 0.4493/tan 61.06 degrees = 0.2009. (Key point: Beta value)

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