--- title: "Digital PI design, plant e^(−2s)/(s+2)" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:pi-design-dead-time-s2" source: https://www.bctengineer.com/q/semester_6/digital_control_system/pi-design-dead-time-s2 site: BCT Engineering --- # Digital PI design, plant e^(−2s)/(s+2) Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2067 Mangsir, 16 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/pi-design-dead-time-s2 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Consider the system shown below. Design a digital PI controller such that the dominant closed loop poles have a damping ratio of 0.55 and number of samples per cycle of damped oscillation is 10. Also find Kᵥ for the designed system. [Figure: unity-feedback loop: R(z) → summing junction → sampler → GD(z) → (1-e⁻ᵀˢ)/s → e⁻²ˢ/(s+2) → C(z)] ## Short answer PI controller G_D(z)=0.8540(z-0.4403)/(z-1) with K_p=0.376, K_i=0.478, assuming T=1 s. Dominant poles at 0.5349±j0.3886 give ζ=0.55 and 10 samples per cycle. K_v=0.239 s⁻¹. ## Answer A digital PI controller $G_D(z) = K_p + \dfrac{K_i}{1-z^{-1}} = K\dfrac{z-a}{z-1}$ is designed with the root-locus angle and magnitude conditions. The sampling period is not given; **$T = 1$ s is assumed**. It makes the 2 s dead time exactly two samples and allows a feasible PI design. #### Plant pulse transfer function With $e^{-2T} = e^{-2} = 0.1353$: $$ \begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+2)}\right] = z^{-2}\,\frac{(1-e^{-2})/2}{z-e^{-2}}\\ &= \frac{0.4323}{z^2(z-0.1353)} \end{aligned} $$ #### Desired dominant poles 10 samples per cycle gives $\omega_dT = 2\pi/10 = 36^\circ$: $$ \begin{aligned} |z| &= \exp\left(-\frac{\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{2\pi}{10}\right) = \exp\left(-\frac{0.55\times0.6283}{0.8352}\right) = 0.6611\\ z_1 &= 0.6611\angle36^\circ = 0.5349 + j0.3886 \end{aligned} $$ #### Angle condition | Pole | Angle at $z_1$ | |---|---| | double pole at 0 | $72.00^\circ$ | | pole at 0.1353 | $44.21^\circ$ | | PI pole at 1 | $140.12^\circ$ | | Total | $256.33^\circ$ | The PI zero must contribute $256.33^\circ - 180^\circ = 76.33^\circ$: $$ a = 0.5349 - \frac{0.3886}{\tan 76.33^\circ} = 0.4403 $$ #### Magnitude condition $$ K = \left|\frac{z_1^2(z_1-0.1353)(z_1-1)}{0.4323(z_1-0.4403)}\right| = 0.8540 $$ $$ G_D(z) = 0.8540\,\frac{z-0.4403}{z-1} = 0.3760 + \frac{0.4780}{1-z^{-1}} $$ So $K_p = Ka = 0.3760$ and $K_i = K(1-a) = 0.4780$. #### Check Characteristic equation: $z^4 - 1.1353z^3 + 0.1353z^2 + 0.3692z - 0.1626 = 0$, with roots $0.5349 \pm j0.3886$ (designed), $0.6435$ and $-0.5780$. Step response: $c(k) = 0, 0, 0, 0.369, 0.626, 0.867, 0.970, 1.020, 1.015, 1.003, \ldots \to 1$. The peak is about 2% and there is zero steady-state error. #### Velocity error constant $$ \begin{aligned} K_v &= \frac{1}{T}\lim_{z\to1}(1-z^{-1})G_D(z)G(z) = \frac{K(1-a)\times0.4323}{T(1-0.1353)}\\ &= \frac{0.4780\times0.4323}{0.8647} = 0.239\ \text{s}^{-1} \end{aligned} $$ The steady-state error to a unit ramp is $1/K_v = 4.18$. **Answer: $G_D(z) = 0.8540\,\dfrac{z-0.4403}{z-1}$ ($K_p = 0.376$, $K_i = 0.478$, with T = 1 s); $K_v = 0.239\ \text{s}^{-1}$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is T = 1 s assumed for the 2 s dead time? A. It makes the dead time exactly two samples and allows a feasible PI design B. It makes the plant pole disappear C. The ZOH requires T = 1 D. It gives a damping ratio of 0.55 Answer: A. A whole number of delay samples (z^-2) keeps G(z) rational, and T = 1 s leaves enough angle for the PI zero. (Key point: Assumed sampling period) 2. With zeta = 0.55 and 10 samples per cycle, |z1| is: A. 0.55 B. exp(-0.55 x 0.6283/0.8352) = 0.6611 C. exp(-0.6283) = 0.5335 D. 1 Answer: B. |z| = exp(-zeta/sqrt(1-zeta^2) x 2pi/10) = 0.6611, at angle 36 degrees. (Key point: Desired pole magnitude) 3. The poles at 0 (double), 0.1353 and 1 total 256.33 degrees at z1. The PI zero contributes: A. 256.33 degrees B. 103.67 degrees C. 76.33 degrees D. 180 degrees Answer: C. The zero must give 256.33 - 180 = 76.33 degrees, which fixes a = 0.4403. (Key point: Angle condition) 4. For G_D(z) = K(z - a)/(z - 1), Kp and Ki are: A. Kp = K(1 - a), Ki = K a B. Kp = K, Ki = a C. Kp = K/a, Ki = K a D. Kp = K a, Ki = K(1 - a) Answer: D. Writing K(z-a)/(z-1) = Ka + K(1-a)/(1 - z^-1) gives Kp = 0.376 and Ki = 0.478. (Key point: PI gains) 5. Kv for this design equals about 0.239 per second. What does the ramp error equal? A. 0.239 B. 4.18 C. 0.376 D. Zero Answer: B. Ramp error is 1/Kv = 4.18; the PI controller makes only step error zero. 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