--- title: "Digital PD controller design for plant 1/s²" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:pd-design-double-integrator" source: https://www.bctengineer.com/q/semester_6/digital_control_system/pd-design-double-integrator site: BCT Engineering --- # Digital PD controller design for plant 1/s² Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2080 Chaitra, 12 marks; 2075 Bhadra, 12 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/pd-design-double-integrator - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Design a digital proportional-plus-derivative controller for the plant as shown in figure below. It is desired that the damping ratio ζ of the dominant closed loop poles be 0.5 and the undamped natural frequency be 4 rad/sec. The sampling period is 0.1 sec. [Figure: R(z) → summing junction → sampler (T) → digital PD controller → ZOH (1-e⁻ᵀˢ)/s → 1/s² → C(z); feedback path H(s) (unity in the 2080 paper)] ## Short answer With T=0.1s, plant G(z)=0.005(z+1)/(z-1)^2. Desired poles are 0.7701±j0.2780. PD controller G_D(z)=42.78(z-0.7706)/z, giving K_P=9.81, K_D=32.97. Difference equation: u(k)=42.78e(k)-32.97e(k-1). ## Answer #### Step 1: Plant with ZOH ($T = 0.1$ s) $$ G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^3}\right] = (1-z^{-1})\frac{T^2z(z+1)}{2(z-1)^3} = \frac{T^2(z+1)}{2(z-1)^2} = \frac{0.005(z+1)}{(z-1)^2} $$ #### Step 2: Desired closed-loop poles With $\zeta = 0.5$ and $\omega_n = 4$: $$ \begin{aligned} \zeta\omega_n &= 2, \qquad \omega_d = 4\sqrt{0.75} = 3.464\ \text{rad/s} \\ |z| &= e^{-\zeta\omega_nT} = e^{-0.2} = 0.8187 \\ \angle z &= \omega_dT = 0.3464\ \text{rad} = 19.85^\circ \\ z_1 &= 0.7701 + j0.2780 \end{aligned} $$ #### Step 3: Digital PD controller $$ G_D(z) = K_P + K_D(1-z^{-1}) = (K_P+K_D)\frac{z - \alpha}{z} = K\frac{z-\alpha}{z}, \qquad \alpha = \frac{K_D}{K_P+K_D} $$ #### Step 4: Angle condition at $z_1$ | Factor | Angle | |---|---| | plant zero at $-1$ | $+8.92^\circ$ | | double plant pole at $1$ | $-2(129.59^\circ) = -259.19^\circ$ | | controller pole at $0$ | $-19.85^\circ$ | | **Sum** | $-270.11^\circ$ | For a total of $-180^\circ$, the controller zero must give $+90.11^\circ$: $$ \alpha = 0.7701 - \frac{0.2780}{\tan 90.11^\circ} = 0.7706 $$ The zero lies almost directly below the desired pole. #### Step 5: Magnitude condition $$ K = \frac{|z_1|\,|z_1-1|^2}{0.005\,|z_1+1|\,|z_1-0.7706|} = \frac{0.8187 \times 0.3607^2}{0.005 \times 1.7918 \times 0.2780} = 42.78 $$ $$ G_D(z) = 42.78\,\frac{z - 0.7706}{z} = 42.78\,(1 - 0.7706z^{-1}) $$ #### Step 6: PD gains $$ K_P + K_D = 42.78, \qquad K_D = K\alpha = 32.97, \qquad K_P = K(1-\alpha) = 9.81 $$ #### Step 7: Check $$ 1 + G_DG = 0 \;\Rightarrow\; z^3 - 1.7861z^2 + 1.0491z - 0.1648 = 0 $$ The roots are $z = 0.7701 \pm j0.2780$ (the desired pair) and $z = 0.2459$. The third pole is much closer to the origin, so it decays fast, and the complex pair is dominant. ✔ **Answer:** $G_D(z) = 42.78\dfrac{z-0.7706}{z}$, i.e. $K_P = 9.81$ and $K_D = 32.97$. The difference equation is $u(k) = 42.78e(k) - 32.97e(k-1)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 1/s^2 plant with ZOH and T = 0.1 s, G(z) is A. 0.005(z+1)/(z-1)^2 B. 0.1/(z-1) C. 0.005/(z-1)^2 D. 0.01(z-1)/(z+1) Answer: A. (1-z^-1)Z[1/s^3] = T^2 (z+1)/(2(z-1)^2) = 0.005(z+1)/(z-1)^2. (Key point: Plant with ZOH) 2. With zeta = 0.5, wn = 4 and T = 0.1, the desired pole is A. 0.8187 at 19.85 degrees = 0.7701 + j0.2780 B. 0.6703 at 39.7 degrees C. 0.9 at 10 degrees D. 0.8187 at 36 degrees Answer: A. |z| = e^-0.2 = 0.8187; angle = wd*T = 3.464*0.1 rad = 19.85 degrees. (Key point: Desired pole) 3. The digital PD controller Kp + KD(1 - z^-1) can be written as A. K (z - alpha)/z with K = Kp + KD and alpha = KD/(Kp+KD) B. K (z - 1)/(z - alpha) C. K z/(z - alpha) D. K (z - alpha)/(z - 1) Answer: A. Combining gives (Kp+KD)(z - KD/(Kp+KD))/z, a zero at alpha and a pole at the origin. (Key point: PD controller form) 4. The controller zero is found from the angle condition to be about A. 0.7706, almost directly below the desired pole B. 0.2459 C. 0 D. -1 Answer: A. It must give +90.11 degrees, so alpha is nearly the real part 0.7701 of the pole. (Key point: Zero placement) 5. The resulting gains are A. Kp = 9.81, KD = 32.97 B. Kp = 32.97, KD = 9.81 C. Kp = 42.78, KD = 0 D. Kp = 0.77, KD = 42.78 Answer: A. K = 42.78, KD = K*alpha = 32.97 and Kp = K(1-alpha) = 9.81. 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