--- title: "Controller design for peak time 1.2 s (ζ = 0.55), Kv" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:design-peak-time-zeta-0-55" source: https://www.bctengineer.com/q/semester_6/digital_control_system/design-peak-time-zeta-0-55 site: BCT Engineering --- # Controller design for peak time 1.2 s (ζ = 0.55), Kv Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2068 Jestha, 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/design-peak-time-zeta-0-55 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Consider the system shown below. Design a digital controller such that the dominant closed loop poles of the system will have damping ratio ζ of 0.55. It is required that the peak time of the response be 1.2 sec. Assume that the sampling period T be 0.2. Also obtain Kᵥ for the designed system. [Figure: unity-feedback loop: r(t)/R(z) → summing junction → sampler → digital controller → (1-e⁻ᵀˢ)/s → 1/s(s+1) → c(t)/C(z)] ## Short answer Lead compensator with zero at 0.8187: G_D(z)=7.582(z-0.8187)/(z-0.3689). Closed-loop poles 0.6134±j0.3542 satisfy ζ=0.55 and tp=1.2 s. Velocity error constant K_v=2.18 s⁻¹. ## Answer A lead-type digital controller $G_D(z)=K\dfrac{z-\alpha}{z-\beta}$ is designed by the root-locus method in the z-plane. Its zero cancels the plant pole at $e^{-T}$. #### Plant pulse transfer function ($T = 0.2$ s) $$ G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.01873(z+0.9355)}{(z-1)(z-0.8187)} $$ #### Desired dominant poles $$ \begin{aligned} \omega_d &= \frac{\pi}{t_p} = \frac{\pi}{1.2} = 2.618\ \text{rad/s}\\ \omega_n &= \frac{2.618}{\sqrt{1-0.55^2}} = 3.135\ \text{rad/s},\quad \zeta\omega_n = 1.724\\ |z| &= e^{-1.724\times0.2} = 0.7083,\quad \angle z = 2.618\times0.2 = 0.5236\ \text{rad} = 30^\circ\\ z_1 &= 0.6134 + j0.3542 \end{aligned} $$ #### Angle condition | Factor | Angle at $z_1$ | |---|---| | zero at $-0.9355$ | $+12.88^\circ$ | | pole at $1$ | $-137.50^\circ$ | | pole at $0.8187$ | $-120.10^\circ$ | | Total | $-244.72^\circ$ | The angle deficiency is $64.72^\circ$, so lead compensation is needed. - Place the controller zero at $\alpha = 0.8187$ to cancel the plant pole. - The controller pole angle must be $\theta_\beta = 12.88^\circ - 137.50^\circ + 180^\circ = 55.38^\circ$: $$ \beta = 0.6134 - \frac{0.3542}{\tan 55.38^\circ} = 0.3689 $$ #### Gain from the magnitude condition $$ K = \left|\frac{(z_1-1)(z_1-0.3689)}{0.01873(z_1+0.9355)}\right| = 7.582 $$ $$ G_D(z) = 7.582\,\frac{z-0.8187}{z-0.3689} $$ #### Check - Closed-loop characteristic equation: $z^2 - 1.2269z + 0.5018 = 0$, roots $0.6134 \pm j0.3542$. - Step response peak $c = 1.126$ at $k = 6$, i.e. $t_p = 1.2$ s. #### Velocity error constant $$ \begin{aligned} K_v &= \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{1}{0.2}\cdot\frac{7.582\times0.01873(1.9355)}{1-0.3689}\\ &= 2.18\ \text{s}^{-1} \end{aligned} $$ So the steady-state error to a unit ramp is $1/K_v = 0.459$. **Answer: $G_D(z) = 7.582\,\dfrac{z-0.8187}{z-0.3689}$; $K_v = 2.18\ \text{s}^{-1}$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For zeta = 0.55, omega_n = 3.135 rad/s and T = 0.2 s, what is |z1|? A. e^(-1.724 x 0.2) = 0.7083 B. 0.55 C. e^(-3.135 x 0.2) = 0.534 D. e^(-2.618 x 0.2) = 0.592 Answer: A. |z| = e^(-zeta omega_n T) with zeta omega_n = 1.724, giving 0.7083. (Key point: Desired pole magnitude) 2. After cancelling the plant pole, the controller pole angle theta_beta satisfies: A. theta_beta = 180 - 12.88 - 137.50 B. theta_beta = 12.88 - 137.50 + 180 = 55.38 degrees C. theta_beta = 64.72 + 12.88 D. theta_beta = 137.50 - 12.88 Answer: B. The angle condition gives 12.88 - 137.50 - theta_beta = -180, so theta_beta = 55.38 degrees. (Key point: Controller pole angle) 3. The gain K is obtained by: A. Setting K = 1 B. Setting K equal to the zero location C. Equating K to the damping ratio D. Applying the magnitude condition |K G_D G| = 1 at z1, giving K = 7.582 Answer: D. At a point on the locus the loop gain magnitude must be 1, which fixes K = 7.582. (Key point: Gain from magnitude condition) 4. What are Kv and the ramp error for this design? A. Kv = 2.18 per second, ramp error 0.459 B. Kv = 7.58 per second, ramp error 0.13 C. Kv = 0.459 per second, ramp error 2.18 D. Kv = 2.18 per second, ramp error 2.18 Answer: A. Kv = 2.18 s^-1 and the steady-state ramp error is 1/Kv = 0.459. (Key point: Velocity error constant) 5. The simulated peak occurs at k = 6. With T = 0.2 s this means: A. tp = 0.6 s B. tp = 1.2 s, meeting the requirement C. tp = 2.4 s D. tp = 3 s Answer: B. tp = 6 x 0.2 = 1.2 s. 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