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1.1. Consider the system in the figure below containing second-order dynamics under feedback control with a proportional-derivative (PD) controller. Using any method you choose, show that the proposed PD control system cannot obtain zero offset for changes in the reference signal R(s).. R(s) Kc(1+ KDS) 1 (s+2)² Y(s) 1.2. Consider a first-order plus dead-time model in standard form such that Y(s) Kpe-0s U(s) TS+1 with positive process gain 1.3. Kp > 0. If the system is controlled using a stable PID feedback controller with G₁(s) = Kc (1 + ±1. 1½ + Kas), complete the table below by circling the best answer indicating (i) The sign that is expected for the controller term. (ii) The effect that an increase in the tuning parameter will have on steady-state offset. TI S (iii) What direction you would change the tuning parameter to increase the controller's aggression. (iv) How the tuning parameters will likely need to be changed if any of the process parameters increase. Tuning response to ↑ in process parameter (circle) Tuning Parameter Sign (circle) Effect on SS offset A for more controller "aggression" (circle) (circle) KP 0 τ Proportional, Kc + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Integral, ti ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Derivative, KD + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ first-order systems under proportional control have extremely fast response times and nearly zero offset as we increase the controller gain Kc → ∞. Succinctly give two reasons why using extremely large values of Kε is not possible in practice.

1.1. Consider the system in the figure below containing second-order dynamics under feedback control with a proportional-derivative (PD) controller. Using any method you choose, show that the proposed PD control system cannot obtain zero offset for changes in the reference signal R(s).. R(s) Kc(1+ KDS) 1 (s+2)² Y(s) 1.2. Consider a first-order plus dead-time model in standard form such that Y(s) Kpe-0s U(s) TS+1 with positive process gain 1.3. Kp > 0. If the system is controlled using a stable PID feedback controller with G₁(s) = Kc (1 + ±1. 1½ + Kas), complete the table below by circling the best answer indicating (i) The sign that is expected for the controller term. (ii) The effect that an increase in the tuning parameter will have on steady-state offset. TI S (iii) What direction you would change the tuning parameter to increase the controller's aggression. (iv) How the tuning parameters will likely need to be changed if any of the process parameters increase. Tuning response to ↑ in process parameter (circle) Tuning Parameter Sign (circle) Effect on SS offset A for more controller "aggression" (circle) (circle) KP 0 τ Proportional, Kc + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Integral, ti ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Derivative, KD + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ first-order systems under proportional control have extremely fast response times and nearly zero offset as we increase the controller gain Kc → ∞. Succinctly give two reasons why using extremely large values of Kε is not possible in practice.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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1.1. Consider the system in the figure below containing second-order dynamics under feedback control with a proportional-derivative (PD) controller. Using any method you choose, show that the proposed PD control system cannot obtain zero offset for changes in the reference signal R(s).. R(s) Kc(1+ KDS) 1 (s+2)² Y(s) 1.2. Consider a first-order plus dead-time model in standard form such that Y(s) Kpe-0s U(s) TS+1 with positive process gain 1.3. Kp > 0. If the system is controlled using a stable PID feedback controller with G₁(s) = Kc (1 + ±1. 1½ + Kas), complete the table below by circling the best answer indicating (i) The sign that is expected for the controller term. (ii) The effect that an increase in the tuning parameter will have on steady-state offset. TI S (iii) What direction you would change the tuning parameter to increase the controller's aggression. (iv) How the tuning parameters will likely need to be changed if any of the process parameters increase. Tuning response to ↑ in process parameter (circle) Tuning Parameter Sign (circle) Effect on SS offset A for more controller "aggression" (circle) (circle) KP 0 τ Proportional, Kc + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Integral, ti ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Derivative, KD + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ first-order systems under proportional control have extremely fast response times and nearly zero offset as we increase the controller gain Kc → ∞. Succinctly give two reasons why using extremely large values of Kε is not possible in practice.
Transcribed Image Text:1.1. Consider the system in the figure below containing second-order dynamics under feedback control with a proportional-derivative (PD) controller. Using any method you choose, show that the proposed PD control system cannot obtain zero offset for changes in the reference signal R(s).. R(s) Kc(1+ KDS) 1 (s+2)² Y(s) 1.2. Consider a first-order plus dead-time model in standard form such that Y(s) Kpe-0s U(s) TS+1 with positive process gain 1.3. Kp > 0. If the system is controlled using a stable PID feedback controller with G₁(s) = Kc (1 + ±1. 1½ + Kas), complete the table below by circling the best answer indicating (i) The sign that is expected for the controller term. (ii) The effect that an increase in the tuning parameter will have on steady-state offset. TI S (iii) What direction you would change the tuning parameter to increase the controller's aggression. (iv) How the tuning parameters will likely need to be changed if any of the process parameters increase. Tuning response to ↑ in process parameter (circle) Tuning Parameter Sign (circle) Effect on SS offset A for more controller "aggression" (circle) (circle) KP 0 τ Proportional, Kc + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Integral, ti ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ Derivative, KD + ↑ (none) ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ first-order systems under proportional control have extremely fast response times and nearly zero offset as we increase the controller gain Kc → ∞. Succinctly give two reasons why using extremely large values of Kε is not possible in practice.
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