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Show that there are a complex conjugate pair of CL poles at GD 2 tiw, where w = /1+ Gp for small enough GD given by 0 < GD < 2√1+Gp - G 4

Show that there are a complex conjugate pair of CL poles at GD 2 tiw, where w = /1+ Gp for small enough GD given by 0 < GD < 2√1+Gp - G 4

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Problem 2 In this problem, we are going to investigate how the feedback controller can have different impact on the closed-loop performance depending on the plant transfer function. For each combination of plant transfer function and feedback controller, we need to first write down the error equation and ascertain the conditions for stability. Finally, we will investigate the steady-state tracking error and evaluate the closed-loop poles. a) Consider the following second-order plant: [ 1 s²+1 with a proportional-derivative (P-D) feedback controller G(s) = Gp + GD S K(s) G(s). We refer to L(s) = G(s)K(s) as the loop gain. The closed-loop system is stable if all the roots of the equation 1+L(s) = 0 are in the left half-plane. Recall that the closed-loop tracking error is given by: = E(s): 1 R(s) 1+ L(s) where R(s) is the reference signal. In the following, we will investigate the impact of the presence of integrators, i.e., terms of the form 1/s, in the feedback controller G(s) on the steady-state tracking error given. Recall that the steady-state tracking error can be computed using the Final Value Theorem: ess= e(∞) = lim s E(s) (only when the closed-loop is system is stable!) s➜0+ E(s) Show that the closed-loop (CL) tracking error is: s² +1 s2 + Gps +1+Gp = ▪ Apply Routh's test to show that the CL system is stable if Gp> 0 and Gp > -1. Under the above conditions, show that the steady-state tracking error is 1/(1+Gp) for unit step reference R(s) = = 1/s. ▪ Show that there are a complex conjugate pair of CL poles at GD 2 for small enough GD given by 0 < GD < 2√1+Gp -R(s) tiw, where w = √₁+Gp_% 1+ GP- 4
Transcribed Image Text:Problem 2 In this problem, we are going to investigate how the feedback controller can have different impact on the closed-loop performance depending on the plant transfer function. For each combination of plant transfer function and feedback controller, we need to first write down the error equation and ascertain the conditions for stability. Finally, we will investigate the steady-state tracking error and evaluate the closed-loop poles. a) Consider the following second-order plant: [ 1 s²+1 with a proportional-derivative (P-D) feedback controller G(s) = Gp + GD S K(s) G(s). We refer to L(s) = G(s)K(s) as the loop gain. The closed-loop system is stable if all the roots of the equation 1+L(s) = 0 are in the left half-plane. Recall that the closed-loop tracking error is given by: = E(s): 1 R(s) 1+ L(s) where R(s) is the reference signal. In the following, we will investigate the impact of the presence of integrators, i.e., terms of the form 1/s, in the feedback controller G(s) on the steady-state tracking error given. Recall that the steady-state tracking error can be computed using the Final Value Theorem: ess= e(∞) = lim s E(s) (only when the closed-loop is system is stable!) s➜0+ E(s) Show that the closed-loop (CL) tracking error is: s² +1 s2 + Gps +1+Gp = ▪ Apply Routh's test to show that the CL system is stable if Gp> 0 and Gp > -1. Under the above conditions, show that the steady-state tracking error is 1/(1+Gp) for unit step reference R(s) = = 1/s. ▪ Show that there are a complex conjugate pair of CL poles at GD 2 for small enough GD given by 0 < GD < 2√1+Gp -R(s) tiw, where w = √₁+Gp_% 1+ GP- 4
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