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system are given below: (input) i(t) = 2u(t) (output) v(oo) (in steady-state) can take any value between 2 to 4 Your cousin suspects the components used in the circuit to be faulty, i.e., their values might be changing over time (e.g., due to temperature at the time of experiment). When your cousin repeated the same experiment with a constant current input of i(t) = 2u(t), they found that the output voltage v(t) could settle down to any value between 2 and 4 during different test runs, i.e., reference R(S) a) Can you use Final Value Theorem to confirm that the resistor is faulty (e.g., temperature-sensitive), with its value varying between 1 and 2? b) Could you help your cousin achieve this by designing a simple proportional controller? feedback controller i(t) = i₁(t) + ₂(t) i2(t) = dv(t) dt G(s) diy(t) v(t) = in(t)R + L dt error input X(s) E(s)=R(S)-Y(s) K(s) output Y(s) Note: In the closed-loop system, X(s) = I(s) is the programmable current input, Y(s) = V(s) is the output voltage, R(s) = 3/s is the reference voltage. You have already identified the plant transfer function K(s). All you need to do is design a proportional gain G, i.e., G(s) = G, such that the steady state relative tracking error is within +10%. For the battery charger to work properly, the output voltage u(t) needs to be maintained within +10% of a target value of 3 unit, i.e., (control objective) v (oo) should stay within +10% of the reference value of 3 unit.

system are given below: (input) i(t) = 2u(t) (output) v(oo) (in steady-state) can take any value between 2 to 4 Your cousin suspects the components used in the circuit to be faulty, i.e., their values might be changing over time (e.g., due to temperature at the time of experiment). When your cousin repeated the same experiment with a constant current input of i(t) = 2u(t), they found that the output voltage v(t) could settle down to any value between 2 and 4 during different test runs, i.e., reference R(S) a) Can you use Final Value Theorem to confirm that the resistor is faulty (e.g., temperature-sensitive), with its value varying between 1 and 2? b) Could you help your cousin achieve this by designing a simple proportional controller? feedback controller i(t) = i₁(t) + ₂(t) i2(t) = dv(t) dt G(s) diy(t) v(t) = in(t)R + L dt error input X(s) E(s)=R(S)-Y(s) K(s) output Y(s) Note: In the closed-loop system, X(s) = I(s) is the programmable current input, Y(s) = V(s) is the output voltage, R(s) = 3/s is the reference voltage. You have already identified the plant transfer function K(s). All you need to do is design a proportional gain G, i.e., G(s) = G, such that the steady state relative tracking error is within +10%. For the battery charger to work properly, the output voltage u(t) needs to be maintained within +10% of a target value of 3 unit, i.e., (control objective) v (oo) should stay within +10% of the reference value of 3 unit.

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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K(s) = S+1 52 +5+1 system are given below: i(t) reference R(s) feedback controller 1₁(t) i(t) i(t) + i2(t) i2(t) = dv(t) dt G(s) (input) i(t) = 2u(t) (output) v(oo) (in steady-state) can take any value between 2 to 4 Your cousin suspects the components used in the circuit to be faulty, i.e., their values might be changing over time (e.g., due to temperature at the time of experiment). When your cousin repeated the same experiment with a constant current input of i(t) = 2 u(t), they found that the output voltage v(t) could settle down to any value between 2 and 4 during different test runs, i.e., din(t) v(t)= i(t) R+ L dt a) Can you use Final Value Theorem to confirm that the resistor is faulty (e.g., temperature-sensitive), with its value varying between 1 and 2? b) Could you help your cousin achieve this by designing a simple proportional controller? 1₂(0)| input X(s) error E(s)=R(S)-Y(s) K(s) v(t) output Y(s) Note: In the closed-loop system, X(s) = I(s) is the programmable current input, Y(s) = V(s) is the output voltage, R(s) = 3/s is the reference voltage. You have already identified the plant transfer function K(s). All you need to do is design a proportional gain G, i.e., G(s) = G, such that the steady state relative tracking error is within 10%. For the battery charger to work properly, the output voltage u(t) needs to be maintained within +10% of a target value of 3 unit, i.e., (control objective) v(oo) should stay within ±10% of the reference value of 3 unit.
Transcribed Image Text:K(s) = S+1 52 +5+1 system are given below: i(t) reference R(s) feedback controller 1₁(t) i(t) i(t) + i2(t) i2(t) = dv(t) dt G(s) (input) i(t) = 2u(t) (output) v(oo) (in steady-state) can take any value between 2 to 4 Your cousin suspects the components used in the circuit to be faulty, i.e., their values might be changing over time (e.g., due to temperature at the time of experiment). When your cousin repeated the same experiment with a constant current input of i(t) = 2 u(t), they found that the output voltage v(t) could settle down to any value between 2 and 4 during different test runs, i.e., din(t) v(t)= i(t) R+ L dt a) Can you use Final Value Theorem to confirm that the resistor is faulty (e.g., temperature-sensitive), with its value varying between 1 and 2? b) Could you help your cousin achieve this by designing a simple proportional controller? 1₂(0)| input X(s) error E(s)=R(S)-Y(s) K(s) v(t) output Y(s) Note: In the closed-loop system, X(s) = I(s) is the programmable current input, Y(s) = V(s) is the output voltage, R(s) = 3/s is the reference voltage. You have already identified the plant transfer function K(s). All you need to do is design a proportional gain G, i.e., G(s) = G, such that the steady state relative tracking error is within 10%. For the battery charger to work properly, the output voltage u(t) needs to be maintained within +10% of a target value of 3 unit, i.e., (control objective) v(oo) should stay within ±10% of the reference value of 3 unit.
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