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For the system given below determine: b. C. Find the mathematical expression of E(s) as a function of R(s) Calculate the Steady-state positional error using the Final Value Theorem (Hint: lim f(t) = lim SF (s)) [-00 $40 What is the "Type" of this system? R(s) + E(s) G(s) C(s) T= S 5+10 1 + $

For the system given below determine: b. C. Find the mathematical expression of E(s) as a function of R(s) Calculate the Steady-state positional error using the Final Value Theorem (Hint: lim f(t) = lim SF (s)) [-00 $40 What is the "Type" of this system? R(s) + E(s) G(s) C(s) T= S 5+10 1 + $

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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### Analysis of a Control System #### Problem Statement For the given system, determine the following: a. Find the mathematical expression of \(E(s)\) as a function of \(R(s)\). b. Calculate the steady-state positional error using the Final Value Theorem (Hint: \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)\)). c. Determine the “Type” of this system. #### System Diagram - **Components**: The system involves \(R(s)\), \(E(s)\), a feedback loop with a transfer function \(G(s)\), and output \(C(s)\). - **Block Diagram**: \(G(s) = \frac{5}{s+10}\). - **Transfer Function**: - Total Transfer Function \(T = \frac{\frac{5}{s+10}}{1 + \frac{5}{s+10}} = \frac{5}{s+15} = \frac{C}{R}\). - Relationship derived: \(\frac{W}{R} = \frac{1}{s+15}\). #### Solutions a. **Expression for \(E(s)\)**: \[ E(s) = R(s) \cdot \left( \frac{5}{s+15} \right) \] This equation highlights the dependence of the error function \(E(s)\) on the reference input \(R(s)\). b. **Steady-State Positional Error**: - Apply the Final Value Theorem: \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)\). - Calculate \(K_p\): \[ K_p = \lim_{s \to 0} G(s) = \lim_{s \to 0} \frac{5}{s+10} = \frac{1}{2} \] - Steady-state error \(e_{ss}\): \[ e_{ss} = \frac{A}{1 + K_p} = \frac{E}{1 + \lim_{s \to 0} 5 \left(\frac{5}{s}\right)} \] \[ e_{ss} = \frac{0.3}{1.5
Transcribed Image Text:### Analysis of a Control System #### Problem Statement For the given system, determine the following: a. Find the mathematical expression of \(E(s)\) as a function of \(R(s)\). b. Calculate the steady-state positional error using the Final Value Theorem (Hint: \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)\)). c. Determine the “Type” of this system. #### System Diagram - **Components**: The system involves \(R(s)\), \(E(s)\), a feedback loop with a transfer function \(G(s)\), and output \(C(s)\). - **Block Diagram**: \(G(s) = \frac{5}{s+10}\). - **Transfer Function**: - Total Transfer Function \(T = \frac{\frac{5}{s+10}}{1 + \frac{5}{s+10}} = \frac{5}{s+15} = \frac{C}{R}\). - Relationship derived: \(\frac{W}{R} = \frac{1}{s+15}\). #### Solutions a. **Expression for \(E(s)\)**: \[ E(s) = R(s) \cdot \left( \frac{5}{s+15} \right) \] This equation highlights the dependence of the error function \(E(s)\) on the reference input \(R(s)\). b. **Steady-State Positional Error**: - Apply the Final Value Theorem: \(\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)\). - Calculate \(K_p\): \[ K_p = \lim_{s \to 0} G(s) = \lim_{s \to 0} \frac{5}{s+10} = \frac{1}{2} \] - Steady-state error \(e_{ss}\): \[ e_{ss} = \frac{A}{1 + K_p} = \frac{E}{1 + \lim_{s \to 0} 5 \left(\frac{5}{s}\right)} \] \[ e_{ss} = \frac{0.3}{1.5
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