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oblem 2. The closed-loop transfer function of the position control system in Figure 3 is T(s)- T(s)+(1+25K,)s+25K, 25K, d values of K, and K, so that the unit step response of the closed-loop system has 5% overshoot and a rise e of about 0.1 second. Motor Amp R(s) E(s) 25 C(s) K1 s(s + 1) K2s Tachometer +

oblem 2. The closed-loop transfer function of the position control system in Figure 3 is T(s)- T(s)+(1+25K,)s+25K, 25K, d values of K, and K, so that the unit step response of the closed-loop system has 5% overshoot and a rise e of about 0.1 second. Motor Amp R(s) E(s) 25 C(s) K1 s(s + 1) K2s Tachometer +

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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**Problem 2.** The closed-loop transfer function of the position control system in Figure 3 is given by: \[ T(s) = \frac{25K_1}{s^2 + (1 + 25K_2)s + 25K_1} \] Find values of \( K_1 \) and \( K_2 \) so that the unit step response of the closed-loop system has 5% overshoot and a rise time of about 0.1 second. --- **Figure 3 Explanation:** Figure 3 depicts a block diagram of an angular position control system which includes a tachometer in the inner loop. - **Input Signal (R(s))**: The reference signal entering the system. - **Summing Junctions**: There are two summing junctions in the system, both using a minus sign to subtract feedback from the input signal, resulting in the error signal \( E(s) \). - **Amplifier (Amp)**: Multiplies the error signal \( E(s) \) by \( K_1 \). - **Motor Dynamics**: The processed signal is sent to the motor, represented by the transfer function \(\frac{25}{s(s+1)}\). - **Feedback Loop (Tachometer)**: The output \( C(s) \) is fed back and multiplied by \( K_2 \) for correction. The system seeks to stabilize the response with specific parameters for performance, namely 5% overshoot and a rise time of approximately 0.1 second.
Transcribed Image Text:**Problem 2.** The closed-loop transfer function of the position control system in Figure 3 is given by: \[ T(s) = \frac{25K_1}{s^2 + (1 + 25K_2)s + 25K_1} \] Find values of \( K_1 \) and \( K_2 \) so that the unit step response of the closed-loop system has 5% overshoot and a rise time of about 0.1 second. --- **Figure 3 Explanation:** Figure 3 depicts a block diagram of an angular position control system which includes a tachometer in the inner loop. - **Input Signal (R(s))**: The reference signal entering the system. - **Summing Junctions**: There are two summing junctions in the system, both using a minus sign to subtract feedback from the input signal, resulting in the error signal \( E(s) \). - **Amplifier (Amp)**: Multiplies the error signal \( E(s) \) by \( K_1 \). - **Motor Dynamics**: The processed signal is sent to the motor, represented by the transfer function \(\frac{25}{s(s+1)}\). - **Feedback Loop (Tachometer)**: The output \( C(s) \) is fed back and multiplied by \( K_2 \) for correction. The system seeks to stabilize the response with specific parameters for performance, namely 5% overshoot and a rise time of approximately 0.1 second.
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