Example 1.16: Consider the following system 10 8-698-4- 0 0 1 ¹0 -2 -31 y=[100] determine the optimal feedback gain matrix k such that the following performance index is minimized: Where J = [™* (x² Qx + u² Ru)dt Q = [100 0 0 001 1 0 0 1 R = 0.01

Example 1.16: Consider the following system 10 8-698-4- 0 0 1 ¹0 -2 -31 y=[100] determine the optimal feedback gain matrix k such that the following performance index is minimized: Where J = [™* (x² Qx + u² Ru)dt Q = [100 0 0 001 1 0 0 1 R = 0.01

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...

See similar textbooks

Similar questions

an operational amplifier is working as differential amplifier it takes input voltage 41 volt at inverting terminal and 5`1 volt at non inverting terminal. by keeping these voltages find following parameters calculate the output voltage for non-inverting terminal if gain is 3 a=163v b=153v c=113v d=123v
Q:- which characteristic of an analog signal are changed to represent the lowpass analog signal in each of the following digital-to-analog conversation? a)ASK b)FSK c)PSK d)QMA.
The common-mode gain of the front-end circuit for the instrumentation amplifier is equal to ACM= 1 V/V. If the instrumentation amplifier is not reducing the common-mode signal, then why is the CMRR of an instrumentation amplifier higher than a differential amplifier, in general?
Design PID controller and calculate the overall transfer function if Kp, Kd, and Ki parameter values are determined using Ziegler-Nichols method. The system is unity feedback shown in Figure 3 (a) and step response of the system is given in Figure 3(b).
Analyze the multiplexer circuit shown below to obtain its output function, f(A,B,A1,A0), in compact min-termform.
(a) An instrumentation amplifier shown in Figure 1 is an anplifier of low-level signals used in process control or measurement applications and conmercially available in single-package units: (i) Show that Vo=R3/R2(1+2R1/Rg)(v2−v1). (ii) Assume R1=50kohms and R2=R3. If Rg is a tunable resistor with resistance ranging from 1000 ohms to 10k ohms, what is the range of variable gain? (iii) Assume R1=10k ohms and R2=R3,∣v2−v1∣<=120mV and Vcc=12 V. In order to keep the op-amp circuit work in linear region, what is the minimum value of Rg ? (iv) State TWO characteristics of this op-amp circuit.
Q: Find the digital equivalent and the resolution for a 3 bit DAC using Successive Approximation of Vin = 0.4 V and Vref = 1 V?
DEGREE: ELECTRICAL ENGINEERING SUBJECT/COURSE: INDUSTRIAL ELECTRONICSTOPIC: FEEDBACK OSCILLATORS NOTE: Please solve in this way. 1. Please have a good handwriting, some of the answers are not readable. Thank you!2. GIVEN. (Include symbols and units)3. ILLUSTRATION (Required).4. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, no skipping, and detailed as possible5. FINAL ANSWERS must be rounded up to two decimal places with corresponding unit. PROBLEM:An amplifier without feedback has an internal gain of -500 with a distortion of 10% and a normal input of 0.5-volt, negative feedback with β = 0.10 was added. Determine the following:A. Gain of amplifier when the feedback was addedB. Percent reduction in distortion when feedback was addedC. The voltage required in the input of the amplifier in order that output voltage is the same as the voltage without feedback
DEGREE: ELECTRICAL ENGINEERING SUBJECT/COURSE: INDUSTRIAL ELECTRONICSTOPIC: FEEDBACK OSCILLATORS NOTE: Please solve in this way. 1. Please have a good handwriting, some of the answers are not readable. Thank you!2. GIVEN. (Include symbols and units)3. ILLUSTRATION (Required).4. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, no skipping, and detailed as possible5. FINAL ANSWERS must be rounded up to two decimal places with corresponding unit. PROBLEM:An amplifier without feedback has an internal gain of -500 with a distortion of 10% and a normal input of 0.5-volt, negative feedback with β = 0.10 was added. Determine the voltage required in the input of the amplifier in order that output voltage is the same as the voltage without feedback
DEGREE: ELECTRICAL ENGINEERING SUBJECT/COURSE: INDUSTRIAL ELECTRONICSTOPIC: FEEDBACK OSCILLATORS NOTE: Please solve in this way. 1. Please have a good handwriting, some of the answers are not readable. Thank you!2. GIVEN. (Include symbols and units)3. ILLUSTRATION (Required).4. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, no skipping, and detailed as possible5. FINAL ANSWERS must be rounded up to two decimal places with corresponding unit. PROBLEM:An amplifier without feedback has an internal gain of -500 with a distortion of 10% and a normal input of 0.5-volt, negative feedback with β = 0.10 was added. Determine the percent reduction in distortion when feedback was added
DEGREE: ELECTRICAL ENGINEERING SUBJECT/COURSE: INDUSTRIAL ELECTRONICSTOPIC: FEEDBACK OSCILLATORS NOTE: Please solve in this way. 1. Please have a good handwriting, some of the answers are not readable. Thank you!2. GIVEN. (Include symbols and units)3. ILLUSTRATION (Required).4. Step-by-step SOLUTION with Formulas and Symbols. No Shortcut, no skipping, and detailed as possible5. FINAL ANSWERS must be rounded up to two decimal places with corresponding unit. PROBLEM:An amplifier without feedback has an internal gain of -500 with a distortion of 10% and a normal input of 0.5-volt, negative feedback with β = 0.10 was added. Determine the gain of amplifier when the feedback was added
I need help figuring out how the following problem would be figured out with the given answer: PROBLEM 4: A pressure transducer has a range of ±120 mm Hg, a sensitivity of 50 μV/mm Hg/V (excitation) and an allowable excitation voltage from 5 to 10 VDC. Estimate the required gain (amplification) needed to maximize the values of the signal. Assume the data acquisition voltage input range is ±10 volts Case 1: 50uv*5v*120mmHg = .030V G = 10/.03 = 333 Case 2: 50uv*10v*120mmHg = .060V G = 10/.06 = 166