A variable resistor, called a potentiometer, is shown in Figure P1.1. The resistance is varied by moving a wiper arm along a fixed resistance. The resistance from A to C is fixed, but the resistance from B to C varies with the position of the wiper arm. If it takes 10 turns to move the wiper arm from A to C, draw a block diagram of the potentiometer showing the input variable, the output variable, and (inside the block) the gain, which is a constant and is the amount by which the input is multiplied to obtain the output. An animation PowerPoint presentation (PPT) demonstrating this system is available for instructors at www.wiley.com/college/nise. See Potentiometer. [Section 1.4: Introduction to a Case Study]
Read moreTable of Contents
1
INTRODUCTION
2
MODELING IN THE FREQUENCY DOMAIN
3
MODELING IN THE TIME DOMAIN
4
TIME RESPONSE
5
REDUCTION OF MULTIPLE SUBSYSTEMS
6
STABILITY
7
STEADY-STATE ERRORS
8
ROOT LOCUS TECHNIQUES
9
DESIGN VIA ROOT LOCUS
10
FREQUENCY RESPONSE TECHNIQUES
11
DESIGN VIA FREQUENCY RESPONSE
12
DESIGN VIA STATE SPACE
13
DIGITAL CONTROL SYSTEMS
Textbook Solutions for Control Systems Engineering
Chapter 1 Problem 20
Question
Solve the following differential equations using classical methods. Assume zero initial conditions. [Review]
a. \(\frac{d x}{d t}+7 x=5 \cos 2 t\)
b. \(\frac{d^2 x}{d t^2}+6 \frac{d x}{d t}+8 x=5 \sin 3 t\)
c. \(\frac{d^2 x}{d t^2}+8 \frac{d x}{d t}+25 x=10 u(t)\)
Solution
Step 1 of 3
a)
The corresponding characteristic equation for the homogeneous solution of the form :
For the particular solution, we assume the solution to be of the form :
Substituting these in the differential equation to get the values of and
On comparing coefficients of like terms:
Solving the above two equations give :
Thus, the solution,
With the initial condition,
Final solution:
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Title
Control Systems Engineering 7
Author
Norman J. Nise
ISBN
9781118170519