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Solutions to Problems 5 Armature voltage current Desired + Controller 9. Desired force 10. Actual 11. 12. Desired position Force Depth 13. di a. L dt + Ri = u(t) Commanded blood pressure + Isoflurane concentration blood pressure - Patient Vaporizer 6 Chap

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Subido en
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Solutions to Problems 5 Armature voltage current Desired + Controller 9. Desired force 10. Actual 11. 12. Desired position Force Depth 13. di a. L dt + Ri = u(t) Commanded blood pressure + Isoflurane concentration blood pressure - Patient Vaporizer 6 Chapter 1: Introduction 29 29 b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB = 1, from which B = 1 . The characteristic equation is LM + R = 0, from which M = - R R . Thus, the total L solution is i(t) = Ae-(R/L)t + 1 . Solving for the arbitrary constants, i(0) = A + 1 = 0. Thus, A = R 1 1 - . The final solution is i(t) = R -- 1 e-(R/L)t = 1 (1 − e−( R/ L)t ). R R R R c. 14. di 1 a. Writing the loop equation, Ri + L dt + C  idt + vC (0 ) = v(t) d2i di b. Differentiating and substituting values, dt 2 + 2 dt + 30i = 0 Writing the characteristic equation and factoring, M2 + 2 M + 30 = M + 1 + i M + 1 - i . The general form of the solution and its derivative is i= e-t cos t A + B sin t e-t di = - A + dt B e-t cos 29 t - 29 A+ B e-t sin 29 t Using i(0) = 0; di (0) = vL (0) = 1 = 2 dt L L i 0 = A =0 di ( 0) =− A + 29 B =2 dt Thus, A = 0 and B = 2 . The solution is 29 29 29 29 Solutions to Problems 7 29 53 53 t i= 2 29 e-t sin t c. i 15. a. Assume a particular solution of Substitute into the differential equation and obtain Equating like coefficients, From which, C = 35 and D = 10 . The characteristic polynomial is Thus, the total solution is 35 35 Solving for the arbitrary constants, x(0) = A +53 = 0. Therefore, A = - 53 . The final solution is b. Assume a particular solution of xp = Asin3t + Bcos3t 29 8 Chapter 1: Introduction dt Substitute into the differential equation and obtain (18A − B)cos(3t) − (A + 18B)sin(3t) = 5sin(3t) Therefore, 18A – B = 0 and –(A + 18B) = 5. Solving for A and B we obtain xp = (-1/65)sin3t + (-18/65)cos3t The characteristic polynomial is M2 + 6 M+ 8 = M+ 4 M+ 2 Thus, the total solution is x= C e- 4 t + D e-2 t + - 18 cos 3 t - 1 sin 3 t 65 Solving for the arbitrary constants, x(0) = C + D − 18 65 65 = 0 . Also, the derivative of the solution is dx = - 3 cos3 t + 54 sin 3 t -4C e- 4 t -2 D e-2 t dt 65 65 Solving for the arbitrary constants, x(0) − − 4C − 2 D = 0 , or C = − and D = . The final solution is x= - 18 cos 3 t - 1 sin 3 t - 3 e-4 t + 15 e-2 t c. Assume a particular solution of xp = A Substitute into the differential equation and obtain 25A = 10, or A = 2/5. The characteristic polynomial is M2 +8 M + 25 = M + 4 +3 i M + 4 - 3 i Thus, the total solution is x = 2 + e- 4 t 5 B sin 3 t + C cos 3 t Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = -2/5. Also, the derivative of the solution is dx = 3 B -4 C cos3 t - 4 B + 3 C sin3 t e-4 t . 3 3 15 65 10 26

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Subido en
28 de mayo de 2025
Número de páginas
649
Escrito en
2024/2025
Tipo
Examen
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Control Systems Engineering 7th Edition
Solutions Manual Nise, Exercises of Control
Systems



Introduction
ANSWERS TO REVIEW QUESTIONS
1. Guided missiles, automatic gain control in radio receivers, satellite tracking antenna
2. Yes - power gain, remote control, parameter conversion; No - Expense, complexity
3. Motor, low pass filter, inertia supported between two bearings
4. Closed-loop systems compensate for disturbances by measuring the response, comparing it to
the input response (the desired output), and then correcting the output response.
5. Under the condition that the feedback element is other than unity
6. Actuating signal
7. Multiple subsystems can time share the controller. Any adjustments to the controller can be
implemented with simply software changes.
8. Stability, transient response, and steady-state error
9. Steady-state, transient
10. It follows a growing transient response until the steady-state response is no longer visible. The
system will either destroy itself, reach an equilibrium state because of saturation in driving
amplifiers, or hit limit stops.
11. Transient response
12. True
13. Transfer function, state-space, differential equations
14. Transfer function - the Laplace transform of the differential equation
State-space - representation of an nth order differential equation as n simultaneous first-order
differential equations
Differential equation - Modeling a system with its differential equation



SOLUTIONS TO PROBLEMS
50 volts
1. Five turns yields 50 v. Therefore K = = 1.59
5 x 2 rad

,2 Chapter 1: Introduction




2.


Desired Temperature Voltage Actual
Fuel
temperature difference difference temperature
flow

+ Amplifier and
Thermostat Heater
valves
-



3.


Desired Input Error Aileron Roll Roll
roll voltage voltage position rate angle
angle
+
controls
-




4.
Input
Speed
Desired voltage Actual
Error Motor
speed speed
+ voltage and
transducer Amplifier drive
system
-
Dancer
Dancer
position
Voltage dynamics
sensor
proportional
to actual speed

5.

Input Power Rod
voltage Error position
Desired voltage Motor Actual
power and power
+ Amplifier
Transducer drive Reactor
system
-

Sensor &
Voltage transducer
proportional
to actual power

, Solutions to Problems 3




6.
Graduating
and
drop-out
rate
Desired Actual
Desired Population student Actual Net rate student
student error rate student - of influx population
population + rate +
Administration Admissions Integrate

-



7.
Voltage Voltage
proportional representing
to desired Volume actual volume Actual
Desired volume volume
error
volume + Volume
Transducer Radio
control circuit
-
Effective
volume
+
Voltage
- proportional
to speed

Transducer
-

Speed

, 4 Chapter 1: Introduction


8.

a.
Fluid input




amplifier
+V
Differential Desired
R level
-V


R
Float -V

Tank
Drain


b.
Flow Actual
voltage level
Desired rate in
in
level + +
Potentiometer Amplifiers Actuator Integrate
and valve
- -




Drain
Flow
rate out

Displacement
voltage
out
Potentiometer Float
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