TEST BANK FOR Control Systems Engineering 7th Edition By Norman S. Nice
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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. Closedloop 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 steadystate error
9. Steadystate, transient
10. It follows a growing transient response until the steadystate 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, statespace, differential equations
14. Transfer function  the Laplace transform of the differential equation
Statespace  representation of an nth order differential equation as n simultaneous firstorder
differential equations
Differential equation  Modeling a system with its differential equation
SOLUTIONS TO PROBLEMS
1. Five turns yields 50 v. Therefore K =
50 volts
5 x 2π rad
= 1.59
2 Chapter 1: Introduction
2.
Thermostat
Amplifier and
valves
Heater
Temperature
difference
Voltage
difference
Fuel
flow
Actual
temperature
Desired
temperature
+

3.
Desired
roll
angle
Input
voltage
+

Pilot
controls
Aileron
position
control
Error
voltage
Aileron
position
Aircraft
dynamics
Roll
rate
Integrate
Roll
angle
Gyro
Gyro voltage
4.
Speed
Error
voltage
Desired
speed
Input
voltage
+

transducer Amplifier
Motor
and
drive
system
Actual
speed
Voltage
proportional
to actual speed
Dancer
position
sensor
Dancer
dynamics
5.
Desired
power
Power
Error
voltage
Input
voltage
+

Transducer Amplifier
Motor
and
drive
system
Voltage
proportional
to actual power
Rod
position
Reactor
Actual
power
Sensor &
transducer
Solutions to Problems 3
6.
Desired
student
population +

Administration
Population
error
Desired
student
rate
Admissions
Actual
student
rate +

Graduating
and
dropout
rate
Net rate
of influx
Integrate
Actual
student
population
7.
Desired
volume +

Transducer
Volume
control circuit
Voltage
proportional
to desired
volume
Volume
error
Radio
Voltage
representing
actual volume Actual
volume

+
Transducer

Speed
Voltage
proportional
to speed
Effective
volume
4 Chapter 1: Introduction
8.
a.
R
+V
V
Differential
amplifier
Desired
level
 +
Power
amplifier
Actuator
Valve
Float
Fluid input
Drain
Tank
R
+V
V
b.
Desired
level
Amplifiers Actuator
and valve
Flow
rate in
Integrate
Actual
level
Flow
rate out
Potentiometer
+

+
Drain
Potentiometer Float

voltage
in
voltage
out
Displacement
Solutions to Problems 5
9.
Desired
force
Transducer Amplifier Valve Actuator
and load
Tire
Load cell
Actual
+ force

Current Displacement Displacement
10.
Commanded
blood pressure
Vaporizer Patient
Actual
blood
+ pressure

Isoflurane
concentration
11.
+

Controller
&
motor
Grinder
Force Feed rate
Integrator
Desired
depth Depth
12.
+

Coil
circuit
Solenoid coil
& actuator
Coil
current Force Armature
&
spool dynamics
Desired
position Depth Transducer
Coil
voltage
LVDT
13.
a. L
di
dt
+ Ri = u(t)
6 Chapter 1: Introduction
b. Assume a steadystate solution iss = B. Substituting this into the differential equation yields RB =
1,
from which B =
1
R
. The characteristic equation is LM + R = 0, from which M = 
R
L
. Thus, the total
solution is i(t) = Ae(R/L)t +
1
R
. Solving for the arbitrary constants, i(0) = A +
1
R
= 0. Thus, A =

1
R
. The final solution is i(t) =
1
R

1
R
e(R/L)t =
1
R
(1 − e−( R/ L)t ).
c.
14.
a. Writing the loop equation, Ri + L
di
dt
+ 1
C
∫ idt + vC (0) = v(t)
b. Differentiating and substituting values,
d2i
dt 2 + 2
di
dt
+ 30i = 0
Writing the characteristic equation and factoring,
M2 + 2 M+ 30 = M+ 1 + 29 i M +1  29 i .
The general form of the solution and its derivative is
i = et cos 29 t A + B sin 29 t e t
di =  A + 29 B et cos 29 t  29 A + B e t sin 29 t
dt
Using i(0) = 0;
di
dt
(0) = vL(0)
L
= 1
L
= 2
i 0 = A =0
di
dt
(0) = −A + 29B=2
Thus, A = 0 and B = 2
29
.
The solution is
Solutions to Problems 7
i =
2
29
29 e t sin 29 t
c.
i
t
[Solved] TEST BANK FOR Control Systems Engineering 7th Edition By Norman S. Nice
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 Submitted On 12 Nov, 2021 08:12:27
 GradeMaster1
 Rating : 1
 Grade : C+
 Questions : 0
 Solutions : 1124
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 Earned : $278.60