O N E
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. 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. Natural response 12. Determine the transient response performance of the system. 13. Determine system parameters to meet the transient response specifications for the system. 14. True 15. Transfer function, statespace, differential equations 16. 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 = 1.59 5 x 2π rad
Copyright ? 2011 by John Wiley & Sons, Inc.
12 Chapter 1: Introduction
2.
Desired temperature
Temperature difference
Voltage difference Amplifier and valves
Fuel flow Heater
Actual temperature
+
Thermostat

3.
Desired roll angle Pilot controls
Input voltage
Error voltage Aileron position control
Aileron position Aircraft dynamics
Roll rate Integrate
Roll angle
+ 
Gyro voltage
Gyro
4.
Desired speed transducer Input voltage Speed Error voltage Amplifier
+ 
Motor and drive system Dancer dynamics
Actual speed
Voltage proportional to actual speed
Dancer position sensor
Copyright ? 2011 by John Wiley & Sons, Inc.
13 Solutions to Problems
5.
Input voltage Desired power Transducer Power Error voltage Amplifier Rod position Motor and drive system Actual power Reactor
+ 
Voltage proportional to actual power
Sensor & transducer
6.
Graduating and dropout rate Actual student rate Admissions
Desired student population
Population error
Desired student rate Administration
+
Net rate of influx Integrate
Actual student population
+ 
7.
Voltage proportional to desired volume Voltage representing actual volume Volume control circuit Radio
Desired volume Transducer
+
Volume error
Actual volume
Effective volume
+ Voltage proportional to speed
Transducer
Speed
Copyright ? 2011 by John Wiley & Sons, Inc.
14 Chapter 1: Introduction
8. a.
Fluid input
Valve Actuator
Power amplifier +V Differential amplifier + +V V
R
Desired level
R
Float Tank V Drain
b.
Desired level Potentiometer voltage in + Amplifiers Flow rate in + Integrate Actual level
Actuator and valve
Drain Flow rate out Displacement Potentiometer Float
voltage out
Copyright ? 2011 by John Wiley & Sons, Inc.
15 Solutions to Problems
9.
Desired force Transducer
+ Amplifier 
Current Valve
Displacement Actuator and load
Displacement Tire
Actual force
Load cell
10.
Commanded blood pressure + Vaporizer 
Isoflurane concentration Patient
Actual blood pressure
11.
Desired depth + 
Controller & motor
Force Grinder
Feed rate Integrator
Depth
12.
Desired position
Coil voltage + Transducer 
Coil circuit
Coil current
Solenoid coil & actuator
Force
Armature & spool dynamics
Depth
LVDT
Copyright ? 2011 by John Wiley & Sons, Inc.
16 Chapter 1: Introduction
13.
a.
Nervous system electrical impulses
Desired Light Intensity
Brain
+
Internal eye muscles
Retina’s Light Intensity
Retina + Optical
b.
Desired Light Intensity
Nervous system electrical impulses
Brain
+
Internal eye muscles
Retina’s Light Intensity
Retina + Optical Nerves
External Light
If the narrow light beam is modulated sinusoidally the pupil’s diameter will also vary sinusoidally (with a delay see part c) in problem) c. If the pupil responded with no time delay the pupil would contract only to the point where a small amount of light goes in. Then the pupil would stop contracting and would remain with a fixed diameter.
Copyright ? 2011 by John Wiley & Sons, Inc.
17 Solutions to Problems
14.
Desired +
HT’s
Amplifier
Actual
Gyroscopic
15.
16.
17. a. L
di + Ri = u(t) dt
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 =  . Thus, the total R L
Copyright ? 2011 by John Wiley & Sons, Inc.
18 Chapter 1: Introduction
1 1 . Solving for the arbitrary constants, i(0) = A + = 0. Thus, A = R R 1 1 1 (R/L)t 1 ?( R/ L) t ). = (1 ? e . The final solution is i(t) = e R R R R
solution is i(t) = Ae(R/L)t +
c.
18. a. Writing the loop equation,
di 1 + idt + vC (0) = v(t) dt C ∫ d 2i di + 2 + 25i = 0 b. Differentiating and substituting values, 2 dt dt Ri + L
Writing the characteristic equation and factoring,
M 2 + 2M + 25 = ( M + 1 + 24i)( M + 1 ? 24i) .
The general form of the solution and its derivative is
i = Ae?t cos( 24t ) + Be? t sin( 24t ) di = (? A + 24 B)e ?t cos( 24t ) ? ( 24 A + B)e? t sin( 24t ) dt di vL (0) 1 = =1 Using i (0) = 0; (0) = dt L L
i 0 = A =0
di (0) = ? A + 24 B =1 dt 1 Thus, A = 0 and B = . 24
The solution is
i=
1 ?t e sin( 24t ) 24
Copyright ? 2011 by John Wiley & Sons, Inc.
19 Solutions to Problems
c.
19. a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
10 35 From which, C = 53 and D = 53 . 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
Copyright ? 2011 by John Wiley & Sons, Inc.
110 Chapter 1: Introduction
xp = Asin3t + Bcos3t 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
18 1 cos 3 t sin 3 t 65 65 18 =0. Solving for the arbitrary constants, x(0) = C + D ? 65 x =C e
4t
+ De
2t
+ 
Also, the derivative of the solution is
dx =  3 cos 3 t + 54 sin 3 t  4 C e  4 t  2 D e  2 t dt 65 65
. Solving for the arbitrary constants, x(0) The final solution is
?
15 3 3 ? 4C ? 2D = 0 , or C = ? and D = . 26 65 10
x =
18 1 3  4 t 15  2 t cos 3 t sin 3 t e + e 65 65 10 26
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 4t +e B sin 3 t + C cos 3 t 5
Solving for the arbitrary constants, x(0) = C + 2/5 = 0. Therefore, C = 2/5. Also, the derivative of the solution is
Copyright ? 2011 by John Wiley & Sons, Inc.
111 Solutions to Problems
dx 4t dt = 3 B 4 C cos 3 t  4 B + 3 C sin 3 t e
. Solving for the arbitrary constants, x(0) = 3B – 4C = 0. Therefore, B = 8/15. The final solution is
x(t) =
20.
2 2 ?8 ? e ?4t ? sin(3t) + cos(3t )? ? 5 5 15
a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
1 1 From which, C =  5 and D =  10 . The characteristic polynomial is
Thus, the total solution is
1 Solving for the arbitrary constants, x(0) = A  5 = 2. Therefore, A = solution is
11 . Also, the derivative of the 5
dx dt
. Solving for the arbitrary constants, x(0) =  A + B  0.2 = 3. Therefore, B = is
3 ? . The final solution 5
1 1 3 ? 11 x(t) = ? cos(2t) ? sin(2t) + e ?t ? cos(t) ? sin(t)? ? 5 10 5 5
b. Assume a particular solution of xp = Ce2t + Dt + E Substitute into the differential equation and obtain
Copyright ? 2011 by John Wiley & Sons, Inc.
112 Chapter 1: Introduction
Equating like coefficients, C = 5, D = 1, and 2D + E = 0. From which, C = 5, D = 1, and E =  2. The characteristic polynomial is
Thus, the total solution is
Solving for the arbitrary constants, x(0) = A + 5  2 = 2 Therefore, A = 1. Also, the derivative of the solution is
dx = (? A + B)e ? t ? Bte ?t ? 10e ?2 t + 1 dt
. Solving for the arbitrary constants, x(0) = B  8 = 1. Therefore, B = 9. The final solution is
c. Assume a particular solution of xp = Ct2 + Dt + E Substitute into the differential equation and obtain
1 Equating like coefficients, C = 4 , D = 0, and 2C + 4E = 0. 1 1 From which, C = 4 , D = 0, and E =  8 . The characteristic polynomial is
Thus, the total solution is
9 1 Solving for the arbitrary constants, x(0) = A  8 = 1 Therefore, A = 8 . Also, the derivative of the solution is
dx dt
. Solving for the arbitrary constants, x(0) = 2B = 2. Therefore, B = 1. The final solution is
Copyright ? 2011 by John Wiley & Sons, Inc.
113 Solutions to Problems
21.
Spring displacement Desired force Input voltage + Input transducer F up Controller Actuator Pantograph dynamics F out
Spring

Sensor
22.
Desired Amount of HIV viruses
RTI
Controller
Amount of HIV viruses
Patient
PI
Copyright ? 2011 by John Wiley & Sons, Inc.
114 Chapter 1: Introduction
23.
a.
Climbing & Rolling Resistances
Motive Force
Inverter Control Command Desired Speed
ECU +
Controlled Voltage
Inverter
Electric Motor
Actual
Vehicle
+
Aerodynamic
Aerodynamic
Speed
Copyright ? 2011 by John Wiley & Sons, Inc.
115 Solutions to Problems
b.
Climbing & Rolling Resistances
Desired
Speed _
Accelerator Displacement ECU
Motive
Actual Vehicle
Accelerator,
+
+
Aerodynamic
Aerodynamic
Speed
Copyright ? 2011 by John Wiley & Sons, Inc.
116 Chapter 1: Introduction
c.
Accelerator Accelerator Speed Error Desired ECU
+
ICE
Climbing & Rolling Resistances Actual
+
Power
Planetary Gear Control
Vehicle
Inverter Control Command
Inverter & Electric Motor
Total Motive Force Aerodynamic Motor Aerodynamic
Speed
Copyright ? 2011 by John Wiley & Sons, Inc.
117 Solutions to Problems
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Copyright ? 2011 by John Wiley & Sons, Inc.