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. 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 1. Five turns yields 50 v. Therefore K =
50 volts = 1.59 5 x 2π rad
2 Chapter 1: Introduction
2.
Desired temperature
Voltage difference
Temperature difference
Actual temperature
Fuel flow
+
Amplifier and valves
Thermostat
Heater
-
3.
-
Roll angle
Roll rate
Aileron position control
+
Pilot controls
Aileron position
Error voltage
Input voltage
Desired roll angle
Aircraft dynamics
Integrate
Gyro
Gyro voltage
4. Input voltage
Desired speed
Speed Error voltage
+
transducer
Amplifier
Motor and drive system
Dancer position sensor
Dancer dynamics
Voltage proportional to actual speed
Actual speed
5. Input voltage
Desired power Transducer
Rod position
Power Error voltage
+
Amplifier
-
Voltage proportional to actual power
Motor and drive system
Sensor & transducer
Actual power Reactor
Solutions to Problems 3
6.
Desired student population
Desired student rate
Population error
+
Graduating and drop-out rate
Administration
Actual student rate +
Actual student population
Net rate of influx Integrate
Admissions
-
7. Voltage proportional to desired volume
Desired volume
+
Transducer
Volume error
Voltage representing actual volume Volume control circuit
Radio
Effective volume
+ -
Transducer
Speed
Voltage proportional to speed
Actual volume
4 Chapter 1: Introduction
8. a. Fluid input
Valve Actuator
Power amplifier +V Differential amplifier + -
R
Desired level
-V
+V
R Float
-V
Tank
Drain
b. Desired level Potentiometer
voltage in + Amplifiers -
Actuator and valve
Actual level
Flow rate in + Integrate -
Drain Flow rate out
voltage out
Displacement Potentiometer
Float
Solutions to Problems 5
9.
Desired force
Current
+ Transducer
Displacement
Amplifier
Actual force
Displacement Actuator and load
Valve
Tire
-
Load cell
10.
Commanded blood pressure +
Actual blood pressure
Isoflurane concentration Vaporizer
Patient
-
11.
Desired depth +
Controller & motor
-
Force
Feed rate Grinder
Depth Integrator
12.
Coil voltage +
Desired position
Coil circuit
Transducer
Coil current
Solenoid coil & actuator
-
LVDT
13. a. L
di + Ri = u(t) dt
Force
Armature & spool dynamics
Depth
6 Chapter 1: Introduction
b. Assume a steady-state solution iss = B. Substituting this into the differential equation yields RB = 1,
1 R . The characteristic equation is LM + R = 0, from which M = - . Thus, the total R L 1 1 solution is i(t) = Ae-(R/L)t + . Solving for the arbitrary constants, i(0) = A + = 0. Thus, A = R R 1 1 1 -(R/L)t 1 −( R / L)t . The final solution is i(t) = -= (1 − e e ). R R R R from which B =
c.
14.
di 1 idt + vC (0) = v(t) + dt C ∫ d 2i di b. Differentiating and substituting values, + 30i = 0 2 +2 dt dt
a. Writing the loop equation, Ri + L
Writing the characteristic equation and factoring, 2
M + 2 M + 30 = M + 1 + 29 i M + 1 -
29 i .
The general form of the solution and its derivative is -t
i = e cos
29 t A + B sin
29 t e
-t
di = - A + 29 B e -t cos 29 t - 29 A + B e- t sin dt v (0) 1 di Using i(0) = 0; (0) = L = =2 L L dt i 0 = A =0
di (0) = − A + 29 B =2 dt 2 . Thus, A = 0 and B = 29 The solution is
29 t
Solutions to Problems 7
i=
2 29
-t
29 t
29 e sin
c.
i
t 15. a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
35 From which, C = 53
10 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 xp = Asin3t + Bcos3t
8 Chapter 1: Introduction
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 2
M +6 M+ 8 = M+ 4 M+ 2 Thus, the total solution is
18 1 cos 3 t sin 3 t 65 65 18 Solving for the arbitrary constants, x(0) = C + D − =0. 65 x =C e
-4t
+ De
-2t
+ -
Also, the derivative of the solution is
dx = - 3 cos 3 t + 54 sin 3 t -4t -2t - 4 C e -2 D e dt 65 65 . 15 3 3 . Solving for the arbitrary constants, x(0) − − 4C − 2D = 0 , or C = − and D =
10
65
26
The final solution is
x =-
18 3 - 4 t 15 - 2 t 1 e e + sin 3 t cos 3 t 26 65 10 65
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 2
M + 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
dx -4t dt = 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e
Solutions to Problems 9
. Solving for the arbitrary constants, x(0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution is
x (t ) =
2 2 ⎛8 − e −4t sin(3 t ) + cos(3t )⎞ ⎝ 15 ⎠ 5 5
16. a. Assume a particular solution of
Substitute into the differential equation and obtain
Equating like coefficients,
1 From which, C = - 5
1 and D = - 10 .
The characteristic polynomial is
Thus, the total solution is
11 1 Solving for the arbitrary constants, x(0) = A - 5 = 2. Therefore, A = . Also, the derivative of the 5
solution is
dx dt . 3 Solving for the arbitrary constants, x(0) = - A + B - 0.2 = -3. Therefore, B = − . The final solution
5
is
1 1 3 −t ⎛ 11 x (t ) = − cos(2t ) − sin(2 t ) + e cos( t ) − sin(t )⎞ ⎝5 ⎠ 5 10 5 b. Assume a particular solution of xp = Ce-2t + Dt + E Substitute into the differential equation and obtain
10 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
Solutions to Problems 11
17. Spring displacement Desired force
Input voltage + Input transducer
F up Controller
Actuator
-
Sensor
Pantograph dynamics
F out Spring
T W O
Modeling in the Frequency Domain SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transfer Functions Finding each transfer function: Vi(s) 10 = ; θi(s) π Vp(s) Pre-Amp: V (s) = K; i Ea(s) 150 Power Amp: V (s) = s+150 p Pot:
2
( ) = 0.25 50 2 Dm =0.01 + 3(250 ) = 0.13
50 Motor: Jm = 0.05 + 5 250
Kt 1 Ra = 5 K tK b 1 Ra = 5 θm(s) Therefore: E (s) = a
Kt RaJm
0.8 = s(s+1.32) K K 1 t b s(s+J (Dm+ R )) m a
θo(s) 1 θm(s) 0.16 And: E (s) = 5 E (s) = s(s+1.32) a a
Transfer Function of a Nonlinear Electrical Network Writing the differential equation, 2
2 0
d(i 0 + δi) 2 + 2(i0 + δi) − 5 = v(t) . Linearizing i2 about i0, dt 2
2 0
(i + δi) - i = 2i ⎮ δi = 2i .δi. Thus, (i + δi) = i + 2i δi. 0
i=i
0
0
0
0
Solutions to Problems 13
dδi Substituting into the differential equation yields, dt + 2i02 + 4i0δi - 5 = v(t). But, the resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since the voltage across the inductor is zero at dc. Hence, 2i02 = 5, or i0 = 1.58. Substituting into the linearized dδi δi(s) 1 differential equation, dt + 6.32δi = v(t). Converting to a transfer function, V(s) = s+6.32 . Using the linearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i2 = 2(i0+δi)2 = 2(i02+2i0δi) = 5+6.32δi. For excursions away from equilibrium, vr(t) - 5 = 6.32δi = δvr(t). δVr(s) 6.32 Therefore, multiplying the transfer function by 6.32, yields, V(s) = s+6.32 as the transfer function about v(t) = 0.
ANSWERS TO REVIEW QUESTIONS 1. Transfer function 2. Linear time-invariant 3. Laplace 4. G(s) = C(s)/R(s), where c(t) is the output and r(t) is the input. 5. Initial conditions are zero 6. Equations of motion 7. Free body diagram 8. There are direct analogies between the electrical variables and components and the mechanical variables
and components. 9. Mechanical advantage for rotating systems 10. Armature inertia, armature damping, load inertia, load damping 11. Multiply the transfer function by the gear ratio relating armature position to load position. 12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select the
equilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform of the linearized differential equation, (6) Find the transfer function.
SOLUTIONS TO PROBLEMS 1.
∞
a. F(s) =
∞
1 1 −st − st ∫0 e dt = − s e 0 = s ∞
e −st − (st + 1) ∞ ∞ b. F(s) = ∫ te dt = 2 ( − st − 1) 0 = s s2 e st 0 0 − st
14 Chapter 2: Modeling in the Frequency Domain
Using L'Hopital's Rule
F(s) t → ∞ =
1 −s = 0. Therefore, F(s) = 2 . 3 st s s e t →∞
∞
∞
e − st ω c. F(s) = ∫ sin ωt e dt = 2 = 2 2 (−s sin ωt − ω cos ωt) 2 s +ω s +ω 0 0 − st
∞
∞
s e − st d. F(s) = ∫ cos ωt e dt = 2 = 2 2 (−s cosω t + ω sin ωt) s +ω2 s +ω 0 0 − st
2.
ω . (s+a)2+ω2 (s+a) b. Using the frequency shift theorem and the Laplace transform of cos ωt, F(s) = . (s+a)2+ω2 a. Using the frequency shift theorem and the Laplace transform of sin ωt, F(s) =
t2 c. Using the integration theorem, and successively integrating u(t) three times, ⌠ dt = t; tdt = ⌠ ⌡ ⌡ 2 ; 2
3
⌠t dt = t , the Laplace transform of t3u(t), F(s) = 6 . 6 ⌡2 s4 3. a. The Laplace transform of the differential equation, assuming zero initial conditions,
5s is, (s+7)X(s) = 2 2 . Solving for X(s) and expanding by partial fractions, s +2
Or,
35 10 35 Taking the inverse Laplace transform, x(t) = - 53 e-7t + (53 cos 2t + 53 sin 2t). b. The Laplace transform of the differential equation, assuming zero initial conditions, is,
(s2+6s+8)X(s) =
15 . s +9 2
Solving for X(s)
X(s) =
15 (s + 9)(s2 + 6s + 8) 2
and expanding by partial fractions,
3 X(s) = − 65
1 9 15 1 3 1 9 + − 2 10 s + 4 26 s + 2 s +9
6s +
Solutions to Problems 15
Taking the inverse Laplace transform,
x(t) = −
18 1 3 15 cos(3t) − sin(3t) − e −4t + e −2t 65 65 10 26
c. The Laplace transform of the differential equation is, assuming zero initial conditions,
(s2+8s+25)x(s) =
10 . Solving for X(s) s
X s =
10 s s 2 + 8 s + 25
and expanding by partial fractions,
2 1 2 X s = 5 s 5
4 9 2 s+4 +9
1 s+4 +
9
Taking the inverse Laplace transform,
x(t) =
2 2 ⎛8 − e −4 t sin(3t) + cos(3t )⎞ ⎝ ⎠ 15 5 5
4.
2 a. Taking the Laplace transform with initial conditions, s2X(s)-2s+3+2sX(s)-4+2X(s) = 2 2 . s +2 Solving for X(s),
2s 3 + s2 + 8s + 6 . X(s) = 2 2 (s + 4)(s + 2s + 2)
Expanding by partial fractions
⎛1 X(s) = − ⎝ ⎞⎠ 5
1 3 4 11(s + 1) − 1 ⎛ 1⎞ 4 1 +⎝ ⎠ 2 2 s +4 5 (s + 1) + 1
s+
Therefore, x(t) = -0.2 cos2t - 0.1 sin2t +e-t (2.2 cost - 0.6 sint). 5 1 b. Taking the Laplace transform with initial conditions, s2X(s)-2s-1+2sX(s)-4+X(s) = s+2 + 2 . s Solving for X(s),
Therefore, x(t) = 5e-2t - e-t + 9te-t - 2 + t. 2 c. Taking the Laplace transform with initial conditions, s2X(s)-s-2+4X(s) = 3 . Solving for X(s), s
16 Chapter 2: Modeling in the Frequency Domain
1 1 9 Therefore, x(t) = 8 cos2t + sin2t - 8 + 4 t2. 5. Program: syms t f=5*t^2*cos(3*t+45); pretty(f) F=laplace(f); F=simple(F); pretty(F) 'b' f=5*t*exp(-2*t)*sin(4*t+60); pretty(f) F=laplace(f); F=simple(F); pretty(F)
Computer response: ans = a 2 5 t cos(3 t + 45) 3 2 s cos(45) - 27 s cos(45) - 9 s sin(45) + 27 sin(45) 10 ----------------------------------------------------2 3 (s + 9) ans = b 5 t exp(-2 t) sin(4 t + 60) sin(60) ((s + 2) sin(60) + 4 cos(60)) (s + 2) -5 ------------- + 10 ------------------------------------2 2 2 (s + 2) + 16 ((s + 2) + 16)
6. Program: syms s 'a' G=(s^2+3*s+7)*(s+2)/[(s+3)*(s+4)*(s^2+2*s+100)]; pretty(G) g=ilaplace(G); pretty(g) 'b' G=(s^3+4*s^2+6*s+5)/[(s+8)*(s^2+8*s+3)*(s^2+5*s+7)]; pretty(G) g=ilaplace(G); pretty(g)
Computer response: ans = a 2