Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Control Systems Engineering 7th Edition Solutions Manual Nise, Exercises of Control Systems

Norman S. Nise instructors solution manual for all chapters in Control Systems Engineering

Typology: Exercises

2020/2021

Uploaded on 05/26/2021

humaira
humaira 🇨🇫

4.8

(126)

274 documents

1 / 649

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
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
5 x 2
π
rad = 1.59
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27
pf28
pf29
pf2a
pf2b
pf2c
pf2d
pf2e
pf2f
pf30
pf31
pf32
pf33
pf34
pf35
pf36
pf37
pf38
pf39
pf3a
pf3b
pf3c
pf3d
pf3e
pf3f
pf40
pf41
pf42
pf43
pf44
pf45
pf46
pf47
pf48
pf49
pf4a
pf4b
pf4c
pf4d
pf4e
pf4f
pf50
pf51
pf52
pf53
pf54
pf55
pf56
pf57
pf58
pf59
pf5a
pf5b
pf5c
pf5d
pf5e
pf5f
pf60
pf61
pf62
pf63
pf64

Partial preview of the text

Download Control Systems Engineering 7th Edition Solutions Manual Nise and more Exercises Control Systems in PDF only on Docsity!

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

5 x 2 π rad

= 1.

2 Chapter 1: Introduction

2.

Thermostat Amplifier andvalves Heater

Temperature difference

Voltage difference Fuelflow temperatureActual

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 voltage Gyro

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

4 Chapter 1: Introduction

8. a.

R

+V -V

Differential amplifier

Desired level

Power amplifier

Actuator

Valve

Float

Fluid input

Tank Drain

R

+V

-V

b. Desired level Amplifiers Actuatorand valve

Flow rate in Integrate

Actual level

Flow rate out

Potentiometer +

Drain

Potentiometer Float

voltage in

voltage out

Displacement

Solutions to Problems 5

9.

Desiredforce Transducer (^) Amplifier Valve Actuatorand load Tire

Load cell

  • Actualforce

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.

circuit^ Coil Solenoid coil & actuator

Coil current Force^ Armature& spool dynamics

Desired position (^) Transducer Depth

Coilvoltage

LVDT

13. a. L di dt

  • Ri = u(t)

Solutions to Problems 7

i = 292 29 e - t^ sin 29 t

c. i

t

15. a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

From which, C =^3553 and D =^1053.

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A +^3553 = 0. Therefore, A = -^3553. 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 M 2 + 6 M + 8 = M + 4 M + 2

Thus, the total solution is x = C e - 4 t^ + D e - 2 t^ + -^1865 cos 3 t - 651 sin 3 t

Solving for the arbitrary constants, x (0) = C + D − 18 65

Also, the derivative of the solution is

dxdt = - 653 cos 3 t +^5465 sin 3 t - 4 C e - 4 t (^) - 2 D e - 2 t

Solving for the arbitrary constants, x

. (0) − 3 65

− 4 C − 2 D = 0 , or C = − 3 10

and D = 15 26

.

The final solution is x = -^1865 cos 3 t - 651 sin 3 t - 103 e - 4 t^ +^1526 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 M 2 + 8 M + 25 = M + 4 + 3 i M + 4 - 3 i

Thus, the total solution is x =^25 + e - 4 t^ 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 dxdt = 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e - 4 t

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 dt

= (− A + B ) e −^ t^ − Btet^ − 10 e −^2 t^ + 1

Solving for the arbitrary constants, x

. (0) = B - 8 = 1. Therefore, B = 9. The final solution is

c. Assume a particular solution of x (^) p = Ct^2 + Dt + E

Substitute into the differential equation and obtain

Equating like coefficients, C =^14 , D = 0, and 2C + 4E = 0. From which, C =^14 , D = 0, and E = -^18.

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A -^18 = 1 Therefore, A =^98. 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.

Input transducer

Desired force Inputvoltage Controller Actuator Pantographdynamics Spring

Fup

Spring displacement Fout

Sensor

Solutions to Problems 13

Substituting into the differential equation yields, ddtδi + 2i 02 + 4i 0 δi - 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, 2i 02 = 5, or i 0 = 1.58. Substituting into the linearized

differential equation, ddtδi + 6.32δi = v(t). Converting to a transfer function, δV(s)i(s) = (^) s+6.32^1. Using the linearized i about i 0 , and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i^2 = 2(i 0 +δi) 2 = 2(i 02 +2i 0 δi) = 5+6.32δi. For excursions away from equilibrium, vr(t) - 5 = 6.32δi = δv (^) r(t).

Therefore, multiplying the transfer function by 6.32, yields, δVr

(s) 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 ) = e −^ st^ dt 0

∞ ∫ = −^

s

est 0

∞ = 1 s

b. F ( s ) = te −^ st^ dt 0

∞ ∫ =^

est s^2

(− st − 1) 0 ∞^ = −( st^ +^ 1) s^2 e st 0

14 Chapter 2: Modeling in the Frequency Domain

Using L'Hopital's Rule

F ( s ) t → ∞ = − s s^3 e st t →∞

= 0. Therefore, F ( s ) = 1 s^2

c. F ( s ) = sin ω t e −^ st^ dt

0

∞ ∫ =^

e −^ st

s^2 +ω 2

(− s sin ω t −ω cos ω t )

0

s^2 +ω 2

d. F ( s ) = cos ω t e −^ st^ dt

0

∞ ∫ =^

e −^ st

s^2 +ω 2

(− s cos ω t +ω sin ω t )

0

∞ = s s^2 +ω 2 2. a. Using the frequency shift theorem and the Laplace transform of sin ωt, F(s) = (^) (s+a)ω 2 +ω 2. b. Using the frequency shift theorem and the Laplace transform of cos ωt, F(s) = (^) (s+a)(s+a) 2 +ω 2.

c. Using the integration theorem, and successively integrating u(t) three times, (^) ⌡⌠dt = t; (^) ⌡⌠tdt = t

2 2 ;

⌠t^2 2 dt =

t^3 6 , the Laplace transform of t^3 u(t), F(s) =

6 s^4. 3. a. The Laplace transform of the differential equation, assuming zero initial conditions,

is, (s+7)X(s) = (^) s 2 5s+2 2. Solving for X(s) and expanding by partial fractions,

Or,

Taking the inverse Laplace transform, x(t) = -^3553 e-7t^ + (^3553 cos 2t +^1053 sin 2t).

b. The Laplace transform of the differential equation, assuming zero initial conditions, is,

(s^2 +6s+8)X(s) = 15 s 2 + 9

.

Solving for X(s) X(s) = 15 (s 2 + 9)(s 2 + 6s + 8) and expanding by partial fractions,

X(s) = − 3 65

6s + 1 9

s 2 + 9

s + 4

s + 2

16 Chapter 2: Modeling in the Frequency Domain

Therefore, x(t) =^98 cos2t + sin2t -^18 +^14 t^2. 5. Program: syms t f=5t^2cos(3t+45); pretty(f) F=laplace(f); F=simple(F); pretty(F) 'b' f=5texp(-2t)sin(4t+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+3s+7)(s+2)/[(s+3)(s+4)(s^2+2s+100)]; pretty(G) g=ilaplace(G); pretty(g) 'b' G=(s^3+4s^2+6s+5)/[(s+8)(s^2+8s+3)(s^2+5*s+7)]; pretty(G) g=ilaplace(G); pretty(g) Computer response: ans = a 2

Solutions to Problems 17

(s + 3 s + 7) (s + 2)

2 (s + 3) (s + 4) (s + 2 s + 100) 11 4681 1/2 1/

  • 7/103 exp(-3 t) + -- exp(-4 t) - ----- exp(-t) 11 sin(3 11 t) 54 61182 4807 1/
  • ---- exp(-t) cos(3 11 t) 5562

ans =

b 3 2 s + 4 s + 6 s + 5


2 2 (s + 8) (s + 8 s + 3) (s + 5 s + 7) 299 1367 1/

  • --- exp(-8 t) + ---- exp(-4 t) cosh(13 t) 93 417 4895 1/2 1/
  • ---- exp(-4 t) 13 sinh(13 t) 5421 232 1/2 1/
  • ----- exp(- 5/2 t) 3 sin(1/2 3 t) 12927 272 1/
  • ---- exp(- 5/2 t) cos(1/2 3 t) 4309

7. The Laplace transform of the differential equation, assuming zero initial conditions, is, (s^3 +3s^2 +5s+1)Y(s) = (s^3 +4s^2 +6s+8)X(s). Solving for the transfer function, Y ( s ) X ( s )

= s

(^3) + 4 s (^2) + 6 s + 8 s^3 + 3 s^2 + 5 s + 1

.

8. a. Cross multiplying, (s^2 +2s+7)X(s) = F(s). Taking the inverse Laplace transform, d

(^2) x dt^2

  • 2 dx dt

  • 7x = f(t). b. Cross multiplying after expanding the denominator, (s^2 +15s+56)X(s) = 10F(s). Taking the inverse Laplace transform, d

(^2) x dt^2

  • 15 dx dt

  • 56x =10f(t).

c. Cross multiplying, (s^3 +8s^2 +9s+15)X(s) = (s+2)F(s). Taking the inverse Laplace transform, d

(^3) x dt^3

  • 8 d

(^2) x dt^2

  • 9 dx dt

  • 15 x = df^ ( t ) dt

+2f(t).

9.

= s

(^5) + 2 s (^4) + 4 s (^3) + s (^2) + 3 s^6 + 7 s^5 + 3 s^4 + 2 s^3 + s^2 + 3

The transfer function is C ( s ). R ( s )

Solutions to Problems 19

Polynomial Transfer function: 5 s^3 + 565 s^2 + 16710 s + 140400


s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s 13. Program: 'Polynomial' Gtf=tf([1 25 20 15 42],[1 13 9 37 35 50]) 'Factored' Gzpk=zpk(Gtf) Computer response: ans = Polynomial Transfer function: s^4 + 25 s^3 + 20 s^2 + 15 s + 42


s^5 + 13 s^4 + 9 s^3 + 37 s^2 + 35 s + 50 ans = Factored Zero/pole/gain: (s+24.2) (s+1.35) (s^2 - 0.5462s + 1.286)


(s+12.5) (s^2 + 1.463s + 1.493) (s^2 - 0.964s + 2.679) 14. Program: numg=[-10 -60]; deng=[0 -40 -30 (roots([1 7 100]))' (roots([1 6 90]))']; [numg,deng]=zp2tf(numg',deng',1e4); Gtf=tf(numg,deng) G=zpk(Gtf) [r,p,k]=residue(numg,deng) Computer response: Transfer function: 10000 s^2 + 700000 s + 6e


s^7 + 83 s^6 + 2342 s^5 + 33070 s^4 + 3.735e005 s^3 + 2.106e006 s^

  • 1.08e007 s

Zero/pole/gain: 10000 (s+60) (s+10)


s (s+40) (s+30) (s^2 + 6s + 90) (s^2 + 7s + 100)

r =

-0.

2.0431 - 2.0385i 2.0431 + 2.0385i -2.3329 + 2.0690i -2.3329 - 2.0690i

p =

20 Chapter 2: Modeling in the Frequency Domain

-40. -30. -3.5000 + 9.3675i -3.5000 - 9.3675i -3.0000 + 9.0000i -3.0000 - 9.0000i 0 k = [] 15. Program: syms s '(a)' Ga=45[(s^2+37s+74)(s^3+28s^2+32s+16)]... /[(s+39)(s+47)(s^2+2s+100)(s^3+27s^2+18s+15)]; 'Ga symbolic' pretty(Ga) [numga,denga]=numden(Ga); numga=sym2poly(numga); denga=sym2poly(denga); 'Ga polynimial' Ga=tf(numga,denga) 'Ga factored' Ga=zpk(Ga) '(b)' Ga=56[(s+14)(s^3+49s^2+62s+53)]... /[(s^2+88s+33)(s^2+56s+77)(s^3+81s^2+76*s+65)]; 'Ga symbolic' pretty(Ga) [numga,denga]=numden(Ga); numga=sym2poly(numga); denga=sym2poly(denga); 'Ga polynimial' Ga=tf(numga,denga) 'Ga factored' Ga=zpk(Ga)

Computer response: ans =

(a)

ans =

Ga symbolic

2 3 2 (s + 37 s + 74) (s + 28 s + 32 s + 16) 45 ----------------------------------------------------------- 2 3 2 (s + 39) (s + 47) (s + 2 s + 100) (s + 27 s + 18 s + 15)

ans =

Ga polynimial

Transfer function:

45 s^5 + 2925 s^4 + 51390 s^3 + 147240 s^2 + 133200 s + 53280