NISE Control Systems Engineering 6th Ed Solutions, Notas de estudo de Engenharia Elétrica

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1-17 Solutions to Problems

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ISBN 13 978-0470-54756-

1-2 Chapter 1: Introduction

Thermostat

Amplifier and valves

Heater

Temperature difference

Voltage difference Fuel flow

Actual temperature

Desired temperature

Desired roll angle

Input voltage

Pilot controls

Aileron position control

Error voltage

Aileron position

Aircraft dynamics

Roll rate

Integrate

Roll angle

Gyro Gyro voltage

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

1-3 Solutions to Problems

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

Desired student population (^) +

Administration

Population error

Desired student rate

Admissions

Actual student rate (^) +

Graduating and drop-out rate Net rate of influx Integrate

Actual student population

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

1-5 Solutions to Problems

Desired force Transducer (^) Amplifier Valve Actuator and load Tire

Load cell

Actual

• force

Current (^) Displacement Displacement

Commanded

blood pressure

Vaporizer Patient

Actual

blood

+ pressure

Isoflurane

concentration

Controller & motor

Grinder

Force Feed rate Integrator

Desired depth (^) Depth

Coil circuit Solenoid coil & actuator

Coil current Force Armature & spool dynamics

Desired position (^) Transducer Depth

Coil voltage

LVDT

1-6 Chapter 1: Introduction

a.

b.

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.

DesiredLight + Intensity

Brain Internal eye muscles

Retina + Optical

Retina’s Light Intensity

Nervous system electrical impulses

Nervous system electrical impulses

Desired Light Intensity

Brain Internal eye muscles

Retina + Optical Nerves

Retina’s Light Intensity

External Light

1-8 Chapter 1: Introduction

solution is i(t) = Ae -(R/L)t +

R

. Solving for the arbitrary constants, i(0) = A +

R

= 0. Thus, A =

R

. The final solution is i(t) =

R

R

e -(R/L)t =

R

(1 − e

−( R / L ) t

c.

a. Writing the loop equation, Ri^ +^ L

di

dt

C

∫ idt^ +^ vC^ (0)=^ v ( t )

b. Differentiating and substituting values,

2

d i di

i

dt dt

Writing the characteristic equation and factoring, 2

M + 2 M + 25 = ( M + 1 + 24 )( i M + 1 − 24 ) i.

The general form of the solution and its derivative is

cos( 24 ) sin( 24 )

t t

i Ae t Be t

− −

( 24 ) cos( 24 ) ( 24 ) sin( 24 )

di t t

A B e t A B e t

dt

− −

Using

(0) 0; (0) L 1

di v

i

dt L L

i 0 = A =

di

A B

dt

Thus, A = 0 and

B =.

The solution is

sin( 24 )

i = e −^ t t

1-9 Solutions to Problems

c.

a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

From which, C =

53 and D =

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A +

53 = 0. Therefore, A = -

53. The final solution is

b. Assume a particular solution of

1-11 Solutions to Problems

= 3 B -4 C cos 3 t - 4 B + 3 C sin 3 t e

dx - 4 t

dt

Solving for the arbitrary constants, x

(0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution is

x ( t ) =

− e

− 4 t^8

sin(3 t ) +

cos(3 t )

a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

From which, C = -

5 and D = -

The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A -

5 = 2. Therefore, A =^

. Also, the derivative of the

solution is

dx

dt

Solving for the arbitrary constants, x

(0) = - A + B - 0.2 = -3. Therefore, B = −

. The final solution

is

x ( t ) = −

cos(2 t ) −

sin(2 t ) + e

− t^11

cos( t ) −

sin( t )

b. Assume a particular solution of

x (^) p = Ce-2t^ + Dt + E

Substitute into the differential equation and obtain

1-12 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

− Bte

− t

− 10 e

− 2 t

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 =

4 , D = 0, and 2C + 4E = 0.

From which, C =

4 , D = 0, and E = -

The characteristic polynomial is

Thus, the total solution is

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

1-14 Chapter 1: Introduction

a.

Speed

Actual

Motive ECU Force

Vehicle

Electric

Motor

Aerodynamic

Climbing &

Rolling

Resistances

Aerodynamic

Speed

Inverter

Control

Command

Controlled

Voltage

Inverter

Desired

1-15 Solutions to Problems

b.

Desired (^) Speed Motive Actual ECU

Accelerator Displacement

Accelerator, Vehicle

Aerodynamic

Climbing &

Rolling

Resistances

Aerodynamic

Speed

_

T W O

Modeling in the

Frequency Domain

SOLUTIONS TO CASE STUDIES CHALLENGES

Antenna Control: Transfer Functions

Finding each transfer function:

Pot:

V (^) i (s) θi (s) =

Pre-Amp:

V (^) p(s) V (^) i (s) = K;

Power Amp:

Ea (s) V (^) p(s) =^

s+

Motor: Jm = 0.05 + 5 (

250 )^

D m =0.01 + 3 (

250 )^

K (^) t Ra =

K (^) t K (^) b Ra =

Therefore:

θm(s) E (^) a (s) =

K (^) t Ra J (^) m

s(s+

Jm(D^ m+

K (^) t K (^) b Ra ))

s(s+1.32)

And:

θo(s) Ea (s) =

θm(s) E (^) a (s) =^

s(s+1.32)

Transfer Function of a Nonlinear Electrical Network

Writing the differential equation,

d(i 0 + δi)

dt

• 2(i 0 +δi) 2

− 5 = v(t). Linearizing i^2 about i0,

(i 0 +δi)

2

• i 0

2

= 2i ⎮

i=i 0

δi = 2i 0

δ. (^) i.Thus, (i 0 +δi)

2 = i 0

2

• 2i 0

δi.

Chapter 2: Modeling in the Frequency Domain 2-

Substituting into the differential equation yields,

dδi dt + 2i^0

(^2) + 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, dδi dt + 6.32δi = v(t). Converting to a transfer function,^

δi(s) 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 v (^) r (t) = 2i^2 =

2(i0+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,

δV (^) r (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

e

− st 0

∞

s

b. F ( s ) = te

− st

dt

0

∞

∫ =^

e − st

s

2 (− st^ −^ 1)^0

∞

−( st + 1)

s

2

e

st 0

∞