Control systems engineering 6th solultions nise

Thermostat Amplifier and valves Heater Temperature difference Voltage difference Fuel flow Actual temperature + -Pilot controls Aileron position control Error voltage Aileron position

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MANUAL

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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 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, state-space, differential equations

16 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

2

Thermostat Amplifier and valves Heater

Temperature difference

Voltage difference Fuel

flow

Actual temperature

+

-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

+ -

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

Desired student rate Admissions

Actual student rate +-

Graduating and drop-out rate Net rate

of influx

Integrate

Actual student population

Voltage proportional

to desired volume

Volume error

Radio

Voltage representing

volume

+

-Transducer

-Speed

Voltage proportional

to speed Effective

volume

8

a

R

+V -V

Differential amplifier

Desired level +

-Power amplifier

Actuator Valve

Float Fluid input

Drain Tank

R

+V -V

Integrate

Actual level

Flow rate out

+

-Drain

Float Potentiometer

-voltage in

voltage out

Displacement

9

Desired

force

Transducer Amplifier Valve Actuator

and load Tire

Load cell

Actual force +

-Isoflurane concentration

LVDT

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

+ Desired

Nervous system electrical impulses

Nervous system electrical impulses

Retina’s Light Intensity

External Light

c

19

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 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

65 = 0 Also, the derivative of the solution is

Solving for the arbitrary constants, x

.(0) − 3

26 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

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

dx dt

Solving for the arbitrary constants, x

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

a Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A - 15 = 2 Therefore, A = 11

5 Also, the derivative of the

solution is

dx dt

Solving for the arbitrary constants, x

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

5 The final solution

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 −10e−2t +1

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

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

Spring displacement

Fout

Sensor

22

Amount of HIV viruses

Motor

Aerodynamic

Climbing &

Rolling Resistances

Aerodynamic

Speed

Inverter Control Command

Controlled Voltage

Inverter Desired

b

ECU

Accelerator Displacement

Vehicle Accelerator,

Aerodynamic

Climbing &

Rolling Resistances

Aerodynamic

Speed

+

+_

Aerodynamic

Climbing &

Rolling Resistances

Inverter Control Command

Inverter

&

ElectricMotor

T W O

Modeling in the Frequency Domain SOLUTIONS TO CASE STUDIES CHALLENGES

Antenna Control: Transfer Functions

Finding each transfer function:

m(Dm+KRtKb

a )) = s(s+1.32)0.8

And: θo(s)

Ea(s) =

15

θm(s)

Ea(s) =

0.16s(s+1.32)

Transfer Function of a Nonlinear Electrical Network

Writing the differential equation, d(i0+ δi)

dt + 2(i0 + δi)2 − 5 = v(t) Linearizing i2 about i0,

Substituting into the differential equation yields, dδi

dt + 2i0 + 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, 2i0 = 5, or i0 = 1.58 Substituting into the linearized

differential equation, dδi

dt + 6.32δi = v(t) Converting to a transfer function, δi(s)

V(s) =

1s+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(i0 +2i0δi) = 5+6.32δi For excursions away from equilibrium, vr(t) - 5 = 6.32δi = δvr(t)

Therefore, multiplying the transfer function by 6.32, yields, δVr(s)

V(s) =

6.32s+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

Using L'Hopital's Rule

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,

1

s + 4 +

15 26 1

s + 2

Taking the inverse Laplace transform,

5203 exp(-t) | cos(3 11 t) - - |

20 exp(-3 t) 7 exp(-4 t) \ 57233 / - - - + -

103 54 5562

b Cross multiplying after expanding the denominator, (s2+21s+110)X(s) = 15F(s)

Taking the inverse Laplace transform, d2

15

Program:

syms s '(a)' Ga=45*[(s^2+37*s+74)*(s^3+28*s^2+32*s+16)]

/[(s+39)*(s+47)*(s^2+2*s+100)*(s^3+27*s^2+18*s+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+49*s^2+62*s+53)]

/[(s^2+88*s+33)*(s^2+56*s+77)*(s^3+81*s^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)

Ga factored

Zero/pole/gain:

17

a

Writing mesh equations

(2s+2)I1(s) –2 I2(s) = Vi(s) -2I1(s) + (2s+4)I2(s) = 0 But from the second equation, I1(s) = (s+2)I2(s) Substituting this in the first equation yields,

(2s+2)(s+2)I2(s) –2 I2(s) = Vi(s)

or

I2(s)/Vi(s) = 1/(2s2 + 4s + 2) But, VL(s) = sI2(s) Therefore, VL(s)/Vi(s) = s/(2s2 + 4s + 2)

2

( ) ( )

(2 1) (2 4 1)

s

V s s

s

sV s s

b Transforming the network yields,

Writing the loop equations,

1 2

'G(s) via Mesh Equations' %Display label

pretty(G) %Pretty print G(s)

G1=collect(G1); %Simplify G1(s)

'G(s) via Nodal Equations' %Display label

Computer response:

2 10 1 ( ) 10

10 ( )

4 10 1 ( ) 1.1 10

0.25 10

4 10

10 27.5 ( ) 6 10

s

s

Z s x

x x

23

Writing the equations of motion, where x2(t) is the displacement of the right member of springr,

(5s2+4s+5)X1(s) -5X2(s) = 0 -5X1(s) +5X2(s) = F(s)

Adding the equations,

2

10F(s)

Solving for X3(s),

θ4(s) = rotation of right - hand side of K2

the equations of motion are

2s(s + 1)

32

Reflecting impedances to θ3,

{[1+2(3)2+16(14)2]s2 + [2+1(3)2+32(14)2]s + 64(14)2} θ2(s) = T(s)(3) Thus,

θ2(s)T(s) =

320s2+13s+4

1300s2+ 4000s + 550

35

Reflecting impedances and applied torque to respective sides of the spring yields the following

Writing the equations of motion,

2θ2(s) -2 θ3(s) = 4.231T(s) -2θ2(s) + (0.955s+2)θ3(s) = 0 Solving for θ3(s),

38

Reflect impedances to the left of J5 to J5 and obtain the following equivalent circuit:

Writing the equations of motion,

[Jeqs2+(Deq+D)s+(K2+Keq)]θ5(s) -[Ds+K2]θ6(s) = 0 -[K2+Ds]θ5(s) + [J6s2+2Ds+K2]θ6(s) = T(s)

From the first equation, θ6(s)

Write the equations of motion from the translational and rotational freebody diagrams,

(Ms2+2fv s+K2)X(s) -fvrsθ(s) = F(s) -fvrsX(s) +(Js2+fvr2s)θ(s) = 0 Solve for θ(s),

where F(s) is the opposing force on J2 due to the translational member and r is the radius of J2 But,

for the translational member,

F(s) = (Ms2+fvs+K2)X(s) = (Ms2+fvs+K2)rθ(s) Substituting F(s) back into the second equation of motion,

(J1s2+K1)θ1(s) - K1θ2(s) = T(s)

-K1θ1(s) + [(J2 + Mr2)s2+(D3 + fvr2)s+(K1 + K2r2)]θ2(s) -D3sθ3(s) = 0

-D3sθ2(s) + (J2s2+D3s)θ3(s) = 0

Notice that the translational components were reflected as equivalent rotational components by the

square of the radius Solving for θ2(s), θ2( s) = K1( J3s2 + D3s)T(s)

Δ , where Δ is the

determinant formed from the coefficients of the three equations of motion Hence,

θ2(s) T(s) = K1(J3s2+ D3s)

θL(s)

Ea(s)=

2 21 38

π 2 π

1 60

θm(s)

Ea(s)=

1 3

For the rotating load, assuming all inertia and damping has been reflected to the load,

(JeqLs2+DeqLs)θL(s) + F(s)r = Teq(s), where F(s) is the force from the translational system, r=2 is the radius of the rotational member, JeqL is the equivalent inertia at the load of the rotational load and the armature, and DeqL is the equivalent damping at the load of the rotational load and the armature Since JeqL = 1(2)2 +1 = 5, and DeqL = 1(2)2 +1 = 5, the equation of motion becomes, (5s2+5s)θL(s) + F(s)r = Teq(s) For the translational system, (s2+s)X(s) = F(s) Since X(s) = 2θL(s), F(s) =

(s2+s)2θL(s) Substituting F(s) into the rotational equation, (9s2+9s) θL(s) = Teq(s) Thus, the

equivalent inertia at the load is 9, and the equivalent damping at the load is 9 Reflecting these back

to the armature, yields an equivalent inertia of 94 and an equivalent damping of 94 Finally, RKt

a = 1;

Kb = 1 Hence, θm(s)

Ea(s) =

49s(s+49(94+1))

=

49s(s+139)

Since θL(s) = 12 θm(s), θL(s)

Ea(s) =

29s(s+139) But

For the series analogy, treating the equations of motion as mesh equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries For the parallel analogy, treating the equations of motion as nodal equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries

48

Writing the equations of motion in terms of angular velocity, Ω(s) yields