Cơ sở tự động: solutions manual feedback control of dynamic systems franklin 5th edition

The equilibrium value of x including the effect of gravity is at x = −mkgand y represents the motion of the mass about that equilibrium point.An alternate solution method, which is applic

Dynamic Models

Problems and Solutions for Section 2.1

1 Write the differential equations for the mechanical systems shown in Fig 2.38.Solution:

The key is to draw the Free Body Diagram (FBD) in order to keep thesigns right For (a), to identify the direction of the spring forces on theobject, let x2 = 0 and Þxed and increase x1 from 0 Then the k1 springwill be stretched producing its spring force to the left and the k2 springwill be compressed producing its spring force to the left also You can usethe same technique on the damper forces and the other mass

(a)

m1x¨1 = −k1x1− b1úx1− k2(x1− x2)

m2x¨2 = −k2(x2− x1) − k3(x2− y) − b2úx2

Figure 2.38: Mechanical systems

3ml2 Assume θ is small enough that sin θ ∼= θ.)

Solution:

Let’s use Eq (2.14)

M = Iα,

Moment about point O.

As we assumed θ is small,

¨

θ +3g2lθ = 0The frequency only depends on the length of the rod

ω2=3g2l

s2l

2π2 = 1.49 m

<Notes>

Figure 2.39: Double pendulum

(a) Compare the formula for the period, T = 2πq

2l 3g with the well knownformula for the period of a point mass hanging with a string withlength l T = 2πq

3 Write the equations of motion for the double-pendulum system shown inFig 2.39 Assume the displacement angles of the pendulums are smallenough to ensure that the spring is always horizontal The pendulumrods are taken to be massless, of length l, and the springs are attached3/4 of the way down

Solution:

3 G

4

3 G

l

l

4 3

If we write the moment equilibrium about the pivot point of the left dulem from the free body diagram,

pen-M = −mgl sin θ1− k34l (sin θ1− sin θ2) cos θ13

4l = ml

2¨θ1

ml2¨θ1+ mgl sin θ1+ 9

16kl

2cos θ1(sin θ1− sin θ2) = 0

Similary we can write the equation of motion for the right pendulem

θ1, sin θ2≈ θ2, cos θ1≈ 1, and cos θ2≈ 1 Finally the linearized equations

of motion becomes,

ml¨θ1+ mgθ1+ 9

16kl (θ1− θ2) = 0ml¨θ2+ mgθ2+ 9

16kl (θ2− θ1) = 0Or

θ1+g

lθ1+

916

k

4 Write the equations of motion for a body of mass M suspended from aÞxed point by a spring with a constant k Carefully deÞne where thebody’s displacement is zero

Solution:

Some care needs to be taken when the spring is suspended vertically inthe presence of the gravity We deÞne x = 0 to be when the spring isunstretched with no mass attached as in (a) The static situation in (b)results from a balance between the gravity force and the spring

From the free body diagram in (b), the dynamic equation results

so if we replace x using y = x +mkg,

The equilibrium value of x including the effect of gravity is at x = −mkgand y represents the motion of the mass about that equilibrium point.

An alternate solution method, which is applicable for any probleminvolving vertical spring motion, is to deÞne the motion to be with respect

to the static equilibrium point of the springs including the effect of gravity,

would deÞne y to be the motion with respect to the equilibrium point,then the FBD in (c) would result directly in

m¨y = −ky

5 For the car suspension discussed in Example 2.2,

(a) write the equations of motion (Eqs (2.10) and (2.11)) in state-variableform Use the state vector x = [ x úx y úy ]T

(b) Plot the position of the car and the wheel after the car hits a “unit

So, for the given sate vector of x = [ x úx y úy ]T, the state-spaceform will be,

¨y

error What passengers feel is the position of the car Some generalrequirements for the smooth ride will be, slow response with smallovershoot and oscillation.

0 0.5 1 1.5 2 0

0.5 1

1.5 Res pons e with b = 2000.0

W heel Car

0 0.5 1 1.5 2 0

0.5 1

1.5 Res pons e with b = 4000.0

W heel Car

From the Þgures, b ≈ 3000 would be acceptable There is too muchovershoot for lower values, and the system gets too fast (and harsh)for larger values

% Problem 2.5 bclear all, close allm1 = 10;

in opposite directions on the wheel axle and the car body

(a) Modify the equations of motion in Example 2.2 to include such trol inputs

con-(b) Is the resulting system linear?

(c) Is it possible to use the forcer, u2,to completely replace the springsand shock absorber? Is this a good idea?

Solution:

(a) The FBD shows the addition of the variable force, u2, and shows b

as in the FBD of Fig 2.5, however, here b is a function of the controlvariable, u1 The forces below are drawn in the direction that wouldresult from a positive displacement of x

m1¨x = b (u1) ( úy − úx) + ks(y − x) − kw(x − r) − u2

m2y¨ = −ks(y − x) − b (u1) ( úy − úx) + u2

(b) The system is linear with respect to u2 because it is additive But

b is not constant so the system is non-linear with respect to u1 cause the control essentially multiplies a state element So if we addcontrollable damping, the system becomes non-linear

be-(c) It is technically possible However, it would take very high forcesand thus a lot of power and is therefore not done It is a much bet-ter solution to modulate the damping coefficient by changing oriÞcesizes in the shock absorber and/or by changing the spring forces byincreasing or decreasing the pressure in air springs These featuresare now available on some cars where the driver chooses between

a soft or stiff ride

7 Modify the equation of motion for the cruise control in Example 2.1,Eq(2.4), so that it has a control law; that is, let

error, Þnd a value of K that you think would result in a control system inwhich the actual speed converges as quickly as possible to the referencespeed with no objectional behavior

Solution:

-The state-variable form of the equations is,

Km

Km

eliminate this error in chapter 4 using integral control, which is contained

in all cruise control systems in use today For this problem, a reasonablecompromise between speed of response and steady state errors would be

K = 1000, where it responds is 5 seconds and the steady state error is 5%

% Problem 2.7clear all, close all

Problems and Solutions for Section 2.2

8 In many mechanical positioning systems there is ßexibility between one

where there is ßexibility of the solar panels Figure 2.40 depicts such asituation, where a force u is applied to the mass M and another mass

m is connected to it The coupling between the objects is often modeled

by a spring constant k with a damping coefficient b, although the actualsituation is usually much more complicated than this

(a) Write the equations of motion governing this system, identify priate state variables, and express these equations in state-variableform

appro-(b) Find the transfer function between the control input, u, and theoutput, y

Solution:

(a) The FBD for the system is

Figure 2.40: Schematic of a system with ßexibility

which results in the equations

m¨x = −k (x − y) − b ( úx − úy)

M ¨y = u + k (x − y) + b ( úx − úy)Let the state-space vector x = [ x úx y úy ]T

k M b

¸

=

·0U

¸

From Cramer’s Rule,

Y

ms2+ bs + k(ms2+ bs + k) (M s2+ bs + k) − (bs + k)2

2+ bs + k

9 For the inverted pendulum, Eqs (2.34),

(a) Try to put the equations of motion into state-variable form using the

state vector x = [ θ úθ x úx ]T Why is it not possible?

(b) Write the equations in the “descriptor” form

E ˙x= F0x+ G0u,and deÞne values for E, F0, and G0 (note that E is a 4 × 4 matrix)

Then show how you would compute F and G for the standard variable description of the equations of motion.]

¨x

10 The longitudinal linearized equations of motion of a Boeing 747 are given

in Eq (9.28) Using MATLAB or other computer aid:

(a) Determine the response of the altitude h for a 2-sec pulse of the vator with a magnitude of 2◦ Note that, since Eq (9.28) represents

ele-a set of lineele-arized equele-ations, the stele-ate vele-ariele-ables ele-actuele-ally represent

example, h represents the amount the altitude of the aircraft differsfrom 20,000 ft

(b) Consider using the feedback law

where the elevator input angle is the sum of a term proportional

to the error in altitude h plus an external input (a disturbance or

elevator causes a negative change in altitude, so that the proposedproportional feedback law has the logical sign to anticipate a stablesystem provided Kh > 0 By trial and error, try to Þnd a value forthe feedback gain Kh such that a 2◦ pulse of 2 sec on δe,ext yields amore stable altitude response

response, try modifying the feedback law to include information onpitch rate q :

Use trial and error to pick appropriate values for both Kh and Kq.Assume the same type of pulse input for δe,ext as in part (b).(d) Show that the further introduction of pitch-angle feedback, θ, suchthat

allows you to decrease the time it takes for the altitude to settle back

to its nominal value, as well as to decrease the value of Kh requiredfor a stable response Note that, although Kh = 0 produces stablealtitude behavior, we require Kh > 0 in order to guarantee that

h → 0 (so there will be no steady-state error)

Solution:

0 1 2 3 4 5 6 7 8 9 10 -350

-300 -250 -200 -150 -100 -50 0

@

+ +

altitude deviation of less than 50ft, plus a settling time of about 30

title(’ 2 deg pulse applied between t = 1 and t =3 sec ’);

Problems and Solutions for Section 2.3

11 A Þrst step toward a realistic model of an op amp is given by the equationsbelow and shown in Fig 2.41

Figure 2.41: Circuit for Problem 11.

Figure 2.42: Circuit for Problem 12

12 Show that the op amp connection shown in Fig 2.42 results in Vo= Vin

if the op amp is ideal Give the transfer function if the op amp has thenon-ideal transfer function of Problem 2.11

Solution:

Figure 2.43: Circuit for Problem 13.

V+6= V−instead,

Vout

Vin =

10 7 s+1

Figure 2.44: Op Amp circuit for Problem 14.

14 A common connection for a motor power ampliÞer is shown in Fig 2.44.The idea is to have the motor current follow the input voltage and theconnection is called a current ampliÞer Assume that the sense resistor,

transfer function from Vin to Ia

Solution:

[Note to Instructors: You might want to assign this problem with Rf = ∞,meaning no feedback from the output voltage.]

This expression shows that, in the steady state when s → 0, the current

is proportional to the input voltage

If fact, the current ampliÞer normally has no feedback form the output

Rin+ Rf for the circuit to remain stable.

Figure 2.45: Op Amp circuit for Problem 15.

Figure 2.46: Lead (a), lag (b), notch (c) circuits

Rfú

Vin− Vout

¶+ C

µ

−RR2inú

2

2 /

17 The very ßexible circuit shown in Fig 2.47 is called a biquad becauseits transfer function can be made to be the ratio of two second-order orquadratic polynomials By selecting different values for Ra, Rb, Rc, and

Rd the circuit can realise a low-pass, band-pass, high-pass, or band-reject(notch) Þlter

(a) Show that if Ra= R, and Rb= Rc= Rd= ∞, the transfer functionfrom Vinto Voutcan be written as the low-pass Þlter

ωn+ 1

ωn = 1, and ζ = 0.1, 0.5, and 1.0

Solution:

Before going in to the speciÞc problem, let’s Þnd the general form of thetransfer function for the circuit

Figure 2.47: Op-amp biquad

Vout with set of equations but for this problem, we will directly solve forthe values we want along with the Laplace Transform

From the Þrst three equations, slove for V1,V2

R2+ Cs

R2+ Cs

· 1

R 1

R 2 + Cs

¸ ·

R 1Vin0

R b −R1a

´1

R a

´1

R2C

R2R2

(b) Step response using MatLab

0 10 20 30 40 50 60 0

R b−R1a

´1

R 2

´

s +¡

−1 R

R 1 1

Problems and Solutions for Section 2.4

19 The torque constant of a motor is the ratio of torque to current and isoften given in ounce-inches per ampere (ounce-inches have dimensionforce-distance where an ounce is 1/16 of a pound.) The electric constant

of a motor is the ratio of back emf to speed and is often given in volts per

1000 rpm In consistent units the two constants are the same for a givenmotor

(a) Show that the units ounce-inches per ampere are proportional tovolts per 1000 rpm by reducing both to MKS (SI) units

(b) A certain motor has a back emf of 25 V at 1000 rpm What is itstorque constant in ounce-inches per ampere?

(c) What is the torque constant of the motor of part (b) in newton-metersper ampere?

Solution:

Before going into the problem, let’s review the units

• Some remarks on non SI units

1oz = 2.835 × 10−2kgOriginally ounce is a unit of mass, but like pounds, it is com-monly used as a unit of force If we translate it as force,

1oz(f) = 2.835 × 10−2 kgf = 2.835 × 10−2× 9.81 N = 0.2778 N

1 in = 2.540 × 10−2m

— RPM (Revolution per Minute)

π

30 rad / s

• Relation between SI units

V olts · Current(amps) = P ower = Energy(joules)/ sec

N ewton − meters/ sec

amps(a) Relation between torque constant and electric constant

20 A simpliÞed sketch of a computer tape drive is given in Fig 2.48

(a) Write the equations of motion in terms of the parameters listed below

K and B represent the spring constant and the damping of tapestretch, respectively, and ω1and ω2are angular velocities A positivecurrent applied to the DC motor will provide a torque on the capstan

in the clockwise direction as shown by the arrow Assume positive

Figure 2.48: Tape drive schematic

angular velocities of the two wheels are in the directions shown bythe arrows.

J1= 5 × 10−5 kg · m2, motor and capstan inertia

B1= 1 × 10−2 N · m · sec, motor damping

(c) Use the values in part (a) and use MATLAB to Þnd the response of

úx1 = r1ω1

J1ωú1 = −Br1( úx1− úx2) − B1ω1− Kr1(x1− x2) + Ktia

úx2 = r2ω2

J2ωú2 = −Br2( úx2− úx1) − B2ω2− Kr2(x2− x1)

By substituting for úx1and úx2in the 2nd and 4th equations, we arrive

at the state-variable form







úx1ú

ω1

úx2ú

(c) Note that this is the perturbation of the tape speed at the head from

its steady state

% Problem 2.20clear all, close all

% dataJ1 = 5e-5;

step(num,den);

21 Assume the driving force on the hanging crane of Fig 2.14 is provided by

a motor mounted on the cab with one of the support wheels connected

directly to the motor’s armature shaft The motor constants are Keand

Kt, and the circuit driving the motor has a resistance Raand negligible

inductance The wheel has a radius r Write the equations of motion

relating the applied motor voltage to the cab position and load angle

where x is the position of the cab, θ is the angle of the load, and u is the

applied force that will be produced by the motor Our task here is to Þnd

motor is

J1¨θm+ b1úθm= Tm= Ktiawhere the current is found from the motor circuit, which reduces to

Raia= Va− Keúθm

for the case where the inductance is negligible However, since the motor

is geared directly to the cab, θm and x are related kinematically by

x = rθmand we can neglect any extra inertia or damping from the motor itselfcompared to the inertia and damping of the large cab Therefore we canrewrite the two motor equations in terms of the force applied by the motor

¡

I + mpl2¢¨

θ + mpgl sin θ = −mpl¨x cos θ(mt+ mp) ¨x + b úx + mpl¨θ cos θ − mpl úθ2sin θ = u

constitute the required relations

22 The electromechanical system shown in Fig 2.49 represents a simpliÞedmodel of a capacitor microphone The system consists in part of a parallelplate capacitor connected into an electric circuit Capacitor plate a isrigidly fastened to the microphone frame Sound waves pass through themouthpiece and exert a force fs(t) on plate b, which has mass M and isconnected to the frame by a set of springs and dampers The capacitance

C is a function of the distance x between the plates, as follows:

x ,where

ε = dielectric constant of the material between the plates,

A = surface area of the plates

The charge q and the voltage e across the plates are related by