Operating Systems Chapter 3 Homework A certain servomechanism system has dynamics dominated by

Using the given information:

00.05 0.1 0.15 0.2

0.4

0.8

1.4

Time (sec)

y(t)

00.05 0.1 0.15 0.2

0.4

0.8

1.4

Time (sec)

y(t)

Step response for Problem 3.27.

Problems and Solutions for Section 3.4: Time–

Domain Specifications

28. Suppose you desire the peak time of a given second-order system to be

less than t0

p. Draw the region in the s-plane that corresponds to values of

the poles that meet the specification tp< t0

p.

Solution:

3062 CHAPTER 3. DYNAMIC RESPONSE

29. A certain servomechanism system has dynamics dominated by a pair of

complex poles and no finite zeros. The time-domain specifications on the

rise time (tr), percent overshoot

(Mp), and settling time (ts) are given by,

tr0:6 sec ,

Mp17%,

ts9:2 sec .

(a) Sketch the region in the s-plane where the poles could be placed

so that the system will meet all three specifications.

(b) Indicate on your sketch the specific locations (denoted by ) that

will have the smallest rise-time and also meet the settling time specification

exactly.

Solution:

(a)-(b)

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s-plane region to meet the specifications.

30. A feedback system has the following response specifications,

Percent overshoot Mp16%.

Settling time ts6:9sec.

Rise time tr1:8sec.

(a) Sketch the region of acceptable closed-loop poles in the s-plane for

the system assuming the transfer function can be approximated as

simple second-order.

(b) What is the expected overshoot if the rise time and settling time

specifications are met exactly?

Solution:

s-plane region to meet the specifications.

31. Suppose you are to design a unity feedback controller for a first-order plant

depicted in Fig. 3.56. (As you will learn in Chapter 4, the configuration

shown is referred to as a proportional-integral controller.) You are to

design the controller so that the closed-loop poles lie within the shaded

regions shown in Fig. 3.57.

(a) What values of !nand correspond to the shaded regions in Fig. 3.57?

Figure 3.57: Desired closed-loop pole locations for Problem 3.31

(c) Prove that no matter what the values of Kand are, the controller

provides enough ‡exibility to place the poles anywhere in the complex

(left-half) plane.

Solution:

(a) The values could be worked out mathematically but working from

the diagram:

From the figure:

3066 CHAPTER 3. DYNAMIC RESPONSE

32. The open-loop transfer function of a unity feedback system is

G(s) = K

s(s+ 2):

The desired system response to a step input is specified as peak time

tp= 1 sec and overshoot Mp= 5%.

(a) Determine whether both specifications can be met simultaneously by

selecting the right value of K.

(b) Sketch the associated region in the s-plane where both specifications

are met, and indicate what root locations are possible for some likely

values of K.

(c) Pick a suitable value for K, and use Matlab to verify that the spec-

ifications are satisfied.

Solution:

(a)

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(b) Now we wish to have:

replace !nand in the system (**):

then with K= 3:02 we will have:

3068 CHAPTER 3. DYNAMIC RESPONSE

s-plane regions.

% Problem 3.32 FPE7e

den=[1, 2, K];

sys=tf(num,den);

t=0:.01:7;

grid on;

0.2

0.4

0.8

1.4

y(t)

Step Response with K=3.02

Max overshoot = 10.97%

Peak time = 2.21

0.2

0.4

0.8

1.4

y(t)

Step Response with K=3.02

Max overshoot = 10.97%

Peak time = 2.21

33. A simple mechanical system is shown in Fig. 3.58 (a). The parameters are

k=spring constant, b=viscous friction constant, m=mass. A step of 2

Newtons force is applied as f= 2 1(t)and the resulting step response

is shown in Fig. 3.58 (b). What are the values of the system parameters

k,b, and m?

Figure 3.58: (a) Mechanical system for Problem 3.33

(b) Step response for Problem 3.33

Solution: The equation of motion is

The transfer function is

In this case

From the plot

34. A mechanical system is shown in Fig. 3.59. The mass M= 20 kg and the

control force, u, is proportional to the reference input, u=Ar.

(a) Derive the transfer function from Rto Y.

(b) Determine the values of the parameters k,b,Asuch that the system

has a rise time of tr= 1 sec and overshoot of Mp= 16%, and zero-steady-

state error to a step in r.

Solution: (a) The equation of motion is

(b)

T(0) = 1 )A

k= 1;

35. The equations of motion for the DC motor shown in Fig. 2.32 were given

in Eqs. (2.62-63) as

Jm•

m+b+KtKe

Ra_

m=Kt

Ra

va:

Assume that

Jm= 0:01 kg m2;

b= 0:001 N msec;

Ke= 0:02 V sec;

Kt= 0:02 N m=A;

Ra= 10 :

(a) Find the transfer function between the applied voltage vaand the

motor speed _

m.

(b) What is the steady-state speed of the motor after a voltage va= 10 V

has been applied?

(c) Find the transfer function between the applied voltage vaand the

shaft angle m.

(d) Suppose feedback is added to the system in part (c) so that it becomes

a position servo device such that the applied voltage is given by

va=K(rm);

where Kis the feedback gain. Find the transfer function between r

and m.

(e) What is the maximum value of Kthat can be used if an overshoot

Mp<20% is desired?

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(f) What values of Kwill provide a rise time of less than 4 sec? (Ignore

the Mpconstraint.)

(g) Use Matlab to plot the step response of the position servo system

for values of the gain K= 0:5, 1, and 2. Find the overshoot and rise

time for each of the three step responses by examining your plots.

Are the plots consistent with your calculations in parts (e) and (f)?

Solution:

(a)

(b) Final Value Theorem

3074 CHAPTER 3. DYNAMIC RESPONSE

(e)

(f)

(g) Matlab

% Problem 3.35 FPE7e

text( 0.5, 0.3, overshootText);

text( 0.5, 0.1, risetimeText);

end

%%%%%%%%%%%%%%%%

% Function for computing rise time

function tr = risetime(t,y)

2

K= 0.4000

2

K= 0.5000

2

K= 1.0000

2

K= 2.0000

0.5

1

1.5

K= 0.0650

Max overshoot = 19.99 %

2

K= 0.4000

2

K= 0.5000

2

K= 1.0000

2

K= 2.0000

0.5

1

1.5

K= 0.0650

Max overshoot = 19.99 %

Problem 3.35: Closed-loop step responses for several values of K.

For part (e) we concluded that K < 6:50 102in order for Mp<

36. You wish to control the elevation of the satellite-tracking antenna shown

in Figs. 3.60 and 3.61. The antenna and drive parts have a moment of

inertia Jand a damping B; these arise to some extent from bearing and

aerodynamic friction, but mostly from the back emf of the DC drive motor.

The equations of motion are

J•

+B_

=Tc;

where Tcis the torque from the drive motor. Assume that

J= 600;000 kgm2B= 20;000 Nmsec:

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Figure 3.60: Satellite Antenna (Courtesy Space Systems/Loral)

(a) Find the transfer function between the applied torque Tcand the

antenna angle .

(b) Suppose the applied torque is computed so that tracks a reference

command raccording to the feedback law

Tc=K(r);

where Kis the feedback gain. Find the transfer function between r

and .

(c) What is the maximum value of Kthat can be used if you wish to

have an overshoot Mp<10%?

(d) What values of Kwill provide a rise time of less than 80 sec? (Ignore

the Mpconstraint.)

(e) Use Matlab to plot the step response of the antenna system for

K= 200, 400, 1000, and 2000. Find the overshoot and rise time

of the four step responses by examining your plots. Do the plots to

confirm your calculations in parts (c) and (d)?

Solution:

J•

+B_

=Tc

(a)

Js2+Bs=Tc(s);

(b)

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(c)

(d)