Control systems engineering 6th solultions nise

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SOLUTION
MANUAL
1-17 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.

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ISBN 13 978-0470-54756-4

Student companion website



Copyright © 2011 by John Wiley & Sons, Inc.
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
1-2 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.

2.

Thermostat
Amplifier and
valves
Heater
Temperature
difference
Voltage
difference
Fuel

flow
Actual
temperature
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
Gyro voltage

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

1-3 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.
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



6.
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

7.
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

1-4 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.
8.
a.
R
+V
-V
Differential
amplifier
Desired
level
+
-
Power
amplifier
Actuator
Valve
Float
Fluid inpu
t
Drain
Tank

R
+V
-V

b.
Desired
level
Amplifiers
Actuator
and valve
Flow
rate in
Integrate
Actual
level
Flow
rate out
Potentiometer
+
-
+
Drain
Float
Potentiometer
-
voltage
in
voltage
out
Displacement

1-5 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.
9.

Desired
force
Transducer
Amplifier Valve
Actuator
and load
Tire
Load cell
Actual
force
+
-
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.

+
-
Coil
circuit
Solenoid coil
& actuator
Coil

current
Force Armature
&
spool dynamics
Desired
position
Depth
Transducer
Coil
voltage
LVDT







1-6 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.

13
.
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.
+
Desired
Light

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

1-7 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.

14.












15.




16.



17.
a. L
di
dt

+ Ri = u(t)

b. Assume a steady-state solution i
ss
= B. Substituting this into the differential equation yields RB =
1,
from which B =
1
R
. The characteristic equation is LM + R = 0, from which M = -
R
L
. Thus, the total
+
Desired
Amplifier
Gyroscopic
Actual
HT’s
1-8 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.
solution is
i(t) = Ae
-(R/L)t
+
1
R
. Solving for the arbitrary constants, i(0) = A +
1
R

= 0. Thus, A =
-
1
R
. The final solution is i(t) =
1
R

1
R
e
-(R/L)t
=
1
R
(1 − e
−( R/L)t
)
.

c.

18.
a. Writing the loop equation,
Ri + L
di
dt
+
1
C

idt + v
C
(0)
∫
= v(t)

b. Differentiating and substituting values,
2
2
2250
di di
i
dt dt
+
+=

Writing the characteristic equation and factoring,
2
225(124)(124)
M
MM iM i++=++ +−
.
The general form of the solution and its derivative is
cos( 24 ) sin( 24 )
tt
iAe t Be t
−−
=+

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

tt
di
A
Be t A Be t
dt
−−
=− + − +

Using
(0) 1
(0) 0; (0) 1
L
di v
i
dt L L
====

i 0 A= =0
(0) 24
di
A
B
dt
=− +
=1
Thus,
A = 0 and
1
24
B = .

The solution is
1
sin( 24 )
24
t
ie t
−
=
1-9 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.
c.

19.
a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,


From which, C =
35
53
and D =
10
53
.
The characteristic polynomial is

Thus, the total solution is


Solving for the arbitrary constants, x(0) = A +
35
53
= 0. Therefore, A = -
35
53
. The final solution is

b. Assume a particular solution of
1-10 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.
x
p
= 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
x

p
= (-1/65)sin3t + (-18/65)cos3t
The characteristic polynomial is

M
2
+6M+8= M+4 M+2

Thus, the total solution is

x=Ce
-4t
+De
-2t
+-
18
65
cos 3 t -
1
65
sin 3 t

Solving for the arbitrary constants,
x(0) = C+ D −
18
65
= 0
.
Also, the derivative of the solution is

=-
3
65
cos 3 t +
54
65
sin 3 t -4Ce
-4t
-2De
-2tdx
dt


Solving for the arbitrary constants, x
.
(0)
−
3
65
− 4C − 2D = 0
, or C =
−
3
10
and D =
15
26
.
The final solution is

x=-
18
65
cos 3 t -
1
65
sin 3 t -
3
10
e
-4t
+
15
26
e
-2t

c. Assume a particular solution of
x
p
= A
Substitute into the differential equation and obtain 25A = 10, or A = 2/5.
The characteristic polynomial is

M
2
+8M+25= M+4+3i M+4-3i

Thus, the total solution is

x=
2
5
+e
-4t
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
1-11 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.
=3B-4Ccos 3 t -4B+3Csin 3 t e
-4t
dx
dt

Solving for the arbitrary constants, x
.
(0) = 3B – 4C = 0. Therefore, B = -8/15. The final solution is

x(t) =
2
5
− e
−4t
8
15
sin(3t)+
2

5
cos(3t)
⎛

⎝
⎞
⎠

20.
a. Assume a particular solution of

Substitute into the differential equation and obtain

Equating like coefficients,


From which, C = -
1
5
and D = -
1
10
.
The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A -
1
5

= 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
is

x(t) =−
1
5
cos(2t) −
1
10
sin(2t) + e
−t
11
5
cos(t)−
3
5

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
Copyright © 2011 by John Wiley & Sons, Inc.

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
x
p
= Ct
2
+ Dt + E
Substitute into the differential equation and obtain

Equating like coefficients, C =
1
4
, D = 0, and 2C + 4E = 0.
From which, C =
1
4
, D = 0, and E = -
1
8

.
The characteristic polynomial is

Thus, the total solution is

Solving for the arbitrary constants, x(0) = A -
1
8
= 1 Therefore, A =
9
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
1-13 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.




21.
+
-
Input

transducer
Desired
force
Input
voltage
Controller Actuator
Pantograph
dynamics
Spring
F
up
Spring
displacement
F
out
Sensor

22.




Amount of
HIV viruses
RTI
PI
Desired
Amount of
HIV viruses
Controller

Patient
1-14 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.


23.
a.










Speed
Actual
Motive
Force
ECU
Vehicle
Electric
Motor
Aerodynamic
Climbing &
Rolling
Resistances
Aerodynamic

Speed
+
+
Inverter
Control
Command
Controlled
Voltage
Inverter
Desired
1-15 Solutions to Problems

Copyright © 2011 by John Wiley & Sons, Inc.


b.













Desired

Speed
Actual
Motive
ECU
Accelerator
Displacement
Vehicle
Accelerator,
Aerodynamic
Climbing &
Rolling
Resistances
Aerodynamic
Speed
+
+
_
1-16 Chapter 1: Introduction
Copyright © 2011 by John Wiley & Sons, Inc.



c.




Speed
Error
Actual

Total
Motive
Force
ECU
Vehicle
Aerodynamic
Climbing &
Rolling
Resistances
Aerodynamic
Speed
+
+
Power
Planetary
Gear
Control
Inverter
Control
Command

Inverter
&
Electric
Motor
Motor
Accelerator
Accelerator
ICE
Desired

Copyright © 2011 by John Wiley & Sons, Inc.
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)
θ
i
(s)
=
10
π
;
Pre-Amp:
V
p
(s)
V
i

(s)
= K;
Power Amp:
E
a
(s)
V
p
(s)
=
150
s+150

Motor: J
m
= 0.05 + 5(
50
250
)
2
= 0.25
D
m
=0.01 + 3(
50
250
)
2
= 0.13

K
t
R
a
=
1
5


K
t
K
b
R
a
=
1
5

Therefore:
θ
m
(s)
E
a
(s)
=
K
t
R

a
J
m
s(s+
1
J
m
(D
m
+
K
t
K
b
R
a
))
=
0.8
s(s+1.32)

And:
θ
o
(s)
E
a
(s)
=
1

5

θ
m
(s)
E
a
(s)
=
0.16
s(s+1.32)


Transfer Function of a Nonlinear Electrical Network

Writing the differential equation,
d(i
0
+
δ
i)
dt
+ 2(i
0
+
δ
i)
2
− 5= v(t) . Linearizing i
2

about i
0
,

(i
0
+
δ
i)
2
-i
0
2
=2i
⎮
i=i
0
δ
i=2i
0
δ
i.
.
Thus, (i
0
+
δ
i)
2
=i

0
2
+2i
0
δ
i.


Chapter 2: Modeling in the Frequency Domain 2-2
Copyright © 2011 by John Wiley & Sons, Inc.
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
0
2
= 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)
=
1
s+6.32
. Using the
linearized i about i
0
, and the fact that v
r
(t) is 5 volts at equilibrium, the linearized v
r
(t) is v
r
(t) = 2i
2
=
2(i
0
+δi)
2
= 2(i
0
2
+2i
0
δi) = 5+6.32δi. For excursions away from equilibrium, v
r
(t) - 5 = 6.32δi = δv
r

(t).
Therefore, multiplying the transfer function by 6.32, yields,
δV
r
(s)
V(s)
=
6.32
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
∞
∫
=−
1
s
e
−st
0
∞
=
1
s


b.
F(s) = te
− st
dt
0
∞
∫
=

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

Solutions to Problems 2-3
Copyright © 2011 by John Wiley & Sons, Inc.
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
(−ssin
ω
t −
ω
cos
ω

t)
0
∞
=
ω
s
2
+
ω
2

d.
F(s) = cos
ω
t e
− st
dt
0
∞
∫
=
e
− st
s
2
+
ω
2
(−scos
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) =
ω
(s+a)
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) =

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

Or,


Taking the inverse Laplace transform, x(t) = -
35
53
e
-7t
+ (
35
53
cos 2t +
10
53
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
9
s
2
+
9
−
3
10
1
s
+

4
+
15
26
1
s
+
2


Chapter 2: Modeling in the Frequency Domain 2-4
Copyright © 2011 by John Wiley & Sons, Inc.
Taking the inverse Laplace transform,

x(t) =−
18
65
cos(3t) −
1
65
sin(3t) −
3
10
e
−4t
+
15
26
e
−2t

c. The Laplace transform of the differential equation is, assuming zero initial conditions,
(s
2
+8s+25)x(s) =
10
s
. Solving for X(s)
Xs=
10
ss
2
+8s+25

and expanding by partial fractions,
Xs=
2
5
1
s
-
2
5
1s+4+
4
9
9
s+4
2
+9

Taking the inverse Laplace transform,

x(t) =
2
5
− e
−4t
8
15
sin(3t)+
2
5
cos(3t)
⎛

⎝
⎞
⎠

4.
a. Taking the Laplace transform with initial conditions, s
2
X(s)-4s+4+2sX(s)-8+2X(s) =
2
s
2
+2
2
.

Solving for X(s),
X(s) =
32
22
4 4 16 18
(4)(22)
ss s
sss
+
++
+
++
.
Expanding by partial fractions

22 2
1
s2
1 1 21(s 1) 2
2
X(s)
5s 2 5 (s1) 1
+
+
+
⎛⎞ ⎛⎞
=− +
⎜⎟ ⎜⎟
+
++

⎝⎠ ⎝⎠

Therefore,
121
() 21 cos sin sin2 cos2
5212
tt
x
tetettt
−−
⎡⎤
=+−−
⎢⎥
⎣⎦


b. Taking the Laplace transform with initial conditions, s
2
X(s)-4s-1+2sX(s)-8+X(s) =
5
s+2
+
1
s
2
.
Solving for X(s),
432
22
41723 2

()
(1)(2)
ssss
Xs
ss s
+
+++
=
++


22
12 11 1 5
()
(1) (1)(2)
Xs
sss s s
=−+ + +
+
++

Therefore
2
() 2 11 5
tt t
x
tt te e e
−
−−
=−+ + + .

c. Taking the Laplace transform with initial conditions, s
2
X(s)-s-2+4X(s) =
2
s
3
. Solving for X(s),
Solutions to Problems 2-5
Copyright © 2011 by John Wiley & Sons, Inc.
43
32
232
()
(4)
ss
Xs
ss
+
+
=
+

23
17 3
*2
1/2 1/8
82
()
4
s

Xs
sss
+
=+−
+


Therefore
2
17 3 1 1
() cos2 sin2
8248
xt t t t=++−
.
5.
Program:
syms t

'a'
theta=45*pi/180
f=8*t^2*cos(3*t+theta);
pretty(f)
F=laplace(f);
F=simple(F);
pretty(F)
'b'
theta=60*pi/180
f=3*t*exp(-2*t)*sin(4*t+theta);
pretty(f)
F=laplace(f);

F=simple(F);
pretty(F)

Computer response:

ans =

a


theta =

0.7854


2 / PI \
8 t cos| + 3 t |
\ 4 /

1/2 2
8 2 (s + 3) (s - 12 s + 9)

2 3
(s + 9)

ans =

Chapter 2: Modeling in the Frequency Domain 2-6
Copyright © 2011 by John Wiley & Sons, Inc.

b


theta =

1.0472


/ PI \

3 t sin| + 4 t | exp(-2 t)
\ 3 /

1/2 2
1/2 1/2 3 3 s
12 s + 6 3 s - 18 3 + + 24
2

2 2

(s + 4 s + 20)

6.
Program:
syms s
'a'
G=(s^2+3*s+10)*(s+5)/[(s+3)*(s+4)*(s^2+2*s+100)];
pretty(G)
g=ilaplace(G);
pretty(g)
'b'
G=(s^3+4*s^2+2*s+6)/[(s+8)*(s^2+8*s+3)*(s^2+5*s+7)];
pretty(G)
g=ilaplace(G);
pretty(g)

Computer response:
ans =

a


2
(s + 5) (s + 3 s + 10)

2
(s + 3) (s + 4) (s + 2 s + 100)

/ 1/2 1/2 \
| 1/2 11 sin(3 11 t) |

5203 exp(-t) | cos(3 11 t) - |
20 exp(-3 t) 7 exp(-4 t) \ 57233 /
- +
103 54 5562
















Solutions to Problems 2-7
Copyright © 2011 by John Wiley & Sons, Inc.
ans =

b


3 2
s + 4 s + 2 s + 6

2 2
(s + 8) (s + 8 s + 3) (s + 5 s + 7)

/ 1/2 1/2 \
| 1/2 4262 13 sinh(13 t) |
1199 exp(-4 t) | cosh(13 t) - |
\ 15587 /
-
417

/ / 1/2 \ \
| 1/2 | 3 t | |
| / 1/2 \ 131 3 sin| | |
/ 5 t \ | | 3 t | \ 2 / |
65 exp| - | | cos| | + |
\ 2 / \ \ 2 / 15 / 266 exp(-8 t)
-
4309 93


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
+
4s
2
+
6s
+
8
s
3
+
3s
2
+
5s
+
1
.
8.
a. Cross multiplying,
(s
2

+5s+10)X(s) = 7F(s).
Taking the inverse Laplace transform,
d
2
x
dt
2
+ 5
dx
dt
+ 10x = 7f.
b. Cross multiplying after expanding the denominator,
(s
2
+21s+110)X(s) = 15F(s).
Taking the inverse Laplace transform,
d
2
x
dt
2
+ 21
dx
dt
+ 110x =15f.
c. Cross multiplying,
(s
3
+11s
2

+12s+18)X(s) = (s+3)F(s).
Taking the inverse Laplace transform,
d
3
x
dt
3
+ 11
d
2
x
dt
2
+ 12
dx
dt
+ 18x = dft/dt + 3f.
9.
The transfer function is
C
(s)
R(s)
=
5432
65432
24 4
732 5
ssss
sssss
++++

+
++++
.
Cross multiplying, (s
6
+7s
5
+3s
4
+2s
3
+s
2
+5)C(s) = (s
5
+2s
4
+4s
3
+s
2
+4)R(s).
Taking the inverse Laplace transform assuming zero initial conditions,
d
6
c
dt
6
+ 7
d

5
c
dt
5
+ 3
d
4
c
dt
4
+ 2
d
3
c
dt
3
+
d
2
c
dt
2
+ 5c =
d
5
r
dt
5
+ 2
d

4
r
dt
4
+ 4
d
3
r
dt
3
+
d
2
r
dt
2
+ 4r.
10.
The transfer function is
C
(s)
R(s)
=
s
4
+
2s
3
+
5s

2
+
s
+
1
s
5
+
3s
4
+
2s
3
+
4s
2
+
5s
+
2
.