Solution of skill Assessment Control Systems Engineering By Norman S.Nise 6th edition

E1SM 11/11/2010 9:29:8 Page 1
Solutions to Skill-Assessment
Exercises
CHAPTER 2
2.1
The Laplace transform of t is
1
s2
using Table 2.1, Item 3. Using Table 2.2, Item 4,
F sð Þ ¼
1
s þ 5ð Þ2
.
2.2
Expanding F(s) by partial fractions yields:
F sð Þ ¼
A
s
þ
B
s þ 2
þ
C
s þ 3ð Þ2
þ
D
s þ 3ð Þ
where,
A ¼
10
s þ 2ð Þ s þ 3ð Þ2

S!0
¼
5
9
B ¼
10
s s þ 3ð Þ2

S!À2
¼ À5
C ¼
10
s s þ 2ð Þ

S!À3
¼
10
3
; and D ¼ s þ 3ð Þ2 dF sð Þ
ds

s!À3
¼
40
9
Taking the inverse Laplace transform yields,
f tð Þ ¼
5
9
À 5eÀ2t
þ
10
3
teÀ3t
þ
40
9
eÀ3t
2.3
Taking the Laplace transform of the differential equation assuming zero initial
conditions yields:
s3
C sð Þ þ 3s2
C sð Þ þ 7sC sð Þ þ 5C sð Þ ¼ s2
R sð Þ þ 4sR sð Þ þ 3R sð Þ
Collecting terms,
s3
þ 3s2
þ 7s þ 5
À Á
C sð Þ ¼ s2
þ 4s þ 3
À Á
R sð Þ
Thus,
C sð Þ
R sð Þ
¼
s2
þ 4s þ 3
s3 þ 3s2 þ 7s þ 5
1

E1SM 11/11/2010 9:29:8 Page 2
2.4
G sð Þ ¼
C sð Þ
R sð Þ
¼
2s þ 1
s2 þ 6s þ 2
Cross multiplying yields,
d2
c
dt2
þ 6
dc
dt
þ 2c ¼ 2
dr
dt
þ r
2.5
C sð Þ ¼ R sð ÞG sð Þ ¼
1
s2
Ã
s
s þ 4ð Þ s þ 8ð Þ
¼
1
s s þ 4ð Þ s þ 8ð Þ
¼
A
s
þ
B
s þ 4ð Þ
þ
C
s þ 8ð Þ
where
A ¼
1

S!0
¼
1
32
B ¼
1
s s þ 8ð Þ

S!À4
¼ À
1
16
; and C ¼
1
s s þ 4ð Þ

S!À8
¼
1
32
Thus,
c tð Þ ¼
1
32
À
1
16
eÀ4t
þ
1
32
eÀ8t
2.6
Mesh Analysis
Transforming the network yields,
+V(s)
I1(s) I2(s)
V1(s)
V2(s)
I9(s)
+
1 1
s s
s
_
_
Now, writing the mesh equations,
s þ 1ð ÞI1 sð Þ À sI2 sð Þ À I3 sð Þ ¼ V sð Þ
ÀsI1 sð Þ þ 2s þ 1ð ÞI2 sð Þ À I3 sð Þ ¼ 0
ÀI1 sð Þ À I2 sð Þ þ s þ 2ð ÞI3 sð Þ ¼ 0
Solving the mesh equations for I2(s),
I2 sð Þ ¼
s þ 1ð Þ V sð Þ À1
Às 0 À1
À1 0 s þ 2ð Þ

s þ 1ð Þ Às À1
Às 2s þ 1ð Þ À1
À1 À1 s þ 2ð Þ

¼
s2
þ 2s þ 1
À Á
V sð Þ
s s2 þ 5s þ 2ð Þ
2 Solutions to Skill-Assessment Exercises

E1SM 11/11/2010 9:29:8 Page 3
But, VL sð Þ ¼ sI2 sð Þ
Hence,
VL sð Þ ¼
s2
þ 2s þ 1
À Á
V sð Þ
s2 þ 5s þ 2ð Þ
or
VL sð Þ
V sð Þ
¼
s2
þ 2s þ 1
s2 þ 5s þ 2
Nodal Analysis
Writing the nodal equations,
1
s
þ 2

V1 sð Þ À VL sð Þ ¼ V sð Þ
ÀV1 sð Þ þ
2
s
þ 1

VL sð Þ ¼
1
s
V sð Þ
Solving for VL(s),
VL sð Þ ¼
1
s
þ 2

V sð Þ
À1
1
s
V sð Þ

1
s
þ 2

À1
À1
2
s
þ 1

¼
s2
þ 2s þ 1
À Á
V sð Þ
s2 þ 5s þ 2ð Þ
or
VL sð Þ
V sð Þ
¼
s2
þ 2s þ 1
s2 þ 5s þ 2
2.7
Inverting
G sð Þ ¼ À
Z2 sð Þ
Z1 sð Þ
¼
À100000
105
=s
¼ Às
Noninverting
G sð Þ ¼
Z1 sð Þ þ Z sð Þ½ Š
Z1 sð Þ
¼
105
s
þ 105
!
105
s
! ¼ s þ 1
2.8
Writing the equations of motion,
s2
þ 3s þ 1
À Á
X1 sð Þ À 3s þ 1ð ÞX2 sð Þ ¼ F sð Þ
À 3s þ 1ð ÞX1 sð Þ þ s2
þ 4s þ 1
À Á
X2 sð Þ ¼ 0
Chapter 2 Solutions to Skill-Assessment Exercises 3

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Solving for X2(s),
X2 sð Þ ¼
s2
þ 3s þ 1
À Á
F sð Þ
À 3s þ 1ð Þ 0

s2
þ 3s þ 1
À Á
À 3s þ 1ð Þ
À 3s þ 1ð Þ s2
þ 4s þ 1
À Á

¼
3s þ 1ð ÞF sð Þ
s s3 þ 7s2 þ 5s þ 1ð Þ
Hence,
X2 sð Þ
F sð Þ
¼
3s þ 1ð Þ
s s3 þ 7s2 þ 5s þ 1ð Þ
2.9
s2
þ s þ 1
À Á
u1 sð Þ À s þ 1ð Þu2 sð Þ ¼ T sð Þ
À s þ 1ð Þu1 sð Þ þ 2s þ 2ð Þu2 sð Þ ¼ 0
where u1 sð Þ is the angular displacement of the inertia.
Solving for u2 sð Þ,
u2 sð Þ ¼
s2
þ s þ 1
À Á
T sð Þ
À s þ 1ð Þ 0

s2
þ s þ 1
À Á
À s þ 1ð Þ
À s þ 1ð Þ 2s þ 2ð Þ

¼
s þ 1ð ÞF sð Þ
2s3 þ 3s2 þ 2s þ 1
From which, after simplification,
u2 sð Þ ¼
1
2s2 þ s þ 1
2.10
Transforming the network to one without gears by reflecting the 4 N-m/rad spring to
the left and multiplying by (25/50)2
, we obtain,
s2
þ s
À Á
u1 sð Þ À sua sð Þ ¼ T sð Þ
Àsu1 sð Þ þ s þ 1ð Þua sð Þ ¼ 0
where u1 sð Þ is the angular displacement of the 1-kg inertia.
Solving for ua sð Þ,
ua sð Þ ¼
s2
þ s
À Á
T sð Þ
Às 0

s2
þ s
À Á
Às
Às s þ 1ð Þ

¼
sT sð Þ
s3 þ s2 þ s

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From which,
ua sð Þ
T sð Þ
¼
1
s2 þ s þ 1
But, u2 sð Þ ¼
1
2
ua sð Þ:
Thus,
u2 sð Þ
T sð Þ
¼
1=2
s2 þ s þ 1
2.11
First find the mechanical constants.
Jm ¼ Ja þ JL
1
5
Ã
1
4
2
¼ 1 þ 400
1
400

¼ 2
Dm ¼ Da þ DL
1
5
Ã
1
4
2
¼ 5 þ 800
1
400

¼ 7
Now find the electrical constants. From the torque-speed equation, set vm ¼ 0 to
find stall torque and set Tm ¼ 0 to find no-load speed. Hence,
Tstall ¼ 200
vnoÀload ¼ 25
which,
Kt
Ra
¼
Tstall
Ea
¼
200
100
¼ 2
Kb ¼
Ea
vnoÀload
¼
100
25
¼ 4
Substituting all values into the motor transfer function,
um sð Þ
Ea sð Þ
¼
KT
RaJm
s s þ
1
Jm

Dm þ
KTKb
Ra
¼
1
s s þ
15
2

where um sð Þ is the angular displacement of the armature.
Now uL sð Þ ¼
1
20
um sð Þ. Thus,
uL sð Þ
Ea sð Þ
¼
1=20
s s þ
15
2

2.12
Letting
u1 sð Þ ¼ v1 sð Þ=s
u2 sð Þ ¼ v2 sð Þ=s

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in Eqs. 2.127, we obtain
J1s þ D1 þ
K
s

v1 sð Þ À
K
s
v2 sð Þ ¼ T sð Þ
À
K
s
v1 sð Þ þ J2s þ D2 þ
K
s

v2 sð Þ
From these equations we can draw both series and parallel analogs by considering
these to be mesh or nodal equations, respectively.
–
+
J1
J1
J2
J2
D1
D1
D2
D2
T(t)
T(t)
w1(t)
w1(t)
w2(t)
w2(t)
1 1
K
1
1
Series analog
Parallel analog
K
2.13
Writing the nodal equation,
C
dv
dt
þ ir À 2 ¼ i tð Þ
But,
C ¼ 1
v ¼ vo þ dv
ir ¼ evr
¼ ev
¼ evoþdv
Substituting these relationships into the differential equation,
d vo þ dvð Þ
dt
þ evoþdv
À 2 ¼ i tð Þ ð1Þ
We now linearize ev
.
The general form is
f vð Þ À f voð Þ %
df
dv

vo
dv
Substituting the function, f vð Þ ¼ ev
, with v ¼ vo þ dv yields,
evoþdv
À evo
%
dev
dv

vo
dv

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Solving for evoþdv
,
evoþdv
¼ evo
þ
dev
dv

vo
dv ¼ evo
þ evo
dv
Substituting into Eq. (1)
ddv
dt
þ evo
þ evo
dv À 2 ¼ i tð Þ ð2Þ
Setting i tð Þ ¼ 0 and letting the circuit reach steady state, the capacitor acts like an
open circuit. Thus, vo ¼ vr with ir ¼ 2. But, ir ¼ evr
or vr ¼ lnir.
Hence, vo ¼ ln 2 ¼ 0:693. Substituting this value of vo into Eq. (2) yields
ddv
dt
þ 2dv ¼ i tð Þ
Taking the Laplace transform,
s þ 2ð Þdv sð Þ ¼ I sð Þ
Solving for the transfer function, we obtain
dv sð Þ
I sð Þ
¼
1
s þ 2
or
V sð Þ
I sð Þ
¼
1
s þ 2
about equilibrium:
CHAPTER 3
3.1
Identifying appropriate variables on the circuit yields
–
+
–
+
C1
iL iC2
iC1
iR
C2
R
L vo(t)v1(t)
Writing the derivative relations
C1
dvC1
dt
¼ iC1
L
diL
dt
¼ vL
C2
dvC2
dt
¼ iC2
ð1Þ

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Using Kirchhoff’s current and voltage laws,
iC1
¼ iL þ iR ¼ iL þ
1
R
vL À vC2
ð Þ
vL ¼ ÀvC1
þ vi
iC2
¼ iR ¼
1
R
vL À vC2
ð Þ
Substituting these relationships into Eqs. (1) and simplifying yields the state
equations as
dvC1
dt
¼ À
1
RC1
vC1
þ
1
C1
iL À
1
RC1
vC2
þ
1
RC1
vi
diL
dt
¼ À
1
L
vC1
þ
1
L
vi
dvC2
dt
¼ À
1
RC2
vC1
À
1
RC2
vC2
1
RC2
vi
where the output equation is
vo ¼ vC2
Putting the equations in vector-matrix form,
_x ¼
À
1
RC1
1
C1
À
1
RC1
À
1
L
0 0
À
1
RC2
0 À
1
RC2
2
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
5
x þ
1
RC1
1
L
1
RC2
2
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
5
vi tð Þ
y ¼ 0 0 1½ Šx
3.2
Writing the equations of motion
s2
þ s þ 1
À Á
X1 sð Þ ÀsX2 sð Þ ¼ F sð Þ
ÀsX1 sð Þ þ s2
þ s þ 1
À Á
X2 sð Þ ÀX3 sð Þ ¼ 0
ÀX2 sð Þ þ s2
þ s þ 1
À Á
X3 sð Þ ¼ 0
Taking the inverse Laplace transform and simplifying,
€x1 ¼ À_x1 À x1 þ _x2 þ f
€x2 ¼ _x1 À _x2 À x2 þ x3
€x3 ¼ À_x3 À x3 þ x2
Defining state variables, zi,
z1 ¼ x1; z2 ¼ _x1; z3 ¼ x2; z4 ¼ _x2; z5 ¼ x3; z6 ¼ _x3

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Writing the state equations using the definition of the state variables and the inverse
transform of the differential equation,
_z1 ¼ z2
_z2 ¼ €x1 ¼ À_x1 À x1 þ _x2 þ f ¼ Àz2 À z1 þ z4 þ f
_z3 ¼ _x2 ¼ z4
_z4 ¼ €x2 ¼ _x1 À _x2 À x2 þ x3 ¼ z2 À z4 À z3 þ z5
_z5 ¼ _x3 ¼ z6
_z6 ¼ €x3 ¼ À_x3 À x3 þ x2 ¼ Àz6 À z5 þ z3
The output is z5. Hence, y ¼ z5. In vector-matrix form,
_z ¼
0 1 0 0 0 0
À1 À1 0 1 0 0
0 0 0 1 0 0
0 1 À1 À1 1 0
0 0 0 0 0 1
0 0 1 0 À1 À1
2
6
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
7
5
z þ
0
1
0
0
0
0
2
6
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
7
5
f tð Þ; y ¼ 0 0 0 0 1 0½ Šz
3.3
First derive the state equations for the transfer function without zeros.
X sð Þ
R sð Þ
¼
1
s2 þ 7s þ 9
Cross multiplying yields
s2
þ 7s þ 9
À Á
X sð Þ ¼ R sð Þ
Taking the inverse Laplace transform assuming zero initial conditions, we get
€x þ 7_x þ 9x ¼ r
Defining the state variables as,
x1 ¼ x
x2 ¼ _x
Hence,
_x1 ¼ x2
_x2 ¼ €x ¼ À7_x À 9x þ r ¼ À9x1 À 7x2 þ r
Using the zeros of the transfer function, we find the output equation to be,
c ¼ 2_x þ x ¼ x1 þ 2x2
Putting all equation in vector-matrix form yields,
_x ¼
0 1
À9 À7
#
x þ
0
1
#
r
c ¼ 1 2½ Šx

E1SM 11/11/2010 9:29:11 Page 10
3.4
The state equation is converted to a transfer function using
G sð Þ ¼ C sI À Að ÞÀ1
B ð1Þ
where
A ¼
À4 À1:5
4 0
!
; B ¼
2
0
!
; and C ¼ 1:5 0:625½ Š:
Evaluating sI À Að Þ yields
sI À Að Þ ¼
s þ 4 1:5
À4 s
!
Taking the inverse we obtain
sI À Að ÞÀ1
¼
1
s2 þ 4s þ 6
s À1:5
4 s þ 4
!
Substituting all expressions into Eq. (1) yields
G sð Þ ¼
3s þ 5
s2 þ 4s þ 6
3.5
Writing the differential equation we obtain
d2
x
dt2
þ 2x2
¼ 10 þ df tð Þ ð1Þ
Letting x ¼ xo þ dx and substituting into Eq. (1) yields
d2
xo þ dxð Þ
dt2
þ 2 xo þ dxð Þ2
¼ 10 þ df tð Þ ð2Þ
Now, linearize x2
.
xo þ dxð Þ2
À x2
o ¼
d x2
À Á
dx

xo
dx ¼ 2xodx
from which
xo þ dxð Þ2
¼ x2
o þ 2xodx ð3Þ
Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives
us the linearized intermediate differential equation,
d2
dx
dt2
þ 4xodx ¼ À2x2
o þ 10 þ df tð Þ ð4Þ
The force of the spring at equilibrium is 10 N. Thus, since F ¼ 2x2
; 10 ¼ 2x2
o from
which
xo ¼
ffiffiffi
5
p

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Substituting this value of xo into Eq. (4) gives us the final linearized differential
equation.
d2
dx
dt2
þ 4
ffiffiffi
5
p
dx ¼ df tð Þ
Selecting the state variables,
x1 ¼ dx
x2 ¼ _dx
Writing the state and output equations
_x1 ¼ x2
_x2 ¼€dx ¼ À4
ffiffiffi
5
p
x1 þ df tð Þ
y ¼ x1
Converting to vector-matrix form yields the final result as
_x ¼
0 1
À4
ffiffiffi
5
p
0
!
x þ
0
1
!
df tð Þ
y ¼ 1 0½ Šx
CHAPTER 4
4.1
For a step input
C sð Þ ¼
10 s þ 4ð Þ s þ 6ð Þ
s s þ 1ð Þ s þ 7ð Þ s þ 8ð Þ s þ 10ð Þ
¼
A
s
þ
B
s þ 1
þ
C
s þ 7
þ
D
s þ 8
þ
E
s þ 10
Taking the inverse Laplace transform,
c tð Þ ¼ A þ BeÀt
þ CeÀ7t
þ DeÀ8t
þ EeÀ10t
4.2
Since a ¼ 50; Tc ¼
1
a
¼
1
50
¼ 0:02s; Ts ¼
4
a
¼
4
50
¼ 0:08 s; and
Tr ¼
2:2
a
¼
2:2
50
¼ 0:044 s.
4.3
a. Since poles are at À6 Æ j 19:08; c tð Þ ¼ A þ BeÀ6t
cos 19:08t þ fð Þ.
b. Since poles are at À78:54 and À11:46; c tð Þ ¼ A þ BeÀ78:54t
þ CeÀ11:4t
.
c. Since poles are double on the real axis at À15 c tð Þ ¼ A þ BeÀ15t
þ CteÀ15t
:
d. Since poles are at Æj 25; c tð Þ ¼ A þ B cos 25t þ fð Þ.
4.4
a. vn ¼
ffiffiffiffiffiffiffiffi
400
p
¼ 20 and 2zvn ¼ 12; ;z ¼ 0:3 and system is underdamped.
b. vn ¼
900
p
¼ 30 and 2zvn ¼ 90; ;z ¼ 1:5 and system is overdamped.
c. vn ¼
225
p
¼ 15 and 2zvn ¼ 30; ;z ¼ 1 and system is critically damped.
d. vn ¼
625
p
¼ 25 and 2zvn ¼ 0; ;z ¼ 0 and system is undamped.

E1SM 11/11/2010 9:29:13 Page 12
4.5
vn ¼
361
p
¼ 19 and 2zvn ¼ 16; ;z ¼ 0:421:
Now, Ts ¼
4
zvn
¼ 0:5 s and Tp ¼
p
vn
ffiffiffiffiffiffiffiffiffiffiffiffiffi
1 À z2
p ¼ 0:182 s.
From Figure 4.16, vnTr ¼ 1:4998. Therefore, Tr ¼ 0:079 s.
Finally, %os ¼ e
Àzpffiffiffi
1
p
À z2
Ã 100 ¼ 23:3%
4.6
a. The second-order approximation is valid, since the dominant poles have a real
part of À2 and the higher-order pole is at À15, i.e. more than five-times further.
b. The second-order approximation is not valid, since the dominant poles have a real
part of À1 and the higher-order pole is at À4, i.e. not more than five-times further.
4.7
a. Expanding G(s) by partial fractions yields G sð Þ ¼
1
s
þ
0:8942
s þ 20
À
1:5918
s þ 10
À
0:3023
s þ 6:5
.
But À0:3023 is not an order of magnitude less than residues of second-order terms
(term 2 and 3). Therefore, a second-order approximation is not valid.
b. Expanding G(s) by partial fractions yields G sð Þ ¼
1
s
þ
0:9782
s þ 20
À
1:9078
s þ 10
À
0:0704
s þ 6:5
.
But 0.0704 is an order of magnitude less than residues of second-order terms
(term 2 and 3). Therefore, a second-order approximation is valid.
4.8
SeeFigure4.31inthetextbookfortheSimulinkblockdiagramandtheoutputresponses.
4.9
a. Since sI À A ¼
s À2
3 s þ 5
!
; sI À Að ÞÀ1
¼
1
s2 þ 5s þ 6
s þ 5 2
À3 s
!
: Also,
BU sð Þ ¼
0
1= s þ 1ð Þ
#
.
The state vector is X sð Þ ¼ sI À Að ÞÀ1
x 0ð Þ þ BU sð Þ½ Š ¼
1
s þ 1ð Þ s þ 2ð Þ s þ 3ð Þ
Â
2 s2
þ 7s þ 7
À Á
s2
À 4s À 6
#
. The output is Y sð Þ ¼ 1 3½ ŠX sð Þ ¼
5s2
þ 2s À 4
¼
À
0:5
s þ 1
À
12
s þ 2
þ
17:5
s þ 3
. Taking the inverse Laplace transform yields y tð Þ ¼
À0:5eÀt
À 12eÀ2t
þ 17:5eÀ3t
.
b. The eigenvalues are given by the roots of jsI À Aj ¼ s2
þ 5s þ 6, or À2 and À3.
4.10
a. Since sI À Að Þ ¼
s À2
2 s þ 5
!
; sI À Að ÞÀ1
¼
1
s2 þ 5s þ 4
s þ 5 2
À2 s
!
. Taking the
Laplace transform of each term, the state transition matrix is given by
F tð Þ ¼
4
3
eÀt
À
1
3
eÀ4t 2
3
eÀt
À
2
3
eÀ4t
À
2
3
eÀt
þ
2
3
eÀ4t
À
1
3
eÀt
þ
4
3
eÀ4t
2
6
6
4
3
7
7
5:

E1SM 11/11/2010 9:29:13 Page 13
b. Since F t À tð Þ ¼
4
3
eÀ tÀtð Þ
À
1
3
eÀ4 tÀtð Þ 2
3
eÀ tÀtð Þ
À
2
3
eÀ4 tÀtð Þ
À
2
3
eÀ tÀtð Þ
þ
2
3
eÀ4 tÀtð Þ
À
1
3
eÀ tÀtð Þ
þ
4
3
eÀ4 tÀtð Þ
2
6
6
4
3
7
7
5 and
Bu tð Þ ¼
0
eÀ2t
!
; F t À tð ÞBu tð Þ ¼
2
3
eÀt
eÀt
À
2
3
e2t
eÀ4t
À
1
3
eÀt
eÀt
þ
4
3
e2t
eÀ4t
2
6
6
4
3
7
7
5:
Thus, x tð Þ ¼ F tð Þx 0ð Þ þ
Rt
0 F t À tð Þ
BU tð Þdt ¼
10
3
eÀt
À eÀ2t
À
4
3
eÀ4t
À
5
3
eÀt
þ eÀ2t
þ
8
3
eÀ4t
2
6
6
4
3
7
7
5
c. y tð Þ ¼ 2 1½ Šx ¼ 5eÀt
À eÀ2t
CHAPTER 5
5.1
Combine the parallel blocks in the forward path. Then, push
1
s
to the left past the
pickoff point.
1
s
s
s
s
s2
+
1
+
–
–
R(s) C(s)
Combine the parallel feedback paths and get 2s. Then, apply the feedback formula,
simplify, and get, T sð Þ ¼
s3
þ 1
2s4 þ s2 þ 2s
.
5.2
Find the closed-loop transfer function, T sð Þ ¼
G sð Þ
1 þ G sð ÞH sð Þ
¼
16
s2 þ as þ 16
, where
and G sð Þ ¼
16
s s þ að Þ
and H sð Þ ¼ 1. Thus, vn ¼ 4 and 2zvn ¼ a, from which z ¼
a
8
.
But, for 5% overshoot, z ¼
Àln
%
100

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p2 þ ln2 %
100
s ¼ 0:69. Since, z ¼
a
8
; a ¼ 5:52.

E1SM 11/11/2010 9:29:13 Page 14
5.3
Label nodes.
++
+–
–
R(s)
s
s
N1 (s) N2 (s) N3 (s) N4 (s)
N6 (s)N5 (s)
N7 (s)
s
C(s)
1
s
1
s
Draw nodes.
R(s) N1(s) N2(s)
N5(s)
N7(s)
N6(s)
N3(s) N4(s) C (s)
Connect nodes and label subsystems.
R(s) C(s)1
1
s
s
−1
ss
1 1
−1
1
1
sN1 (s) N2 (s)
N5 (s) N6 (s)
N7 (s)
N3 (s) N4 (s)
Eliminate unnecessary nodes.
R(s) C(s)1 ss
1
s
1
s
–s
–1

E1SM 11/11/2010 9:29:14 Page 15
5.4
Forward-path gains are G1G2G3 and G1G3.
Loop gains are ÀG1G2H1; ÀG2H2; and ÀG3H3.
Nontouching loops are ÀG1G2H1½ Š ÀG3H3½ Š ¼ G1G2G3H1H3 and
ÀG2H2½ Š ÀG3H3½ Š ¼ G2G3H2H3.
Also, D ¼ 1 þ G1G2H1 þ G2H2 þ G3H3 þ G1G2G3H1H3 þ G2G3H2H3:
Finally, D1 ¼ 1 and D2 ¼ 1.
Substituting these values into T sð Þ ¼
C sð Þ
R sð Þ
¼
P
k
TkDk
D
yields
T sð Þ ¼
G1 sð ÞG3 sð Þ 1 þ G2 sð Þ½ Š
1 þ G2 sð ÞH2 sð Þ þ G1 sð ÞG2 sð ÞH1 sð Þ½ Š 1 þ G3 sð ÞH3 sð Þ½ Š
5.5
The state equations are,
_x1 ¼ À2x1 þ x2
_x2 ¼ À3x2 þ x3
_x3 ¼ À3x1 À 4x2 À 5x3 þ r
y ¼ x2
Drawing the signal-flow diagram from the state equations yields
1
s
1
s
1
s11 1
1
–5
–4
–2–3
–3
r x1x2x3 y
5.6
From G sð Þ ¼
100 s þ 5ð Þ
s2 þ 5s þ 6
we draw the signal-flow graph in controller canonical form
and add the feedback.
1
–5
–6
100
500
–1
y
r
1
s
1
sx1 x2

E1SM 11/11/2010 9:29:14 Page 16
Writing the state equations from the signal-flow diagram, we obtain
x ¼
À105 À506
1 0
#
x þ
1
0
#
r
y ¼ 100 500½ Šx
5.7
From the transformation equations,
PÀ1
¼
3 À2
1 À4
!
Taking the inverse,
P ¼
0:4 À0:2
0:1 À0:3
!
Now,
PÀ1
AP ¼
3 À2
1 À4
#
1 3
À4 À6
#
0:4 À0:2
0:1 À0:3
#
¼
6:5 À8:5
9:5 À11:5
#
PÀ1
B ¼
3 À2
1 À4
#
1
3
#
¼
À3
À11
#
CP ¼ 1 4½ Š
0:4 À0:2
0:1 À0:3
#
¼ 0:8 À1:4½ Š
Therefore,
_z ¼
6:5 À8:5
9:5 À11:5
#
z þ
À3
À11
#
u
y ¼ 0:8 À1:4½ Šz
5.8
First find the eigenvalues.
jlI À Aj ¼
l 0
0 l
!
À
1 3
À4 À6
!

¼
l À 1 À3
4 l þ 6

¼ l2
þ 5l þ 6
From which the eigenvalues are À2 and À3.
Now use Axi ¼ lxi for each eigenvalue, l.
Thus,
1 3
À4 À6
!
x1
x2
!
¼ l
x1
x2
!
For l ¼ À2,
3x1 þ 3x2 ¼ 0
À4x1 À 4x2 ¼ 0

E1SM 11/11/2010 9:29:15 Page 17
Thus x1 ¼ Àx2
For l ¼ À3
4x1 þ 3x2 ¼ 0
À4x1 À 3x2 ¼ 0
Thus x1 ¼ Àx2 and x1 ¼ À0:75x2; from which we let
P ¼
0:707 À0:6
À0:707 0:8
!
Taking the inverse yields
PÀ1
¼
5:6577 4:2433
5 5
!
Hence,
D ¼ PÀ1
AP ¼
5:6577 4:2433
5 5
#
1 3
À4 À6
#
0:707 À0:6
À0:707 0:8
#
¼
À2 0
0 À3
#
PÀ1
B ¼
5:6577 4:2433
5 5
#
1
3
#
¼
18:39
20
#
CP ¼ 1 4½ Š
0:707 À0:6
À0:707 0:8
#
¼ À2:121 2:6½ Š
Finally,
_z ¼
À2 0
0 À3
#
z þ
18:39
20
#
u
y ¼ À2:121 2:6½ Šz
CHAPTER 6
6.1
Make a Routh table.
s7
3 6 7 2
s6
9 4 8 6
s5
4.666666667 4.333333333 0 0
s4
À4.35714286 8 6 0
s3
12.90163934 6.426229508 0 0
s2
10.17026684 6 0 0
s1
À1.18515742 0 0 0
s0
6 0 0 0
Since there are four sign changes and no complete row of zeros, there are four right
half-plane poles and three left half-plane poles.
6.2
Make a Routh table. We encounter a row of zeros on the s3
row. The even polynomial
is contained in the previous row as À6s4
þ 0s2
þ 6. Taking the derivative yields

E1SM 11/11/2010 9:29:15 Page 18
À24s3
þ 0s. Replacing the row of zeros with the coefficients of the derivative yields
the s3
row. We also encounter a zero in the first column at the s2
row. We replace the
zero with e and continue the table. The final result is shown now as
s6
1 À6 À1 6
s5
1 0 À1 0
s4
6 0 6 0
s3
À24 0 0 0 ROZ
s2
e 6 0 0
s1
144=e 0 0 0
s0
6 0 0 0
There is one sign change below the even polynomial. Thus the even polynomial
(4th
order) has one right half-plane pole, one left half-plane pole, and 2 imaginary
axis poles. From the top of the table down to the even polynomial yields one sign
change. Thus, the rest of the polynomial has one right half-plane root, and one left
half-plane root. The total for the system is two right half-plane poles, two left half-
plane poles, and 2 imaginary poles.
6.3
Since G sð Þ ¼
K s þ 20ð Þ
; T sð Þ ¼
G sð Þ
1 þ G sð Þ
¼
K s þ 20ð Þ
s3 þ 5s2 þ 6 þ Kð Þs þ 20K
Form the Routh table.
s3
1 6 þ Kð Þ
s2
5 20K
s1 30 À 15K
5
s0
20K
From the s1
row, K 2. From the s0
row, K 0. Thus, for stability, 0 K 2.
6.4
First find
sI À Aj j ¼
s 0 0
0 s 0
0 0 s
2
6
4
3
7
5 À
2 1 1
1 7 1
À3 4 À5
2
6
4
3
7
5

¼
s À 2ð Þ À1 À1
À1 s À 7ð Þ À1
3 À4 s þ 5ð Þ

¼ s3
À 4s2
À 33s þ 51
Now form the Routh table.
s3
1 À33
s2
À4 51
s1
À20:25
s0
51
There are two sign changes. Thus, there are two rhp poles and one lhp pole.

E1SM 11/11/2010 9:29:15 Page 19
CHAPTER 7
7.1
a. First check stability.
T sð Þ ¼
G sð Þ
1 þ G sð Þ
¼
10s2
þ 500s þ 6000
s3 þ 70s2 þ 1375s þ 6000
¼
10 s þ 30ð Þ s þ 20ð Þ
s þ 26:03ð Þ s þ 37:89ð Þ s þ 6:085ð Þ
Poles are in the lhp. Therefore, the system is stable. Stability also could be checked
via Routh-Hurwitz using the denominator of T(s). Thus,
15u tð Þ : estep 1ð Þ ¼
15
1 þ lim
s!0
G sð Þ
¼
15
1 þ 1
¼ 0
15tu tð Þ : eramp 1ð Þ ¼
15
lim
s!0
sG sð Þ
¼
15
10Ã
20Ã
30
25Ã
35
¼ 2:1875
15t2
u tð Þ : eparabola 1ð Þ ¼
15
lim
s!0
s2
G sð Þ
¼
30
0
¼ 1; since L 15t2
Â Ã
¼
30
s3
b. First check stability.
T sð Þ ¼
G sð Þ
1 þ G sð Þ
¼
10s2
þ 500s þ 6000
s5 þ 110s4 þ 3875s3 þ 4:37e04s2 þ 500s þ 6000
¼
10 s þ 30ð Þ s þ 20ð Þ
s þ 50:01ð Þ s þ 35ð Þ s þ 25ð Þ s2 À 7:189e À 04s þ 0:1372ð Þ
From the second-order term in the denominator, we see that the system is unstable.
Instability could also be determined using the Routh-Hurwitz criteria on the
denominator of T(s). Since the system is unstable, calculations about steady-state
error cannot be made.
7.2
a. The system is stable, since
T sð Þ ¼
G sð Þ
1 þ G sð Þ
¼
1000 s þ 8ð Þ
s þ 9ð Þ s þ 7ð Þ þ 1000 s þ 8ð Þ
¼
1000 s þ 8ð Þ
s2 þ 1016s þ 8063
and is of Type 0. Therefore,
Kp ¼ lim
s!0
G sð Þ ¼
1000Ã
8
7Ã
9
¼ 127; Kv ¼ lim
s!0
sG sð Þ ¼ 0;
and Ka ¼ lim
s!0
s2
G sð Þ ¼ 0
b.
estep 1ð Þ ¼
1
1 þ lim
s!0
G sð Þ
¼
1
1 þ 127
¼ 7:8e À 03
eramp 1ð Þ ¼
1
lim
s!0
sG sð Þ
¼
1
0
¼ 1
eparabola 1ð Þ ¼
1
lim
s!0
s2
G sð Þ
¼
1
0
¼ 1

E1SM 11/11/2010 9:29:16 Page 20
7.3
System is stable for positive K. System is Type 0. Therefore, for a step input
estep 1ð Þ ¼
1
1 þ Kp
¼ 0:1. Solving for Kp yields Kp ¼ 9 ¼ lim
s!0
G sð Þ ¼
12K
14Ã
18
; from
which we obtain K ¼ 189.
7.4
System is stable. Since G1 sð Þ ¼ 1000, and G2 sð Þ ¼
s þ 2ð Þ
s þ 4ð Þ
,
eD 1ð Þ ¼ À
1
lim
s!0
1
G2 sð Þ
þ lim G1
s!0
sð Þ
¼
1
2 þ 1000
¼ À9:98e À 04
7.5
System is stable. Create a unity-feedback system, where He sð Þ ¼
1
s þ 1
À 1 ¼
Às
s þ 1
The system is as follows:
+
–
–
R(s) Ea(s) C(s)100
s + 4
−s
s + 1
Thus,
Ge sð Þ ¼
G sð Þ
1 þ G Sð ÞHe sð Þ
¼
100
s þ 4ð Þ
1 À
100s
¼
100 s þ 1ð Þ
S2
À 95s þ 4
Hence, the system is Type 0. Evaluating Kp yields
Kp ¼
100
4
¼ 25
The steady-state error is given by
estep 1ð Þ ¼
1
1 þ Kp
¼
1
1 þ 25
¼ 3:846e À 02
7.6
Since G sð Þ ¼
K s þ 7ð Þ
s2 þ 2s þ 10
; e 1ð Þ ¼
1
1 þ Kp
¼
1
1 þ
7K
10
¼
10
10 þ 7K
.
Calculating the sensitivity, we get
Se:K ¼
K
e
@e
@K
¼
K
10
10 þ 7K

À10ð Þ7
10 þ 7Kð Þ2
¼ À
7K
10 þ 7K

E1SM 11/11/2010 9:29:16 Page 21
7.7
Given
A ¼
0 1
À3 À6
!
; B ¼
0
1
!
; C ¼ 1 1½ Š; R sð Þ ¼
1
s
:
Using the final value theorem,
estep 1ð Þ ¼ lim
s!0
sR sð Þ 1 À C sI À Að ÞÀ1
B
h i
¼ lim
s!0
1 À 1 1½ Š
s À1
3 s þ 6
#À1
0
1
#2
4
3
5
¼ lim
s!0
1 À 1 1½ Š
s þ 6 1
À3s s
#
s2 þ 6s þ 3
0
1
#
2
6
6
6
6
4
3
7
7
7
7
5
¼ lim
s!0
s2
þ 5s þ 2
s2 þ 6s þ 3
¼
2
3
Using input substitution,
step 1ð Þ ¼ 1 þ CAÀ1
B ¼ 1 À 1 1½ Š
0 1
À3 À6
#À1
0
1
#
¼ 1 þ 1 1½ Š
À6 À1
3 0
#
3
0
1
#
¼ 1 þ 1 1½ Š
À
1
3
0
2
4
3
5 ¼
2
3
CHAPTER 8
8.1
a. F À7 þ j9ð Þ ¼
À7 þ j9 þ 2ð Þ À7 þ j9 þ 4ð Þ0:0339
À7 þ j9ð Þ À7 þ j9 þ 3ð Þ À7 þ j9 þ 6ð Þ
¼
À5 þ j9ð Þ À3 þ j9ð Þ
À7 þ j9ð Þ À4 þ j9ð Þ À1 þ j9ð Þ
¼
À66 À j72ð Þ
944 À j378ð Þ
¼ À0:0339 À j0:0899 ¼ 0:096 À110:7
b. The arrangement of vectors is shown as follows:
jw
s
s-plane
X X
–2–4 –1–3–6 –5–7
X
M1 M2 M3 M4 M5
(–7+j9)
0

E1SM 11/11/2010 9:29:16 Page 22
From the diagram,
F À7 þ j9ð Þ ¼
M2M4
M1M3M5
¼
À3 þ j9ð Þ À5 þ j9ð Þ
À1 þ j9ð Þ À4 þ j9ð Þ À7 þ j9ð Þ
¼
À66 À j72ð Þ
944 À j378ð Þ
¼ À0:0339 À j0:0899 ¼ 0:096 ; À110:7
8.2
a. First draw the vectors.
jw
s
s-plane
X
X
–2–3 0–1
j1
j2
j3
–j1
–j2
–j3
From the diagram,
P
angles ¼ 180
À tanÀ1 À3
À1

À tanÀ1 À3
1

¼ 180
À 108:43
þ 108:43
¼ 180
:
b. Since the angle is 180
, the point is on the root locus.
c. K ¼
P pole lengths
P zero lengths
¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
12
þ 32
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
12
þ 32
p
1
¼ 10
8.3
First, find the asymptotes.
sa ¼
P
poles À
P
zeros
# poles À # zeros
¼
À2 À 4 À 6ð Þ À 0ð Þ
3 À 0
¼ À4
ua ¼
2k þ 1ð Þp
3
¼
p
3
; p;
5p
3

E1SM 11/11/2010 9:29:17 Page 23
Next draw root locus following the rules for sketching.
–8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3
–5
–4
–3
–2
–1
0
1
2
3
4
5
Real Axis
ImagAxis
8.4
a.
j3
s
jw
s-plane
X
X
O
–2 2
–j3
0
b. Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function.
T sð Þ ¼
G sð Þ
1 þ G sð Þ
¼
K s þ 2ð Þ
s2 þ K À 4ð Þs þ 2K þ 13ð Þ
Using the denominator of T(s), make a Routh table.
We get a row of zeros for K ¼ 4. From the s2
row with K ¼ 4; s2
þ 21 ¼ 0. From
which we evaluate the imaginary axis crossing at
ffiffiffiffiffi
21
p
.
c. From part (b), K ¼ 4.
d. Searching for the minimum gain to the left of À2 on the real axis yields À7 at a
gain of 18. Thus the break-in point is at À7.
s2
1 2K þ 13
s1
K À 40 0
s0
2K þ 13 0

E1SM 11/11/2010 9:29:17 Page 24
e. First, draw vectors to a point e close to the complex pole.
jw
s
s-plane
X
X
–2 20
j3
–j3
At the point e close to the complex pole, the angles must add up to zero. Hence,
angle from zero – angle from pole in 4th
quadrant – angle from pole in 1st
quadrant ¼
180
, or tanÀ1 3
4

À 90
À u ¼ 180
. Solving for the angle of departure,
u ¼ À233:1.
8.5
a.
jw
4–3
X
X
s-plane
o
0
z= 0.5
j4
–j4
s
2
o

E1SM 11/11/2010 9:29:18 Page 25
b. Search along the imaginary axis and find the 180
point at s ¼ Æj4:06.
c. For the result in part (b), K ¼ 1.
d. Searching between 2 and 4 on the real axis for the minimum gain yields the break-in
at s ¼ 2:89.
e. Searching along z ¼ 0:5 for the 180
point we find s ¼ À2:42 þ j4:18.
f. For the result in part (e), K ¼ 0:108.
g. Using the result from part (c) and the root locus, K 1.
8.6
a.
s
jw
z = 0.591
–2–4–6
XXX
0
s-plane
b. Searching along the z ¼ 0:591 (10% overshoot) line for the 180
point yields
À2:028 þ j2:768 with K ¼ 45:55.
c. Ts ¼
4
jRej
¼
4
2:028
¼ 1:97 s; Tp ¼
p
jImj
¼
p
2:768
¼ 1:13 s; vnTr ¼ 1:8346 from the
rise-time chart and graph in Chapter 4. Since vn is the radial distance to the pole,
vn ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2:0282
þ 2:7682
p
¼ 3:431. Thus, Tr ¼ 0:53 s; since the system is Type 0,
Kp ¼
K
2Ã
4Ã
6
¼
45:55
48
¼ 0:949. Thus,
estep 1ð Þ ¼
1
1 þ Kp
¼ 0:51:
d. Searching the real axis to the left of À6 for the point whose gain is 45.55, we find
À7:94. Comparing this value to the real part of the dominant pole, À2:028, we find
that it is not five times further. The second-order approximation is not valid.
8.7
Find the closed-loop transfer function and put it the form that yields pi as the root
locus variable. Thus,
T sð Þ ¼
G sð Þ
1 þ G sð Þ
¼
100
s2 þ pis þ 100
¼
100
s2 þ 100ð Þ þ pis
¼
100
s2 þ 100
1 þ
pis
s2 þ 100

E1SM 11/11/2010 9:29:18 Page 26
Hence, KG sð ÞH sð Þ ¼
pis
s2 þ 100
. The following shows the root locus.
s
jw
–j10X
O
s-plane
0
X j10
8.8
Following the rules for plotting the root locus of positive-feedback systems, we
obtain the following root locus:
s
jw
–2–4
X
s-plane
–1
XXo
0–3
8.9
The closed-loop transfer function is T sð Þ ¼
K s þ 1ð Þ
s2 þ K þ 2ð Þs þ K
. Differentiating the
denominator with respect to K yields
2s
@s
@K
þ K þ 2ð Þ
@s
@K
þ s þ 1ð Þ ¼ 2s þ K þ 2ð Þ
@s
@K
þ s þ 1ð Þ ¼ 0
Solving for
@s
@K
, we get
@s
@K
¼
À s þ 1ð Þ
2s þ K þ 2ð Þ
. Thus, Ss:K ¼
K
s
@s
@K
¼
ÀK s þ 1ð Þ
s 2s þ K þ 2ð Þ
:
Substituting K ¼ 20 yields Ss:K ¼
À10 s þ 1ð Þ
s s þ 11ð Þ
.
Now find the closed-loop poles when K ¼ 20. From the denominator of T sð Þ;
s1;2 ¼ À21:05; À 0:95, when K ¼ 20.
For the pole at À21:05,
Ds ¼ s Ss:Kð Þ
DK
K
¼ À12:05
À10 À21:05 þ 1ð Þ
À21:05 À21:05 þ 11ð Þ

0:05 ¼ À0:9975:

E1SM 11/11/2010 9:29:19 Page 27
For the pole at À0:95,
Ds ¼ s Ss:Kð Þ
DK
K
¼ 0:95
À10 À0:95 þ 1ð Þ
À0:95 À0:95 þ 11ð Þ

0:05 ¼ À0:0025:
CHAPTER 9
9.1
a. Searching along the 15% overshoot line, we find the point on the root locus at
À3:5 þ j5:8 at a gain of K ¼ 45:84. Thus, for the uncompensated system,
Kv ¼ lim
s!0
sG sð Þ ¼ K=7 ¼ 45:84=7 ¼ 6:55.
Hence, eramp uncompensated 1ð Þ ¼ 1=Kv ¼ 0:1527.
b. Compensator zero should be 20x further to the left than the compensator pole.
Arbitrarily select Gc sð Þ ¼
s þ 0:2ð Þ
s þ 0:01ð Þ
.
c. Insert compensator and search along the 15% overshoot line and find the root
locus at À3:4 þ j5:63 with a gain, K ¼ 44:64. Thus, for the compensated system,
Kv ¼
44:64 0:2ð Þ
7ð Þ 0:01ð Þ
¼ 127:5 and eramp compensated 1ð Þ ¼
1
Kv
¼ 0:0078.
d.
eramp uncompensated
eramp compensated
¼
0:1527
0:0078
¼ 19:58
9.2
À3:5 þ j5:8 at a gain of K ¼ 45:84. Thus, for the uncompensated system,
Ts ¼
4
jRej
¼
4
3:5
¼ 1:143 s:
b. The real part of the design point must be three times larger than the un-
compensated pole’s real part. Thus the design point is 3 À3:5ð Þþ j 3 5:8ð Þ ¼
À10:5 þ j 17:4. The angular contribution of the plant’s poles and compensator
zero at the design point is 130:8
. Thus, the compensator pole must contribute
180
À 130:8
¼ 49:2
. Using the following diagram,
–pc
s
jw
s-plane
–10.5
j17.4
49.2°
we find
17:4
Pc À 10:5
¼ tan 49:2
, from which, pc ¼ 25:52. Adding this pole, we find
the gain at the design point to be K ¼ 476:3. A higher-order closed-loop pole is
found to be at À11:54. This pole may not be close enough to the closed-loop zero
at À10. Thus, we should simulate the system to be sure the design requirements
have been met.

E1SM 11/11/2010 9:29:19 Page 28
9.3
À3:5 þ 6:83 at a gain of K ¼ 58:9. Thus, for the uncompensated system,
Ts ¼
4
jRej
¼
4
3:5
¼ 1:143 s:
b. For the uncompensated system, Kv ¼ lim
s!0
sG sð Þ ¼ K=7 ¼ 58:9=7 ¼ 8:41. Hence,
eramp uncompensated 1ð Þ ¼ 1=Kv ¼ 0:1189.
c. In order to decrease the settling time by a factor of 2, the design point is twice the
uncompensated value, or À7 þ j 13:66. Adding the angles from the plant’s poles
and the compensator’s zero at À3 to the design point, we obtain À100:8
. Thus,
the compensator pole must contribute 180
À 100:8
¼ 79:2
. Using the following
diagram,
–pc
s
jw
s-plane
79.2°
–7
j13.66
we find
13:66
Pc À 7
¼ tan79:2
, from which, pc ¼ 9:61. Adding this pole, we find the
gain at the design point to be K ¼ 204:9.
Evaluating Kv for the lead-compensated system:
Kv ¼ lim
s!0
sG sð ÞGlead ¼ K 3ð Þ= 7ð Þ 9:61ð Þ½ Š ¼ 204:9ð Þ 3ð Þ= 7ð Þ 9:61ð Þ½ Š ¼ 9:138:
Kv for the uncompensated system was 8.41. For a 10x improvement in steady-state
error, Kv must be 8:41ð Þ 10ð Þ ¼ 84:1. Since lead compensation gave us Kv ¼ 9:138,
we need an improvement of 84:1=9:138 ¼ 9:2. Thus, the lag compensator zero
should be 9.2x further to the left than the compensator pole. Arbitrarily select
Gc sð Þ ¼
s þ 0:092ð Þ
s þ 0:01ð Þ
.
Using all plant and compensator poles, we find the gain at the design point to
be K ¼ 205:4. Summarizing the forward path with plant, compensator, and gain
yields
Ge sð Þ ¼
205:4 s þ 3ð Þ s þ 0:092ð Þ
s s þ 7ð Þ 9:61ð Þ s þ 0:01ð Þ
:
Higher-order poles are found at À0:928 and À2:6. It would be advisable to simulate
the system to see if there is indeed pole-zero cancellation.

E1SM 11/11/2010 9:29:19 Page 29
9.4
The configuration for the system is shown in the figure below.
1
s(s+ 7)(s +10)
R(s) C(s)+
K
+
––
Kfs
Minor-Loop Design:
For the minor loop, G sð ÞH sð Þ ¼
Kf
. Using the following diagram, we
find that the minor-loop root locus intersects the 0.7 damping ratio line at
À8:5 þ j 8:67. The imaginary part was found as follows: u ¼ cosÀ1
z ¼ 45:57
. Hence,
Im
8:5
¼ tan45:57
, from which Im ¼ 8:67.
s
jw
s-plane
–7
z = 0.7
X X
−10 −8.5
(-8.5 + j8.67)
q
Im
The gain, Kf , is found from the vector lengths as
Kf ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:52
þ 8:672
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:52
þ 8:672
p
¼ 77:42
Major-Loop Design:
Using the closed-loop poles of the minor loop, we have an equivalent forward-path
transfer function of
Ge sð Þ ¼
K
s s þ 8:5 þ j8:67ð Þ s þ 8:5 À j8:67ð Þ
¼
K
s s2 þ 17s þ 147:4ð Þ
:
Using the three poles of Ge sð Þ as open-loop poles to plot a root locus, we search
along z ¼ 0:5 and find that the root locus intersects this damping ratio line at À4:34 þ
j7:51 at a gain, K ¼ 626:3.

E1SM 11/11/2010 9:29:20 Page 30
9.5
a. An active PID controller must be used. We use the circuit shown in the following
figure:
+
–
Z1(s)
Z2(s)
I1(s)
V1(s)
Vo(s)
Vi(s)
Ia(s)
I2(s)
where the impedances are shown below as follows:
C1
R1
Z1(s) Z2(s)
C2R2
Matching the given transfer function with the transfer function of the PID
controller yields
Gc sð Þ ¼
s þ 0:1ð Þ s þ 5ð Þ
s
¼
s2
þ 5:1s þ 0:5
s
¼ s þ 5:1 þ
0:5
8
¼ À
R2
R1
þ
C1
C2

þ R2C1s þ
1
R1C2
s
2
6
6
4
3
7
7
5
Equating coefficients
1
R1C2
¼ 0:5 ð1Þ
R2C1 ¼ 1 ð2Þ
R2
R1
þ
C1
C2

¼ 5:1 ð3Þ
In Eq. (2) we arbitrarily let C1 ¼ 10À5
. Thus, R2 ¼ 105
. Using these values along with
Eqs. (1) and (3) we find C2 ¼ 100mF and R1 ¼ 20 kV.
b. The lag-lead compensator can be implemented with the following passive net-
work, since the ratio of the lead pole-to-zero is the inverse of the ratio of the lag
pole-to-zero:
R1
C1
R2
C2
+
–
+
–
vo(t)vi(t)

E1SM 11/11/2010 9:29:20 Page 31
Matchingthegiventransferfunctionwiththetransferfunctionofthepassivelag-lead
compensator yields
Gc sð Þ ¼
s þ 0:01ð Þ s þ 20ð Þ
¼
s2 þ 20:01s þ 0:2
¼
s þ
1
R1C1

s þ
1
R2C2

s2 þ
1
R1C1
þ
1
R2C2
þ
1
R2C1

s þ
1
R1R2C1C2
Equating coefficients
1
R1C1
¼ 0:1 ð1Þ
1
R2C2
¼ 0:1 ð2Þ
1
R1C1
þ
1
R2C2
þ
1
R2C1

¼ 20:01 ð3Þ
Substituting Eqs. (1) and (2) in Eq. (3) yields
1
R2C1
¼ 17:91 ð4Þ
Arbitrarily letting C1 ¼ 100 mF in Eq. (1) yields R1 ¼ 100 kV.
Substituting C1 ¼ 100 mF into Eq. (4) yields R2 ¼ 558 kV.
Substituting R2 ¼ 558 kV into Eq. (2) yields C2 ¼ 900 mF.
CHAPTER 10
10.1
a.
G sð Þ ¼
1
; G jvð Þ ¼
1
8 þ v2ð Þ þ j6v
M vð Þ ¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
8 À v2ð Þ
2
þ 6vð Þ2
q
For v
ffiffiffi
8
p
, f vð Þ ¼ ÀtanÀ1 6v
8 À v2

:
For v
ffiffiffi
8
p
, f vð Þ ¼ À p þ tanÀ1 6v
8 À v2
!
:
b.
Frequency (rad/sec)
Phase(deg);Magnitude(dB)
Bode Diagrams
10–1
100
101 102
–200
–150
–100
–50
0
–100
–80
–60
–40
–20
0

E1SM 11/11/2010 9:29:20 Page 32
c.
Real Axis
ImaginaryAxis
Nyquist Diagrams
–0.05 0 0.05 0.1 0.15 0.2
–0.08
–0.06
–0.04
–0.02
0
0.02
0.04
0.06
0.08
10.2
20logM
–120
–100
–80
–60
–40
0.1 1 10 100 1000
–40 dB/dec
–20 dB/dec
–20 dB/dec
–40 dB/dec
Frequency (rad/s)
Actual
Asymptotic
Frequency (rad/s)
Phase(degrees)
–150
–100
–50
–200
–45o
/dec
–90o
/dec
–45o
/dec
–90o
/dec
–45o/dec
–45o
/dec
0.1 1 10 100
Actual
Asymptotic
1000

E1SM 11/11/2010 9:29:21 Page 33
10.3
The frequency response is 1/8 at an angle of zero degrees at v ¼ 0. Each pole rotates
90
in going from v ¼ 0 to v ¼ 1. Thus, the resultant rotates À180
while its
magnitude goes to zero. The result is shown below.
Re
Im
0 1
8
w = ∞ w = 0
10.4
a. The frequency response is 1/48 at an angle of zero degrees at v ¼ 0. Each pole
rotates 90
in going from v ¼ 0 to v ¼ 1. Thus, the resultant rotates À270
while
its magnitude goes to zero. The result is shown below.
Im
Re
1
48
w = 0w = ∞
0
w = 6.63
1
480
–
b. Substituting jv into G sð Þ ¼
1
¼
1
s3 þ 12s2 þ 44s þ 48
and
simplifying, we obtain G jvð Þ ¼
48 À 12v2
À Á
À j 44v À v3
À Á
v6 þ 56v4 þ 784v2 þ 2304
. The Nyquist diagram
crosses the real axis when the imaginary part of G jvð Þ is zero. Thus, the Nyquist
diagram crosses the real axis at v2
¼ 44; or v ¼
ffiffiffiffiffi
44
p
¼ 6:63 rad=s. At this fre-
quency G jvð Þ ¼ À
1
480
. Thus, the system is stable for K 480.

E1SM 11/11/2010 9:29:22 Page 34
10.5
If K ¼ 100, the Nyquist diagram will intersect the real axis at À100=480. Thus,
GM ¼ 20 log
480
100
¼ 13:62 dB. From Skill-Assessment Exercise Solution 10.4, the
180
frequency is 6.63 rad/s.
10.6
a.
0001001011
–180
–160
–140
–120
–100
–80

E1SM 11/11/2010 9:29:22 Page 35
10.8
For both parts find that G jvð Þ ¼
160
27
Ã
6750000 À 101250v2
À Á
þ j1350 v2
À 1350
À Á
v
v6 þ 2925v4 þ 1072500v2 þ 25000000
.
For a range of values for v, superimpose G jvð Þ on the a. M and N circles, and on the
b. Nichols chart.
a.
M = 1.3
1.4
1.5
1.6
1.8
2.0
M = 0.7
0.6
0.5
0.4
M = 1.0
–2
–1
1
2
3
Im
Re
1 2–1–2–3
–3
Φ = 20°
25°
30°
–20°
–40°
–50°
–30°
–70°
40°
50°
70°
–25°
G-plane
–4
b.
Open-Loop Phase (deg)
Open-LoopGain(dB)
Nichols Charts
–350 –300 –250 –200 –150 –100 –50 0
–200
–150
–100
–50
0 6 dB
3 dB
1 dB
0.5 dB
0.25 dB
0 dB
–1 dB
–3 dB
–6 dB
–12 dB
–20 dB
–40 dB
–60 dB
–80 dB
–100 dB
–120 dB
–140 dB
–160 dB
–180 dB
–200 dB
–220 dB
–240 dB

E1SM 11/11/2010 9:29:22 Page 36
Plotting the closed-loop frequency response from a. or b. yields the following plot:
–120
–100
–80
–60
–40
–20
0
Frequency (rad/s)
20logM
0001001011
Frequency (rad/s)
1 10 100 1000
–300
–250
–200
–150
–100
–50
0
Phase(degrees)
10.9
The open-loop frequency response is shown in the following figure:
Frequency (rad/sec)
Bode Diagrams
–40
–20
0
20
40
10–1
100
101
102
–160
–140
–120
–100

E1SM 11/11/2010 9:29:24 Page 37
The open-loop frequency response is À7 at v ¼ 14:5 rad=s. Thus, the estimated
bandwidth is vWB ¼ 14:5 rad=s. The open-loop frequency response plot goes
through zero dB at a frequency of 9.4 rad/s, where the phase is 151:98
.
Hence, the phase margin is 180
À 151:98
¼ 28:02
. This phase margin corre-
sponds to
z ¼ 0:25: Therefore; %OS ¼ eÀ zp=
1Àz2
pÀ Á
Â 100 ¼ 44:4%
Ts ¼
4
vBWz
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 À 2z2ð Þ þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4z4 À 4z2 þ 2
pq
¼ 1:64 s and
Tp ¼
p
vBW
1 À z2
p
1 À 2z2ð Þ þ
4z4 À 4z2 þ 2
pq
¼ 0:33 s
10.10
The initial slope is 40 dB/dec. Therefore, the system is Type 2. The initial slope
intersects 0 dB at v ¼ 9:5 rad=s. Thus, Ka ¼ 9:52
¼ 90:25 and Kp ¼ Kv ¼ 1.
10.11
a. Without delay, G jvð Þ ¼
10
jv jv þ 1ð Þ
¼
10
v Àv þ jð Þ
, from which the zero dB fre-
quency is found as follows: M ¼
10
v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v2 þ 1
p ¼ 1. Solving for v; v
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
v2 þ 1
p
¼ 10, or
after squaring both sides and rearranging, v4
þ v2
À 100 ¼ 0. Solving for the
roots, v2
¼ À10:51; 9:51. Taking the square root of the positive root, we find the
0 dB frequency to be 3.08 rad/s. At this frequency, the phase angle, f ¼
Àff Àv þ jð Þ ¼ Àff À3:08 þ jð Þ ¼ À162
. Therefore the phase margin is 180
À162
¼ 18
.
b. With a delay of 0.1 s,
f ¼ Àff Àv þ jð Þ À vT ¼ Àff À3:08 þ jð Þ À 3:08ð Þ 0:1ð Þ 180=pið Þ
¼ À162 À 17:65 ¼ À179:65
Therefore the phase margin is 180
À 179:65
¼ 0:35
. Thus, the system is table.
c. With a delay of 3 s,
f ¼ Àff Àv þ jð Þ À vT ¼ Àff À3:08 þ jð Þ À 3:08ð Þ 3ð Þ 180=pið Þ ¼ À162
À 529:41
¼ À 691:41
¼ 28:59 deg:
Therefore the phase margin is 28:59 À 180 ¼ À151:41 deg. Thus, the system is
unstable.
10.12
Drawing judicially selected slopes on the magnitude and phase plot as shown below
yields a first estimate.

E1SM 11/11/2010 9:29:24 Page 38
Experimental
Frequency (rad/sec)
–40
–30
–20
–10
0
10
20
1 2 3 4 5 6 7 8 10 20 30 40 50 70 100 200 300 500 1000
–95
–90
–85
–80
–75
–70
–65
–60
–55
–50
–45
Phase(deg)Gain(dB)
We see an initial slope on the magnitude plot of À20 dB/dec. We also see a final
À20 dB/dec slope with a break frequency around 21 rad/s. Thus, an initial estimate is
G1 sð Þ ¼
1
s s þ 21ð Þ
. Subtracting G1 sð Þ from the original frequency response yields the
frequency response shown below.
40
50
60
70
80
90
1 2 3 4 5 6 7 8 10 20 30 40 50 70 100 200 300 500 1000
0
20
40
60
80
100
Experimental Minus 1/s(s+21)
Frequency (rad/sec)
Gain(dB)Phase(deg)

E1SM 11/11/2010 9:29:24 Page 39
Drawing judicially selected slopes on the magnitude and phase plot as shown
yields a final estimate. We see first-order zero behavior on the magnitude and
phase plots with a break frequency of about 5.7 rad/s and a dc gain of about
44 dB ¼ 20log 5:7Kð Þ, or K ¼ 27:8. Thus, we estimate G2 sð Þ ¼ 27:8 s þ 7ð Þ. Thus,
G sð Þ ¼ G1 sð ÞG2 sð Þ ¼
27:8 s þ 5:7ð Þ
s s þ 21ð Þ
. It is interesting to note that the original problem
was developed from G sð Þ ¼
30 s þ 5ð Þ
s s þ 20ð Þ
.
CHAPTER 11
11.1
The Bode plot for K ¼ 1 is shown below.
Frequency (rad/sec)
Bode Diagrams
–180
–160
–140
–120
–100
–80
–60
10–1 100 101 102 103
–250
–200
–150
–100
A 20% overshoot requires z ¼
Àlog
%
100

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p2 þ log2 %
100
s ¼ 0:456. This damping ratio implies a
phase margin of 48.10, which is obtained when the phase angle ¼ À1800 þ 48:10
¼ 131:9
. This phase angle occurs at v ¼ 27:6 rad=s. The magnitude at this frequency
is 5:15 Â 10À6
. Since the magnitude must be unity K ¼
1
5:15 Â 10À6
¼ 194; 200.
11.2
To meet the steady-state error requirement, K ¼ 1; 942; 000. The Bode plot for this
gain is shown below.

E1SM 11/11/2010 9:29:25 Page 40
Frequency (rad/sec)
Bode Diagrams
–40
–20
0
20
40
60
10–1 100 101 102 103
–250
–200
–150
–100
Àlog
%
100

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p2 þ log2 %
100
s ¼ 0:456. This damping ratio
implies a phase margin of 48:1
. Adding 10
to compensate for the phase angle
contribution of the lag, we use 58:1
. Thus, we look for a phase angle of
À180
þ 58:1
¼ À129:9
. The frequency at which this phase occurs is 20.4 rad/s.
At this frequency the magnitude plot must go through zero dB. Presently, the
magnitude plot is 23.2 dB. Therefore draw the high frequency asymptote of the lag
compensator at À23:2 dB. Insert a break at 0:1 20:4ð Þ ¼ 2:04 rad=s. At this frequency,
draw À23:2 dB/dec slope until it intersects 0 dB. The frequency of intersection
will be the low frequency break or 0.141 rad/s. Hence the compensator is
Gc sð Þ ¼ Kc
s þ 2:04ð Þ
s þ 0:141ð Þ
, where the gain is chosen to yield 0 dB at low frequencies,
or Kc ¼ 0:141=2:04 ¼ 0:0691. In summary,
Gc sð Þ ¼ 0:0691
s þ 2:04ð Þ
s þ 0:141ð Þ
and G sð Þ ¼
1;942;000
11.3
Àlog
%
100

p2 þ log2 %
100
s ¼ 0:456. The required bandwidth
is then calculated as vBW ¼
4
Tsz
1 À 2z2ð Þ þ
4z4 À 4z2 þ 2
pq
¼ 57:9 rad/s. In order
to meet the steady-state error requirement of Kv ¼ 50 ¼
K
50ð Þ 120ð Þ
, we calculate
K ¼ 300; 000. The uncompensated Bode plot for this gain is shown below.

E1SM 11/11/2010 9:29:25 Page 41
Frequency (rad/sec)
Bode Plot for K = 300000
–60
–40
–20
0
20
40
10–1
100
101
102
103
–250
–200
–150
–100
The uncompensated system’s phase margin measurement is taken where the
magnitude plot crosses 0 dB. We find that when the magnitude plot crosses 0 dB,
the phase angle is À144:8
. Therefore, the uncompensated system’s phase margin is
À180
þ 144:8
¼ 35:2
. The required phase margin based on the required damping
ratio is FM ¼ tanÀ1 2z
À2z2 þ
1 þ 4z4
pq ¼ 48:1
. Adding a 10
correction factor, the
required phase margin is 58:1
. Hence, the compensator must contribute
fmax ¼ 58:1
À 35:2
¼ 22:9
. Using fmax ¼ sinÀ1 1 À b
1 þ b
, b ¼
1 À sinfmax
1 þ sinfmax
¼ 0:44.
The compensator’s peak magnitude is calculated as Mmax ¼
1
ffiffiffi
b
p ¼ 1:51. Now find
the frequency at which the uncompensated system has a magnitude 1=Mmax, or
À3:58 dB. From the Bode plot, this magnitude occurs at vmax ¼ 50 rad=s. The
compensator’s zero is at zc ¼
1
T
. vmax ¼
1
T
ffiffiffi
b
p Therefore, zc ¼ 33:2.
The compensator’s pole is at Pc ¼
1
bT
¼
zc
b
¼ 75:4. The compensator gain is
chosen to yield unity gain at dc.
Hence, Kc ¼ 75:4=33:2 ¼ 2:27. Summarizing, Gc sð Þ ¼ 2:27
s þ 33:2ð Þ
s þ 75:4ð Þ
, and
G sð Þ
300;000
.

E1SM 11/11/2010 9:29:26 Page 42
11.4
Àlog
%
100

p2 þ log2 %
100
s ¼ 0:591. The required bandwidth
is then calculated as vBW ¼
p
Tp
1 À z2
p
1 À 2z2ð Þ þ
4z4 À 4z2 þ 2
pq
¼ 7:53 rad=s:
In order to meet the steady-state error requirement of Kv ¼ 10 ¼
K
8ð Þ 30ð Þ
, we
calculate K ¼ 2400. The uncompensated Bode plot for this gain is shown below.
Frequency (rad/sec)
Bode Diagrams
–100
–80
–60
–40
–20
0
20
40
10–1
100 101 102 103
–250
–200
–150
–100
Let us select a new phase-margin frequency at 0:8vBW ¼ 6:02 rad=s. The
required phase margin based on the required damping ratio is FM ¼ tanÀ1
2z
À2z2 þ
1 þ 4z4
pq ¼ 58:6
. Adding a 5
correction factor, the required phase
margin is 63:6
. At 6.02 rad/s, the new phase-margin frequency, the phase angle
is–which represents a phase margin of 180
À 138:3
¼ 41:7
. Thus, the lead com-
pensator must contribute fmax ¼ 63:6
À 41:7
¼ 21:9
.
Using fmax ¼ sinÀ1 1 À b
1 þ b
, b ¼
1 À sinfmax
1 þ sinfmax
¼ 0:456.
We now design the lag compensator by first choosing its higher break frequency one
decade below the new phase-margin frequency, that is, zlag ¼ 0:602 rad=s. The lag
compensator’s pole is plag ¼ bzlag ¼ 0:275. Finally, the lag compensator’s gain is
Klag ¼ b ¼ 0:456.

E1SM 11/11/2010 9:29:27 Page 43
Now we design the lead compensator. The lead zero is the product of the new
phase margin frequency and
ffiffiffi
b
p
, or zlead ¼ 0:8vBW
ffiffiffi
b
p
¼ 4:07. Also, plead ¼
zlead
b
¼ 8:93. Finally, Klead ¼
1
b
¼ 2:19. Summarizing,
Glag ¼ sð Þ ¼ 0:456
s þ 0:602ð Þ
s þ 0:275ð Þ
; Glead sð Þ ¼ 2:19
s þ 4:07ð Þ
s þ 8:93ð Þ
; and k ¼ 2400:
CHAPTER 12
12.1
We first find the desired characteristic equation. A 5% overshoot requires
z ¼
Àlog
%
100

p2 þ log2 %
100
s ¼ 0:69. Also, vn ¼
p
Tp
1 À z2
p ¼ 14:47 rad=s. Thus, the char-
acteristic equation is s2
þ 2zvns þ v2
n ¼ s2
þ 19:97s þ 209:4. Adding a pole at À10 to
cancel the zero at À10 yields the desired characteristic equation,
s2
þ 19:97s þ 209:4
À Á
s þ 10ð Þ ¼ s3
þ 29:97s2
þ 409:1s þ 2094. The compensated sys-
tem matrix in phase-variable form is
A À BK ¼
0 1 0
0 0 1
À k1ð Þ À 36 þ k2ð Þ À 15 þ k3ð Þ
2
6
4
3
7
5. The characteristic equation for this
system is jsI À A À BKð Þj ¼ s3
þ 15 þ k3ð Þs2
þ 36 þ k2ð Þs þ k1ð Þ. Equating coeffi-
cients of this equation with the coefficients of the desired characteristic equation
yields the gains as
K ¼ k1 k2 k3½ Š ¼ 2094 373:1 14:97½ Š:
12.2
The controllability matrix is CM ¼ B AB A2
B
Â Ã
¼
2 1 1
1 4 À9
1 À1 16
2
6
4
3
7
5. Since
jCMj ¼ 80, CM is full rank, that is, rank 3. We conclude that the system is controllable.
12.3
First check controllability. The controllability matrix is CMz ¼ B AB A2
B
Â Ã
¼
0 0 1
0 1 À17
1 À9 81
2
6
4
3
7
5. Since jCMzj ¼ À1, CMz is full rank, that is, rank 3. We conclude that
the system is controllable. We now find the desired characteristic equation. A 20%
overshoot requires z ¼
Àlog
%
100

p2 þ log2 %
100
s ¼ 0:456. Also, vn ¼
4
zTs
¼ 4:386 rad=s.

E1SM 11/11/2010 9:29:27 Page 44
Thus, the characteristic equation is s2
þ 2zvns þ v2
n ¼ s2
þ 4s þ 19:24. Adding a pole
at À6 to cancel the zero at À6 yields the resulting desired characteristic equation,
s2
þ 4s þ 19:24
À Á
s þ 6ð Þ ¼ s3
þ 10s2
þ 43:24s þ 115:45:
Since G sð Þ ¼
s þ 6ð Þ
¼
s þ 6
s3 þ 24s2 þ 191s þ 504
, we can write the phase-
variable representation as Ap ¼
0 1 0
0 0 1
À504 À191 À24
2
6
4
3
7
5 ; Bp ¼
0
0
1
2
6
4
3
7
5; Cp ¼
6 1 0½ Š. The compensated system matrix in phase-variable form is Ap À BpKp ¼
0 1 0
0 0 1
À 504 þ k1ð Þ À 191 þ k2ð Þ À 24 þ k3ð Þ
2
6
4
3
7
5. The characteristic equation for this
system is jsI À Ap À BpKp
À Á
j ¼ s3
þ 24 þ k3ð Þs2
þ 191 þ k2ð Þs þ 504 þ k1ð Þ. Equat-
ing coefficients of this equation with the coefficients of the desired characteristic
equation yields the gains as Kp ¼ k1 k2 k3½ Š ¼ À388:55 À147:76 À14½ Š. We
now develop the transformation matrix to transform back to the z-system.
CMz ¼ Bz AzBz A2
zB z
h i
¼
0 0 1
0 1 À17
1 À9 81
2
6
4
3
7
5 and
CMp ¼ Bp ApBp A2
pBp
h i
¼
0 0 1
0 1 À24
1 À24 385
2
6
4
3
7
5:
Therefore,
P ¼ CMzCÀ1
Mx ¼
0 0 1
0 1 À17
1 À9 81
2
6
4
3
7
5
191 24 1
24 1 0
1 0 0
2
6
4
3
7
5 ¼
1 0 0
7 1 0
56 15 1
2
6
4
3
7
5
Hence, Kz ¼ KpPÀ1
¼ À388:55 À147:76 À14½ Š
1 0 0
À7 1 0
49 À15 1
2
6
4
3
7
5
¼ À40:23 62:24 À14½ Š.
12.4
For the given system e_x ¼ A À LCð Þex ¼
À 24 þ l1ð Þ 1 0
À 191 þ l2ð Þ 0 1
À 504 þ l3ð Þ 0 0
2
6
4
3
7
5ex. The characteristic
polynomial is given by j sI À A À LCð Þ½ Šj ¼ s3
þ 24 þ l1ð Þs2
þ 191 þ l2ð Þs þ
504 þ l3ð Þ. Now we find the desired characteristic equation. The dominant poles
from Skill-Assessment Exercise 12.3 come from s2
þ 4s þ 19:24
À Á
. Factoring yields
À2 þ j3:9ð Þ and À2 À j3:9ð Þ. Increasing these poles by a factor of 10 and adding a
third pole 10 times the real part of the dominant second-order poles yields the

E1SM 11/11/2010 9:29:28 Page 45
desired characteristic polynomial, s þ 20 þ j39ð Þ s þ 20 À j39ð Þ s þ 200ð Þ ¼ s3
þ 240s2
þ 9921s þ 384200. Equating coefficients of the desired characteristic equation to the
system’s characteristic equation yields L ¼
216
9730
383696
2
6
4
3
7
5.
12.5
The observability matrix is OM ¼
C
CA
CA2
2
6
4
3
7
5 ¼
4 6 8
À64 À80 À78
674 848 814
2
6
4
3
7
5, where
A2
¼
25 28 32
À7 À4 À11
77 95 94
2
6
4
3
7
5. The matrix is of full rank, that is, rank 3, since
jOMj ¼ À1576. Therefore the system is observable.
12.6
The system is represented in cascade form by the following state and output
equations:
_z ¼
À7 1 0
0 À8 1
0 0 À9
2
6
4
3
7
5z þ
0
0
1
2
6
4
3
7
5u
y ¼ 1 0 0½ Šz
The observability matrix is OMz ¼
Cz
CzAz
CzA2
z
2
6
4
3
7
5 ¼
1 0 0
À7 1 0
49 À15 1
2
6
4
3
7
5,
where A2
z ¼
49 À15 1
0 64 À17
0 0 81
2
6
4
3
7
5. Since G sð Þ ¼
1
¼
1
s3 þ 24s2 þ 191s þ 504
, we can write the observable canonical form as
_x ¼
À24 1 0
À191 0 1
À504 0 0
2
6
4
3
7
5x þ
0
0
1
2
6
4
3
7
5u
y ¼ 1 0 0½ Šx
The observability matrix for this form is OMx ¼
Cx
CxAx
CxA2
x
2
6
4
3
7
5 ¼
1 0 0
À24 1 0
385 À24 1
2
6
4
3
7
5,

E1SM 11/11/2010 9:29:28 Page 46
where
A2
x ¼
385 À24 1
4080 À191 0
12096 504 0
2
6
4
3
7
5:
We next find the desired characteristic equation. A 10% overshoot requires
z ¼
Àlog
%
100

p2 þ log2 %
100
s ¼ 0:591. Also, vn ¼
4
zTs
¼ 67:66 rad=s. Thus, the characteristic
equation is s2
þ 2zvns þ v2
n ¼ s2
þ 80s þ 4578:42. Adding a pole at À400, or 10 times
the real part of the dominant second-order poles, yields the resulting desired character-
istic equation, s2
þ 80s þ 4578:42
À Á
s þ 400ð Þ ¼ s3
þ 480s2
þ 36580s þ 1:831x106
. For
the system represented in observable canonical form e_x ¼ Ax À LxCxð Þ ex ¼
À 24 þ l1ð Þ 1 0
À 191 þ l2ð Þ 0 1
À 504 þ l3ð Þ 0 0
2
6
4
3
7
5ex. The characteristic polynomial is given by
j sI À Ax À LxCxð Þ½ Šj ¼ s3
þ 24 þ l1ð Þs2
þ 191 þ l2ð Þs þ 504 þ l3ð Þ. Equating coeffi-
cients of the desired characteristic equation to the system’s characteristic equation
yields Lx ¼
456
36; 389
1; 830; 496
2
6
4
3
7
5.
Now, develop the transformation matrix between the observer canonical and
cascade forms.
P ¼ OÀ1
MzOMx ¼
1 0 0
À7 1 0
49 À15 1
2
6
6
6
4
3
7
7
7
5
À1
1 0 0
À24 1 0
385 À24 1
2
6
6
6
4
3
7
7
7
5
¼
1 0 0
7 1 0
56 15 1
2
6
6
6
4
3
7
7
7
5
1 0 0
À24 1 0
385 À24 1
2
6
6
6
4
3
7
7
7
5
¼
1 0 0
À17 1 0
81 À9 1
2
6
6
6
4
3
7
7
7
5
Finally,
Lz ¼ PLx ¼
1 0 0
À17 1 0
81 À9 1
2
6
4
3
7
5
456
36; 389
1; 830; 496
2
6
4
3
7
5 ¼
456
28; 637
1; 539; 931
2
6
4
3
7
5 %
456
28; 640
1; 540; 000
2
6
4
3
7
5.

E1SM 11/11/2010 9:29:29 Page 47
12.7
We first find the desired characteristic equation. A 10% overshoot requires
z ¼
Àlog
%
100

p2 þ log2 %
100
s ¼ 0:591
Also, vn ¼
p
Tp
1 À z2
p ¼ 1:948 rad=s. Thus, the characteristic equation is s2
þ
2zvns þ v2
n ¼ s2
þ 2:3s þ 3:79. Adding a pole at À4, which corresponds to the
original system’s zero location, yields the resulting desired characteristic equation,
s2
þ 2:3s þ 3:79
À Á
s þ 4ð Þ ¼ s3
þ 6:3s2
þ 13s þ 15:16.
Now,
_x
_xN
!
¼
A À BKð Þ BKe
À C 0
!
x
xN
!
þ
0
1
!
r; and y ¼ C 0½ Š
x
xN
!
,
where
A À BK ¼
0 1
À7 À9
!
À
0
1
!
k1 k2½ Š ¼
0 1
À7 À9
!
À
0 0
k1 k2
!
¼
0 1
À 7 þ k1ð Þ À 9 þ k2ð Þ
!
C ¼ 4 1½ Š
Bke ¼
0
1
!
ke ¼
0
ke
!
Thus,
_x1
_x2
_xN
2
6
4
3
7
5 ¼
0 1 0
À 7 þ k1ð Þ À 9 þ k2ð Þ ke
À4 À1 0
2
6
4
3
7
5
x1
x2
xN
2
6
4
3
7
5 þ
0
1
!
r; y ¼ 4 1 0½ Š
x1
x2
xN
2
6
4
3
7
5.
Finding the characteristic equation of this system yields
sI À
A À BKð Þ BKe
ÀC 0
!

¼
s 0 0
0 s 0
0 0 s
2
6
4
3
7
5 À
0 1 0
À 7 þ k1ð Þ À 9 þ k2ð Þ ke
À4 À1 0
2
6
4
3
7
5

¼
s À1 0
7 þ k1ð Þ s þ 9 þ k2ð Þ Àke
4 1 s
2
6
4
3
7
5

¼ s3
þ 9 þ k2ð Þs2
þ 7 þ k1 þ keð Þs þ 4ke
Equating this polynomial to the desired characteristic equation,
s3
þ 6:3s2
þ 13s þ 15:16 ¼ s3
þ 9 þ k2ð Þs2
þ 7 þ k1 þ keð Þs þ 4ke
Solving for the k’s,
K ¼ 2:21 À2:7½ Š and ke ¼ 3:79:

E1SM 11/11/2010 9:29:30 Page 48
CHAPTER 13
13.1
f tð Þ ¼ sin vkTð Þ; fÃ
tð Þ ¼
P1
k¼0
sin vkTð Þd t À kTð Þ;
FÃ
sð Þ ¼
X1
k¼0
sin vkTð ÞeÀkTs
¼
X1
k¼0
ejvkT
À eÀjvkT
À Á
eÀkTs
2j
¼
1
2j
X1
k¼0
eT sÀjvð Þ
Àk
À eT sþjvð Þ
Àk
But,
P1
k¼0
xÀk
¼
1
1 À xÀ1
Thus,
FÃ
sð Þ ¼
1
2j
1
1 À eÀT sÀjvð Þ
À
1
1 À eÀT sþjvð Þ
!
¼
1
2j
eÀTs
ejvT
À eÀTs
ejvT
1 À eÀTsejvT À eÀTsejvTð Þ þ eÀ2Ts
!
¼ eÀTs sin vTð Þ
1 À eÀTs2cos vTð Þ þ eÀ2Ts
!
¼
zÀ1
sin vTð Þ
1 À 2zÀ1cos vTð Þ þ zÀ2
13.2
F zð Þ ¼
z z þ 1ð Þ z þ 2ð Þ
z À 0:5ð Þ z À 0:7ð Þ z À 0:9ð Þ
F zð Þ
z
¼
z z þ 1ð Þ z þ 2ð Þ
z À 0:5ð Þ z À 0:7ð Þ z À 0:9ð Þ
¼ 46:875
z
z À 0:5
À 114:75
1
z À 0:7
þ 68:875
z
z À 0:9
F zð Þ ¼ 46:875
z
z À 0:5
À 114:75
z
z À 0:7
þ 68:875
z
z À 0:9
;
f kTð Þ ¼ 46:875 0:5ð Þk
À 114:75 0:7ð Þk
þ 68:875 0:9ð Þk
13.3
Since G sð Þ ¼ 1 À eÀTs
À Á 8
s s þ 4ð Þ
,
G zð Þ ¼ 1 À zÀ1
À Á
z
8
s s þ 4ð Þ
'
¼
z À 1
z
z
A
s
þ
B
s þ 4
'
¼
z À 1
z
z
2
s
þ
2
s þ 4
'
:
Let G2 sð Þ ¼
2
s
þ
2
s þ 4
. Therefore, g2 tð Þ ¼ 2 À 2eÀ4t
, or g2 kTð Þ ¼ 2 À 2eÀ4kT
.
Hence, G2 zð Þ ¼
2z
z À 1
À
2z
z À eÀ4T
¼
2z 1 À eÀ4T
À Á
z À 1ð Þ z À eÀ4Tð Þ
.

E1SM 11/11/2010 9:29:30 Page 49
Therefore, G zð Þ ¼
z À 1
z
G2 zð Þ ¼
2 1 À eÀ4T
À Á
z À eÀ4Tð Þ
.
For T ¼
1
4
s, G zð Þ ¼
1:264
z À 0:3679
.
13.4
Add phantom samplers to the input, feedback after H(s), and to the output. Push
G1(s)G2(s), along with its input sampler, to the right past the pickoff point and obtain
the block diagram shown below.
H(s)G1(s)G2(s)
G1(s)G2(s)
R(s) C(s)+
–
Hence, T zð Þ ¼
G1G2 zð Þ
1 þ HG1G2 zð Þ
.
13.5
Let G sð Þ ¼
20
s þ 5
. Let G2 sð Þ ¼
G sð Þ
s
¼
20
s s þ 5ð Þ
¼
4
s
À
4
s þ 5
. Taking the inverse
Laplace transform and letting t ¼ kT, g2 kTð Þ ¼ 4 À 4eÀ5kT
. Taking the z-transform
yields G2 zð Þ ¼
4z
z À 1
À
4z
z À eÀ5T
¼
4z 1 À eÀ5T
À Á
z À 1ð Þ z À eÀ5Tð Þ
.
Now, G zð Þ ¼
z À 1
z
À G2 zð Þ ¼
4 1 À eÀ5T
À Á
z À eÀ5Tð Þ
.
Finally, T zð Þ ¼
G zð Þ
1 þ G zð Þ
¼
4 1 À eÀ5T
À Á
z À 5eÀ5T þ 4
.
The pole of the closed-loop system is at 5eÀ5T
À 4. Substituting values of T, we find
that the pole is greater than 1 if T 0:1022 s. Hence, the system is stable for
0 T 0:1022 s.
13.6
Substituting z ¼
s þ 1
s À 1
into D zð Þ ¼ z3
À z2
À 0:5z þ 0:3, we obtain D sð Þ ¼ s3
À 8s2
À27s À 6. The Routh table for this polynomial is shown below.
s3
1 À27
s2
À8 À6
s1
À27:75 0
s0
À6 0
Since there is one sign change, we conclude that the system has one pole outside the
unit circle and two poles inside the unit circle. The table did not produce a row of
zeros and thus, there are no jv poles. The system is unstable because of the pole
outside the unit circle.

E1SM 11/11/2010 9:29:31 Page 50
13.7
Defining G(s) as G1(s) in cascade with a zero-order-hold,
G sð Þ ¼ 20 1 À eÀTs
À Á s þ 3ð Þ
!
¼ 20 1 À eÀTs
À Á 3=20
s
þ
1=4
s þ 4ð Þ
À
2=5
s þ 5ð Þ
!
:
Taking the z-transform yields
G zð Þ ¼ 20 1 À zÀ1
À Á 3=20ð Þz
z À 1
þ
1=4ð Þz
z À eÀ4T
À
2=5ð Þz
z À eÀ5T
!
¼ 3 þ
5 z À 1ð Þ
z À eÀ4T
À
8 z À 1ð Þ
z À eÀ5T
:
Hence for T ¼ 0:1 second, Kp ¼ lim
z!1
G zð Þ ¼ 3, and Kv ¼
1
T
lim
z!1
z À 1ð ÞG zð Þ ¼ 0, and
Ka ¼
1
T2
lim
z!1
z À 1ð Þ2
G zð Þ ¼ 0. Checking for stability, we find that the system is
stable for T ¼ 0:1 second, since T zð Þ ¼
G zð Þ
1 þ G zð Þ
¼
1:5z À 1:109
z2 þ 0:222z À 0:703
has poles
inside the unit circle at À0:957 and þ 0:735. Again, checking for stability, we find that
the system is unstable for T ¼ 0:5 second, since T zð Þ ¼
G zð Þ
1 þ G zð Þ
¼
3:02z À 0:6383
z2 þ 2:802z À 0:6272
has poles inside and outside the unit circle at þ0:208 and
À3:01, respectively.
13.8
Draw the root locus superimposed over the z ¼ 0:5 curve shown below. Searching
along a 54:3
line, which intersects the root locus and the z ¼ 0:5 curve, we find the
point 0:587ﬀ54:3
¼ 0:348 þ j 0:468ð Þ and K ¼ 0:31.
–3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2
–1.5
–1
–0.5
0
0.5
1
1.5
Real Axis
ImagAxis
z-Plane Root Locus
54.3°
(0.348 + j0.468)
K = 0.31
13.9
Let
Ge sð Þ ¼ G sð ÞGc sð Þ ¼
100K
2:38 s þ 25:3ð Þ
s þ 60:2ð Þ
¼
342720 s þ 25:3ð Þ
s s þ 36ð Þ s þ 100ð Þ s þ 60:2ð Þ
:

E1SM 11/11/2010 9:29:31 Page 51
The following shows the frequency response of Ge jvð Þ.
Frequency (rad/sec)
Bode Diagrams
–60
–40
–20
0
20
40
10–1
100
101
102
103
–250
–200
–150
–100
We find that the zero dB frequency, vFM
, for Ge jvð Þ is 39 rad/s. Using Astrom’s
guideline the value of T should be in the range, 0:15=vFM
¼ 0:0038 second to
0:5=vFM
¼ 0:0128 second. Let us use T ¼ 0:001 second. Now find the Tustin
transformation for the compensator. Substituting s ¼
2 z À 1ð Þ
T z À 1ð Þ
into Gc sð Þ ¼
2:38 s þ 25:3ð Þ
s þ 60:2ð Þ
with T ¼ 0:001 second yields
Gc zð Þ ¼ 2:34
z À 0:975ð Þ
z À 0:9416ð Þ
:
13.10
Gc zð Þ ¼
X zð Þ
E zð Þ
¼
1899z2
À 3761z þ 1861
z2 À 1:908z þ 0:9075
. Cross-multiply and obtain z2
À 1:908z þ
À
0:9075X zð Þ ¼ 1899z2
À 3761z þ 1861
À Á
E zð Þ. Solve for the highest power of z
operating on the output, X(z), and obtain z2
X zð Þ ¼ 1899z2
À 3761z þ 1861
À Á
E zð Þ À À1:908z þ 0:9075ð ÞX zð Þ. Solving for X(z) on the left-hand side yields

E1SM 11/11/2010 9:29:31 Page 52
X zð Þ ¼ 1899 À 3761zÀ1
þ 1861zÀ2
À Á
E zð ÞÀ À1:908zÀ1
þ 0:9075zÀ2
À Á
X zð Þ. Finally,
we implement this last equation with the following flow chart:
e*(t) x*(t)
e*(t-0.1) x*(t-0.1)
x*(t-0.2)e*(t-0.2)
Delay
0.1 second
1899
–3761
1861 0.9075
–1.9.08
+
+
+
+
–
–
Delay
0.1 second
Delay
0.1 second
Delay
0.1 second

E1SM 11/11/2010 9:29:32 Page 53
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E1SM 11/11/2010 9:29:32 Page 54

Solution of skill Assessment Control Systems Engineering By Norman S.Nise 6th edition

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