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Solutions to Skill-Assessment
Exercises
CHAPTER 2
The Laplace transform of t is
s 2
using Table 2.1, Item 3. Using Table 2.2, Item 4,
F sð Þ ¼
ð s þ 5 Þ
2
Expanding F(s) by partial fractions yields:
F sð Þ ¼
A
s
þ
B
s þ 2
þ
C
ðs þ 3 Þ
2
þ
D
ðs þ 3 Þ
where,
A ¼
ð s þ 2 Þ ðs þ 3 Þ
2 S! 0
B ¼
s sð þ 3 Þ
2 S! 2
C ¼
s sð þ 2 Þ S! 3
; and D ¼ ðs þ 3 Þ
2 dF sð Þ
ds s! 3
Taking the inverse Laplace transform yields,
f tð Þ ¼
5 e
2 t þ
te
3 t þ
e
3 t
Taking the Laplace transform of the differential equation assuming zero initial
conditions yields:
s
3 C sð Þ þ 3 s
2 C sð Þ þ 7 sC sð Þ þ 5 C sð Þ ¼ s
2 R sð Þ þ 4 sR sð Þ þ 3 R sð Þ
Collecting terms,
s
3 þ 3 s
2 þ 7 s þ 5
C sð Þ ¼ s
2 þ 4 s þ 3
R sð Þ
Thus,
C sð Þ
R sð Þ
s
2 þ 4 s þ 3
s^3 þ 3 s^2 þ 7 s þ 5
G sð Þ ¼
C sð Þ
R sð Þ
2 s þ 1
s 2 þ 6 s þ 2
Cross multiplying yields,
d
2 c
dt 2
þ 6
dc
dt
þ 2 c ¼ 2
dr
dt
þ r
C sð Þ ¼ R sð ÞG sð Þ ¼
s 2
s
ðs þ 4 Þ ðs þ 8 Þ
s sð þ 4 Þ ðs þ 8 Þ
A
s
þ
B
ðs þ 4 Þ
þ
C
ðs þ 8 Þ
where
A ¼
ð s þ 4 Þ ðs þ 8 Þ S! 0
B ¼
s sð þ 8 Þ S! 4
; and C ¼
s sð þ 4 Þ S! 8
Thus,
c tð Þ ¼
e
4 t þ
e
8 t
Mesh Analysis
Transforming the network yields,
V(s)
I 1 (s) I 2 (s)
V 1 (s)
V 2 (s)
I 9 (s)
(^1 )
s s
s
_
_
Now, writing the mesh equations,
ð s þ 1 ÞI 1 ð Þ s sI 2 ð Þ s I 3 ð Þ ¼s V sð Þ
sI 1 ð Þ þs ð 2 s þ 1 ÞI 2 ð Þ s I 3 ð Þ ¼s 0
I 1 ð Þ s I 2 ð Þ þs ðs þ 2 ÞI 3 ð Þ ¼s 0
Solving the mesh equations for I 2 (s),
I 2 ð Þ ¼s
ð s þ 1 Þ V sð Þ 1
s 0 1
1 0 ð s þ 2 Þ
ð s þ 1 Þ s 1
s ð 2 s þ 1 Þ 1
1 1 ðs þ 2 Þ
s
2 þ 2 s þ 1
V sð Þ
s s 2 ð þ 5 s þ 2 Þ
2 Solutions to Skill-Assessment Exercises
Solving for X 2 (s),
X 2 ð Þ ¼s
s
2 þ 3 s þ 1
F sð Þ
ð 3 s þ 1 Þ 0
s
2 þ 3 s þ 1
ð 3 s þ 1 Þ
ð 3 s þ 1 Þ s
2 þ 4 s þ 1
ð 3 s þ 1 ÞF sð Þ
s s 3 þ 7 s 2 ð þ 5 s þ 1 Þ
Hence,
X 2 ð Þs
F sð Þ
ð 3 s þ 1 Þ
s s 3 þ 7 s 2 ð þ 5 s þ 1 Þ
Writing the equations of motion,
s
2 þ s þ 1
u 1 ð Þ s ðs þ 1 Þu 2 ð Þ ¼s T sð Þ
ðs þ 1 Þu 1 ð Þ þs ð 2 s þ 2 Þu 2 ð Þ ¼s 0
where u 1 ð Þs is the angular displacement of the inertia.
Solving for u 2 ð Þs,
u 2 ð Þ ¼s
s 2 þ s þ 1
T sð Þ
ðs þ 1 Þ 0
s
2 þ s þ 1
ðs þ 1 Þ
ðs þ 1 Þ ð 2 s þ 2 Þ
ðs þ 1 ÞF sð Þ
2 s 3 þ 3 s 2 þ 2 s þ 1
From which, after simplification,
u 2 ð Þ ¼s
2 s 2 þ s þ 1
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,
Writing the equations of motion,
s
2 þ s
u 1 ð Þ s sua ð Þ ¼s T sð Þ
su 1 ð Þ þs ðs þ 1 Þua ð Þ ¼s 0
where u 1 ð Þs is the angular displacement of the 1-kg inertia.
Solving for ua ð Þs,
ua ð Þ ¼s
s
2 þ s
T sð Þ
s 0
s
2 þ s
s
s ðs þ 1 Þ
sT sð Þ
s 3 þ s 2 þ s
4 Solutions to Skill-Assessment Exercises
From which,
ua ð Þs
T sð Þ
s^2 þ s þ 1
But, u 2 ð Þ ¼s
ua ð Þs:
Thus,
u 2 ð Þs
T sð Þ
s 2 þ s þ 1
First find the mechanical constants.
Jm ¼ Ja þ JL
2
¼ 1 þ 400
Dm ¼ D (^) a þ DL
2
¼ 5 þ 800
Now find the electrical constants. From the torque-speed equation, set vm ¼ 0 to
find stall torque and set T (^) m ¼ 0 to find no-load speed. Hence,
T (^) stall ¼ 200
vnoload ¼ 25
which,
K (^) t
R (^) a
T (^) stall
Ea
K (^) b ¼
E (^) a
vnoload
Substituting all values into the motor transfer function,
um ð Þs
Ea ð Þ s
K T
R (^) a J (^) m
s s þ
J (^) m
Dm þ
K (^) T K (^) b
R (^) a
s s þ
where um ð Þs is the angular displacement of the armature.
Now uL ð Þ ¼s
um ð Þs. Thus,
uL ð Þs
Ea ð Þ s
s s þ
Letting
u 1 ð Þ ¼s v 1 ð Þs=s
u 2 ð Þ ¼s v 2 ð Þs=s
Chapter 2 Solutions to Skill-Assessment Exercises (^5)
Solving for e v (^) oþdv ,
e
v (^) oþdv ¼ e
v (^) o þ
de
v
dv v (^) o
dv ¼ e
vo þ e
v (^) o dv
Substituting into Eq. (1)
ddv
dt
þ e
vo þ e
v (^) o dv 2 ¼ i tð Þ ð^2 Þ
Setting i tð Þ ¼ 0 and letting the circuit reach steady state, the capacitor acts like an
open circuit. Thus, v (^) o ¼ v (^) r with ir ¼ 2. But, ir ¼ e
v (^) r or v (^) r ¼ lni (^) r.
Hence, v (^) o ¼ ln 2 ¼ 0 :693. Substituting this value of v (^) o into Eq. (2) yields
ddv
dt
þ 2 dv ¼ i tð Þ
Taking the Laplace transform,
ð s þ 2 Þdv sð Þ ¼ I sð Þ
Solving for the transfer function, we obtain
dv sð Þ
I sð Þ
s þ 2
or
V sð Þ
I sð Þ
s þ 2
about equilibrium:
CHAPTER 3
Identifying appropriate variables on the circuit yields
C 1
iL i^ C 2
iC 1
i (^) R
C 2
R
L v 1 (t) v (^) o(t)
Writing the derivative relations
C 1
dv (^) C 1
dt
¼ iC 1
L
di (^) L
dt
¼ v (^) L
C 2
dv (^) C 2
dt
¼ iC 2
ð 1 Þ
Chapter 3 Solutions to Skill-Assessment Exercises (^7)
Using Kirchhoff’s current and voltage laws,
i (^) C 1 ¼ iL þ i (^) R ¼ iL þ
R
v (^) L v (^) C 2 ð Þ
v (^) L ¼ v (^) C 1 þ v (^) i
i (^) C 2 ¼ iR ¼
R
v (^) L v (^) C 2 ð Þ
Substituting these relationships into Eqs. (1) and simplifying yields the state
equations as
dv (^) C 1
dt
RC 1
v (^) C 1 þ
C 1
i (^) L
RC 1
v (^) C 2 þ
RC 1
v (^) i
di (^) L
dt
L
v (^) C 1 þ
L
v (^) i
dv (^) C 2
dt
RC 2
v (^) C 1
RC 2
v (^) C 2
RC 2
v (^) i
where the output equation is
v (^) o ¼ v (^) C 2
Putting the equations in vector-matrix form,
x^ _ ¼
RC 1
C 1
RC 1
L
RC 2
RC 2
x þ
RC 1
L
RC 2
v (^) i ð Þt
y ¼ ½ 0 0 1 x
Writing the equations of motion
s 2 þ s þ 1
X 1 ð Þs sX 2 ð Þs ¼ F sð Þ
sX 1 ð Þ þs s 2 þ s þ 1
X 2 ð Þs X 3 ð Þs ¼ 0
X 2 ð Þ þs s
2 þ s þ 1
X 3 ð Þ ¼s 0
Taking the inverse Laplace transform and simplifying,
x € 1 ¼ x_ 1 x 1 þ x_ 2 þ f
x € 2 ¼ x_ 1 x_ 2 x 2 þ x 3
x € 3 ¼ x_ 3 x 3 þ x 2
Defining state variables, z (^) i,
z 1 ¼ x 1 ; z 2 ¼ x_ 1 ; z 3 ¼ x 2 ; z 4 ¼ x_ 2 ; z 5 ¼ x 3 ; z 6 ¼ x_ 3
8 Solutions to Skill-Assessment Exercises
The state equation is converted to a transfer function using
G sð Þ ¼ C ðs I AÞ
1 B ð^1 Þ
where
A ¼
; B ¼
; and C ¼ ½ 1 : 5 0 : 625 :
Evaluating ðs I AÞ yields
ð s I AÞ ¼
s þ 4 1 : 5
4 s
Taking the inverse we obtain
ð s I AÞ
1 ¼
s 2 þ 4 s þ 6
s 1 : 5
4 s þ 4
Substituting all expressions into Eq. (1) yields
G sð Þ ¼
3 s þ 5
s^2 þ 4 s þ 6
Writing the differential equation we obtain
d
2 x
dt 2
þ 2 x
2 ¼ 10 þ df tð Þ ð^1 Þ
Letting x ¼ x (^) o þ dx and substituting into Eq. (1) yields
d
2 ðx (^) o þ dxÞ
dt 2
þ 2 ðx (^) o þ dxÞ
2 ¼ 10 þ df tð Þ ð^2 Þ
Now, linearize x
2 .
ð x (^) o þ dxÞ
2 x
2 o
d x
2
dx x (^) o
dx ¼ 2 x (^) odx
from which
ð x (^) o þ dxÞ
2 ¼ x
2 o þ^2 x^ odx^ ð 3 Þ
Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives
us the linearized intermediate differential equation,
d
2 dx
dt 2
þ 4 x (^) odx ¼ 2 x
2 o þ 10 þ df tð Þ ð^4 Þ
The force of the spring at equilibrium is 10 N. Thus, since F ¼ 2 x
2 ; 10 ¼ 2 x
2 o from
which
x (^) o ¼
ffiffiffi 5
p
10 Solutions to Skill-Assessment Exercises
Substituting this value of x (^) o into Eq. (4) gives us the final linearized differential
equation.
d
2 dx
dt 2
þ 4
ffiffiffi 5
p dx ¼ df tð Þ
Selecting the state variables,
x 1 ¼ dx
x 2 ¼ d_x
Writing the state and output equations
x _ 1 ¼ x 2
x _ 2 ¼€dx ¼ 4
ffiffiffi 5
p x 1 þ df tð Þ
y ¼ x 1
Converting to vector-matrix form yields the final result as
x _ ¼
ffiffiffi 5
p 0
x þ
df tð Þ
y ¼ ½ 1 0 x
CHAPTER 4
For a step input
C sð Þ ¼
10 ðs þ 4 Þ ðs þ 6 Þ
s sð þ 1 Þ ðs þ 7 Þ ðs þ 8 Þ ðs þ 10 Þ
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 Be
t þ Ce
7 t þ De
8 t þ Ee
10 t
Since a ¼ 50 ; T (^) c ¼
a
¼ 0 :02s; T (^) s ¼
a
¼ 0 :08 s; and
T (^) r ¼
a
¼ 0 :044 s.
a. Since poles are at 6 j 19 : 08 ; c tð Þ ¼ A þ Be 6 t cos 19ð : 08 t þ fÞ.
b. Since poles are at 78 :54 and 11 : 46 ; c tð Þ ¼ A þ Be 78 : 54 t þ Ce 11 : 4 t .
c. Since poles are double on the real axis at 15 c tð Þ ¼ A þ Be 15 t þ Cte 15 t :
d. Since poles are at j 25 ; c tð Þ ¼ A þ B cos 25ð t þ fÞ.
a. (^) v n ¼^
ffiffiffiffiffiffiffiffi 400
p ¼ 20 and 2zvn ¼ 12 ; ;z ¼ 0 :3 and system is underdamped.
b. (^) vn ¼
ffiffiffiffiffiffiffiffi 900
p ¼ 30 and 2zvn ¼ 90 ; ;z ¼ 1 :5 and system is overdamped.
c. (^) vn ¼
ffiffiffiffiffiffiffiffi 225
p ¼ 15 and 2zvn ¼ 30 ; ;z ¼ 1 and system is critically damped.
d. (^) vn ¼
ffiffiffiffiffiffiffiffi 625
p ¼ 25 and 2zvn ¼ 0 ; ;z ¼ 0 and system is undamped.
Chapter 4 Solutions to Skill-Assessment Exercises (^11)
b. Since F ðt tÞ ¼
e
ðt tÞ
e
4 ðt tÞ 2
3
e
ðt tÞ
e
4 ðt tÞ
e
ðt tÞ þ
e
4 ðt tÞ
e
ðt tÞ þ
e
4 ðt tÞ
and
Bu ð Þ ¼t
e 2 t
; F ðt tÞBu ð Þ ¼t
e
t e
t
e
2 t e
4 t
e
t e
t þ
e
2 t e
4 t
Thus, x ð Þ ¼t F ð Þtx ð Þ þ 0
R
t 0
F ðt tÞ
BU ð Þtdt ¼
e
t e
2 t
e
4 t
e
t þ e
2 t þ
e
4 t
c. y tð Þ ¼ ½ 2 1 x ¼ 5 e
t e
2 t
CHAPTER 5
Combine the parallel blocks in the forward path. Then, push
s
to the left past the
pickoff point.
1
s
s
s
s
s 2
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
s 3 þ 1
2 s 4 þ s 2 þ 2 s
Find the closed-loop transfer function, T sð Þ ¼
G sð Þ
1 þ G sð ÞH sð Þ
s 2 þ as þ 16
, where
and G sð Þ ¼
s sð þ aÞ
and H sð Þ ¼ 1. Thus, vn ¼ 4 and 2zvn ¼ a, from which z ¼
a
But, for 5% overshoot, z ¼
ln
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p 2 þ ln
s ¼ 0 :69. Since, z ¼
a
; a ¼ 5 :52.
Chapter 5 Solutions to Skill-Assessment Exercises (^13)
Label nodes.
R ( s )
s
s N 1 ( s ) N 2 ( s ) N 3 ( s ) N 4 ( s )
N 5 ( s ) N 6 ( s )
N 7 ( s )
s
C ( s )
1 s
1 s
Draw nodes.
R ( s ) N 1 ( s )^ N 2 ( s )
N 5 ( s )
N 7 ( s )
N 6 ( s )
N 3 ( s ) N 4 ( s ) (^) C ( s )
Connect nodes and label subsystems.
R ( s ) 1 C ( s )
1 s
s
− 1
s^ s
(^1 )
− 1
1
1 s N 1 ( s ) N 2 ( s )
N 5 ( s ) N 6 ( s )
N 7 ( s )
N 3 ( s ) (^) N 4 ( s )
Eliminate unnecessary nodes.
R ( s ) (^1) s s C ( s )
1 s
1 s
14 Solutions to Skill-Assessment Exercises
Writing the state equations from the signal-flow diagram, we obtain
x ¼
x þ
r
y ¼ ½ 100 500 x
From the transformation equations,
P
1 ¼
Taking the inverse,
P ¼
Now,
P
1 AP ¼
P
1 B ¼
CP ¼ ½ 1 4
Therefore,
_z ¼
z þ
u
y ¼ ½ 0 : 8 1 : 4 z
First find the eigenvalues.
jlI Aj ¼
l 0
0 l
l 1 3
4 l þ 6
¼ l
2 þ 5 l þ 6
From which the eigenvalues are 2 and 3.
Now use Ax (^) i ¼ lx (^) i for each eigenvalue, l.
Thus,
x 1
x 2
¼ l
x 1
x 2
For l ¼ 2,
3 x 1 þ 3 x 2 ¼ 0
4 x 1 4 x 2 ¼ 0
16 Solutions to Skill-Assessment Exercises
Thus x 1 ¼ x 2
For l ¼ 3
4 x 1 þ 3 x 2 ¼ 0
4 x 1 3 x 2 ¼ 0
Thus x 1 ¼ x 2 and x 1 ¼ 0 : 75 x 2 ; from which we let
P ¼
Taking the inverse yields
P
1 ¼
Hence,
D ¼ P
1 AP ¼
P
1 B ¼
CP ¼ ½ 1 4
Finally,
z _ ¼
z þ
u
y ¼ ½ 2 : 121 2 : 6 z
CHAPTER 6
Make a Routh table.
s
7 3 6 7 2
s
6 9 4 8 6
s 5 4.666666667 4.333333333 0 0
s
4 4.35714286 8 6 0
s
3 12.90163934 6.426229508 0 0
s
2 10.17026684 6 0 0
s
1 1.18515742 0 0 0
s
0 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.
Make a Routh table. We encounter a row of zeros on the s
3 row. The even polynomial
is contained in the previous row as 6 s 4 þ 0 s 2 þ 6. Taking the derivative yields
Chapter 6 Solutions to Skill-Assessment Exercises (^17)
CHAPTER 7
a. First check stability.
T sð Þ ¼
G sð Þ
1 þ G sð Þ
10 s
2 þ 500 s þ 6000
s 3 þ 70 s 2 þ 1375 s þ 6000
10 ðs þ 30 Þ ðs þ 20 Þ
ð s þ 26 : 03 Þ ðs þ 37 : 89 Þ ðs þ 6 : 085 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,
15 u tð Þ : e (^) step ð 1 Þ ¼
1 þ lim s! 0
G sð Þ
1 þ 1
15 tu tð Þ : e (^) ramp ð 1 Þ ¼
lim s! 0
sG sð Þ
20
30
35
15 t
2 u tð Þ : e (^) parabola ð 1 Þ ¼
lim s! 0
s
2 G sð Þ
¼ 1; since L 15 t
2 ¼
s 3
b. First check stability.
T sð Þ ¼
G sð Þ
1 þ G sð Þ
10 s
2 þ 500 s þ 6000
s 5 þ 110 s 4 þ 3875 s 3 þ 4 : 37 e 04 s 2 þ 500 s þ 6000
10 ðs þ 30 Þ ðs þ 20 Þ
ð s þ 50 : 01 Þ ðs þ 35 Þ ðs þ 25 Þ s 2 ð 7 : 189 e 04 s þ 0 : 1372 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.
a. The system is stable, since
T sð Þ ¼
G sð Þ
1 þ G sð Þ
1000 ðs þ 8 Þ
ð s þ 9 Þ ðs þ 7 Þ þ 1000 ðs þ 8 Þ
1000 ðs þ 8 Þ
s 2 þ 1016 s þ 8063
and is of Type 0. Therefore,
K (^) p ¼ lim s! 0
G sð Þ ¼
8
9
¼ 127 ; K (^) v ¼ lim s! 0
sG sð Þ ¼ 0 ;
and K (^) a ¼ lim s! 0
s
2 G sð Þ ¼ 0
b.
estep ð 1 Þ ¼
1 þ lim s! 0
G sð Þ
1 þ 127
¼ 7 : 8 e 03
e (^) ramp ð 1 Þ ¼
lim s! 0
sG sð Þ
e (^) parabola ð 1 Þ ¼
lim s! 0
s
2 G sð Þ
Chapter 7 Solutions to Skill-Assessment Exercises (^19)
System is stable for positive K. System is Type 0. Therefore, for a step input
e (^) step ð 1 Þ ¼
1 þ K (^) p
¼ 0 :1. Solving for Kp yields K (^) p ¼ 9 ¼ lim s! 0
G sð Þ ¼
12 K
18
; from
which we obtain K ¼ 189.
System is stable. Since G 1 ð Þ ¼s 1000, and G 2 ð Þ ¼s
ðs þ 2 Þ
ð s þ 4 Þ
e (^) D ð 1 Þ ¼
lim s! 0
G 2 ð Þ s
þ lim G 1 s! 0
ð Þ s
2 þ 1000
¼ 9 : 98 e 04
System is stable. Create a unity-feedback system, where H (^) e ð Þ ¼s
s þ 1
s
s þ 1
The system is as follows:
R ( s ) E^ a ( s )^100 C ( s )
s + 4
− s
s + 1
Thus,
G (^) e ð Þ ¼s
G sð Þ
1 þ G Sð ÞH (^) e ð Þ s
ð s þ 4 Þ
100 s
ðs þ 1 Þ ðs þ 4 Þ
100 ðs þ 1 Þ
S
2 95 s þ 4
Hence, the system is Type 0. Evaluating Kp yields
K (^) p ¼
The steady-state error is given by
e (^) step ð 1 Þ ¼
1 þ K (^) p
1 þ 25
¼ 3 : 846 e 02
Since G sð Þ ¼
K sð þ 7 Þ
s 2 þ 2 s þ 10
; e ð 1 Þ ¼
1 þ K (^) p
1 þ
7 K
10 þ 7 K
Calculating the sensitivity, we get
S (^) e:K ¼
K
e
@e
@K
K
10 þ 7 K
ð 10 Þ 7
ð 10 þ 7 KÞ
2
7 K
10 þ 7 K
20 Solutions to Skill-Assessment Exercises
