Solutions to Skill-Assessment Exercises
CHAPTER 2 2.1
The Laplace transform of t is is
ð Þ ¼ ð þ1 Þ2. s 5
1 s2
using Table 2.1, Item 3. Using Table 2.2, Item 4,
F s 2.2
Expanding F ( s) by partial fractions yields:
ð Þ ¼ A s þ s þB 2 þ ð þC Þ2 þ ð s þD 3Þ s 3
F s
where, A
¼ ð þ Þð þ Þ ¼ ð þ Þ ¼ 10
2 s
s
C
2
3
!0
S
10
2
s s
!3
S
¼
5 9
B
10 ; and D 3
¼ ð þ Þ 10
s s
3
2
!2
S
ðÞ
dF s 32 ds
¼ ðs þ Þ
Taking the inverse Laplace transform yields,
ð Þ ¼ 59 5e2 þ 103 te3 þ 409 e3 t
f t
t
¼ 5
s
!3
¼ 409
t
2.3
Taking the Laplace transform of the differential equation assuming zero initial conditions yields: s3 C s
ð Þ þ 3 s2 CðsÞ þ 7 sCðsÞ þ 5CðsÞ ¼ s2RðsÞ þ 4 sRðsÞ þ 3RðsÞ
Collecting terms, s3
Thus,
þ 3 s2 þ 7 s þ 5 CðsÞ ¼
s2
þ 4 s þ 3 RðsÞ
ð Þ ¼ s2 þ 4 s þ 3 ð Þ s3 þ 3 s2 þ 7 s þ 5
C s R s
1
2
Solutions to Skill-Assessment Exercises 2.4
ð Þ ¼ CRððssÞÞ ¼ s2 2þ s 6þ s 1þ 2
G s
Cross multiplying yields, d2 c dt 2
dr þ 6 dc þ 2c ¼ 2 þ r dt dt
2.5
ð Þ ¼ RðsÞGðsÞ ¼ s12 ð s þ 4Þsðs þ 8Þ ¼ sðs þ 41Þðs þ 8Þ ¼ A s þ ð s þB 4Þ þ ð s þC 8Þ
C s
where A
¼ ð þ Þð þ Þ 1
s
4 s
8
!0
S
1 32
¼
B
Thus,
¼ ð þ Þ 1
8
s s
¼
!4
S
¼ ð þ Þ
1 1 ; and C 16 s s 4
ð Þ ¼ 321 161 e4 þ 321 e8 t
c t
t
2.6
Mesh Analysis
Transforming the network yields, s
1
V(s)
V 1(s)
I 9(s)
+ _
1
+ s I 1(s)
s
V 2(s) _
I2(s)
Now, writing the mesh equations,
ð s þ 1Þ I 1ðsÞ sI 2 ðsÞ I 3ðsÞ ¼ VðsÞ sI 1 ðsÞ þ ð2 s þ 1Þ I 2ðsÞ I 3ðsÞ ¼ 0 I 1 ðsÞ I 2ðsÞ þ ðs þ 2Þ I 3ðsÞ ¼ 0 Solving the mesh equations for I 2( s),
ð þ Þ ð Þ ð þ Þ þ þ ðÞ ðÞ¼ ð þ Þ ¼ ð þ Þ ð þ þ Þ ðþ Þ 1
s
V s
0 0
s
I 2 s
s
1 1
s
1
s
2 s
1
1
1 1 s 2 1 1 s 2
s2
s s2
2 s
1 V s 5 s 2
!8
S
¼ 321
2
Solutions to Skill-Assessment Exercises 2.4
ð Þ ¼ CRððssÞÞ ¼ s2 2þ s 6þ s 1þ 2
G s
Cross multiplying yields, d2 c dt 2
dr þ 6 dc þ 2c ¼ 2 þ r dt dt
2.5
ð Þ ¼ RðsÞGðsÞ ¼ s12 ð s þ 4Þsðs þ 8Þ ¼ sðs þ 41Þðs þ 8Þ ¼ A s þ ð s þB 4Þ þ ð s þC 8Þ
C s
where A
¼ ð þ Þð þ Þ 1
s
4 s
8
!0
S
1 32
¼
B
Thus,
¼ ð þ Þ 1
8
s s
¼
!4
S
¼ ð þ Þ
1 1 ; and C 16 s s 4
ð Þ ¼ 321 161 e4 þ 321 e8 t
c t
t
2.6
Mesh Analysis
Transforming the network yields, s
1
V(s)
V 1(s)
I 9(s)
+ _
1
+ s I 1(s)
s
V 2(s) _
I2(s)
Now, writing the mesh equations,
ð s þ 1Þ I 1ðsÞ sI 2 ðsÞ I 3ðsÞ ¼ VðsÞ sI 1 ðsÞ þ ð2 s þ 1Þ I 2ðsÞ I 3ðsÞ ¼ 0 I 1 ðsÞ I 2ðsÞ þ ðs þ 2Þ I 3ðsÞ ¼ 0 Solving the mesh equations for I 2( s),
ð þ Þ ð Þ ð þ Þ þ þ ðÞ ðÞ¼ ð þ Þ ¼ ð þ Þ ð þ þ Þ ðþ Þ 1
s
V s
0 0
s
I 2 s
s
1 1
s
1
s
2 s
1
1
1 1 s 2 1 1 s 2
s2
s s2
2 s
1 V s 5 s 2
!8
S
¼ 321
Chapter Chapt er 2 Solut Solutions ions to to Skill-Asse Skill-Assessmen ssmentt Exercises Exercises
But, V L s Hence,
ð Þ ¼ sI 2ðsÞ
s2
þ 2 s þ 1 V ðsÞ V ðsÞ ¼ ð s2 þ 5 s þ 2Þ L
or
ð Þ ¼ s2 þ 2 s þ 1 ð Þ s2 þ 5 s þ 2
V L s V s
Nodal Analysis
Writing the nodal equations, 1
þ
2 V 1 s
s
ð Þ V ðsÞ ¼ VðsÞ L
ðÞþ þ
V 1 s
2
ð Þ ¼ s1 V ðsÞ
1 V L s
s
Solving Solvi ng for V L( s),
þ ð Þ ð Þ þ þ ð Þ ¼ ¼ ð þ þ Þð Þ þ þ 1
2
s
V s
1
1
V L s
1
s
s
2
s2
2
s
2 s
s2
1
1
or
V s
1 V s 5 s 2
1
ð Þ ¼ s2 þ 2 s þ 1 ð Þ s2 þ 5 s þ 2
V L s V s 2.7
Inverting
100000 ¼ s 2 ðsÞ ð Þ ¼ Z ¼ 5 Z 1 ðsÞ
G s
10 = s
Noninverting
ð Þ ¼ ½Z 1ðsZ Þ1þðsÞZðsÞ ¼
G s
5
10 s
þ 105
! 105
!
¼ s þ 1
s
2.8
Writing the equations of motion, s2
þ 3 s þ 1 X 1ðsÞ ð3 s þ 1Þ X 2ðsÞ ¼ F ðsÞ ð3 s þ 1Þ X 1ðsÞ þ s2 þ 4 s þ 1 X 2ðsÞ ¼ 0
3
4
Solutions to Skill-Assessment Exercises
Solving Solvin g for X 2( s), s2
þ þ ð Þ ð Þ ¼ ð þ Þ ¼ ð ðþ þ þÞ ð Þþ Þ þð þþ Þ ðþ þþ Þ 3 s
3 s
X 2 s
s2
3 s
3 s
Hence,
1
F s
1
0
1
s
3 s 1 2 4 s 1 s
1
s3
3 s 1 F s 7 s2 5 s
1
ðÞ¼ ð3 s þ 1Þ 3 ð Þ sðs þ 7 s2 þ 5 s þ 1Þ
X 2 s F s
2.9
Writing the equations of motion, s2 s 1 u 1 s s 1 u 2 s T s 2 s 2 u 2 s 0 s 1 u 1 s
þ þ ð Þð þ Þ ð Þ ¼ ð Þ ð þ Þ ð Þ þ ð þ Þ ð Þ ¼
where u 1 s is the angular displacement of the inertia. Solving for u 2 s ,
ðÞ
ðÞ
s2
þ þ ð Þ ð Þ ¼ þð þþ Þ ð þ Þ ¼ ð þ Þ ð þ Þ
u 2 s
s
1
s
1 1
s s
2
s
T s
0
1 2
s
1
2 s
ðs þ 1ÞF ðsÞ 2 s3 þ 3 s2 þ 2 s þ 1
From which, after simplification,
ð Þ ¼ 2 s2 þ1 s þ 1
u 2 s 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,
Writing the equations of motion, s2
þ ð Þ
ð Þ ¼ T ðsÞ su 1ðsÞ þ ðs þ 1Þu ðsÞ ¼ 0 s u 1 s
su a s a
where u 1 s is the angular displacement of the 1-kg inertia. Solving for u a s ,
ðÞ
ðÞ
s2
þ ð Þ ð Þ ¼ þ ¼ ð þ Þ
u a s
s2
s
T s
0
s s
s
s
s
1
ðÞ þ þ
sT s s3 s2 s
Chapter Chapt er 2 Solut Solutions ions to to Skill-Asse Skill-Assessmen ssmentt Exercises Exercises
From which,
ð Þ ¼ 1 T ðsÞ s2 þ s þ 1
u a s
ð Þ ¼ 12 u ðsÞ:
But, u 2 s
a
Thus,
ð Þ ¼ 1=2 T ðsÞ s2 þ s þ 1
u 2 s
2.11
First find the mechanical constants.
¼ J
J m
a
¼ D
Dm
2
þ ¼ þ ¼ þ ¼ þ ¼ 1 1 J L 5 4
400
1 400
2
1 1 5 4
DL
a
1
5
800
2
1 400
7
Now find the electrical constants. From the torque-speed equation, set vm find stall torque and set T m 0 to find no-load speed. Hence,
¼ 0 to
¼
¼ 200 ¼ 25
T stall stall vno
load
which,
¼ T E ¼ 200 ¼2 100
K t t Ra
stall stall a
¼ v E ¼ 100 ¼4 25 a
K b
no load
Substituting all values into the motor transfer function, K T T Ra J m
ðÞ¼ 1 E ðsÞ s s þ J u m s
a
m
Dm
1
þ K RK T T
b
a
¼ þ s s
where u m s is the angular displacement of the armature. 1 u m s . Thus, Now u L s 20
ðÞ ðÞ¼
ðÞ
ð Þ ¼ 1=20 15 E ðsÞ s s þ 2 u L s a
2.12
Letting
ð Þ ¼ v1 ðsÞ= s u 2 ðsÞ ¼ v2 ðsÞ = s u 1 s
15 2
5
6
Solutions to Skill-Assessment Exercises
in Eqs. 2.127, we obtain
J 1 s
þ D1 þ
K v1 s s
ð Þþ
K v1 s s
J 2 s
ð Þ K v2 ðsÞ ¼ T ðsÞ s
K v2 s s
þ D2 þ
ðÞ
From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively. J 1
T(t)
D1
J 2
1
– +
K
w 1(t )
D2
w 2(t )
Series analog 1
w 1(t )
1
J 1
T(t)
w 2(t )
K
J 2
D1
1 D2
Parallel analog
2.13
Writing the nodal equation, dv C dt
þ i 2 ¼ iðt Þ r
But,
¼1 v ¼ v þ dv i ¼ e ¼ e ¼ e C
o
vr
r
v
þ
vo dv
Substituting these relationships into the differential equation,
ð þ Þþe
d vo dv dt
þ
vo dv
2 ¼ iðt Þ
We now linearize e v. The general form is
ð Þ f ðv Þ
f v
o
v
Substituting the function, f v
df dv
ð Þ ¼ e , with v ¼ v þ
vo dv
e
þ
dv
vo
o
vo
e
dev dv
dv yields,
dv
vo
ð1Þ
Chapter 3 Solutions to Skill-Assessment Exercises
Solving for evo þdv,
þ
vo dv
e
vo
¼e þ
dev dv
Substituting into Eq. (1) ddv dt
vo
vo
þe þe
dv vo
dv
vo
vo
¼e þe
dv
ð2Þ
2 ¼ iðt Þ
Setting i t 0 and letting the circuit reach steady state, the capacitor acts like an open circuit. Thus, v o vr with ir 2. But, i r evr or vr lnir . Hence, v o ln 2 0:693. Substituting this value of v o into Eq. (2) yields
ðÞ¼ ¼
¼ ¼
¼
¼
ddv dt
¼
þ 2dv ¼ iðt Þ
Taking the Laplace transform,
ð s þ 2ÞdvðsÞ ¼ I ðsÞ Solving for the transfer function, we obtain
ðÞ¼ 1 I ðsÞ s þ 2
dv s
or
ð Þ ¼ 1 about equilibrium: ð Þ s þ 2
V s I s
CHAPTER 3 3.1
Identifying appropriate variables on the circuit yields C 1
R i R
iC
1
v1(t)
+
L
–
+
C 2
vo(t) –
iC
i L
2
Writing the derivative relations dvC 1 iC 1 dt diL L vL dt dvC 2 C 2 iC 2 dt
¼
C 1
¼
¼
ð1Þ
7
8
Solutions to Skill-Assessment Exercises
Using Kirchhoff’s current and voltage laws,
¼ i þ i ¼ i þ 1R ðv v Þ v ¼ v þ v 1 i ¼ i ¼ ðv v Þ R
iC 1
L
R
L
L
C 1
C 2
L
C 2
i
R
L
C 2
Substituting these relationships into Eqs. (1) and simplifying yields the state equations as dvC 1 dt diL dt dvC 2 dt
1 1 1 1 ¼ RC v þ i v þ v C 1 RC 1 RC 1 1 C 1
L
C 2
¼ L1 v þ 1L v 1 1 ¼ RC v v RC 2 2 C 1
i
i
C 1
1
C 2
RC 2
vi
where the output equation is vo
¼ v
C 2
Putting the equations in vector-matrix form,
x_
1
1
RC 1
C 1
RC 1
0
0
0
1 RC 2
2 66 ¼6 64
1 L
1
RC 2
y
¼ ½0
1
3 2 77 66 77 þ 66 5 4 x
1 RC 1
1
3 77 77 ð Þ 5 vi t
L
1 RC 2
0 1 x
3.2
Writing the equations of motion s2
sX 2ðsÞ ¼ F ðsÞ þ s þ 1 X 1 ðsÞ sX 1ðsÞ þ s2 þ s þ 1 X 2ðsÞ X 3 ðsÞ ¼ 0 X 2 ðsÞ þ s2 þ s þ 1 X 3ðsÞ ¼ 0
Taking the inverse Laplace transform and simplifying,
¼ x_ 1 x1 þ x _ 2 þ f x€2 ¼ x_ 1 x_ 2 x2 þ x3 x€3 ¼ x_ 3 x3 þ x2 x€1
Defining state variables, zi,
¼ x1; z2 ¼ x_ 1; z3 ¼ x2; z4 ¼ x_ 2; z5 ¼ x3; z6 ¼ x_ 3
z1
Chapter 3 Solutions to Skill-Assessment Exercises
Writing the state equations using the definition of the state variables and the inverse transform of the differential equation, z_ 1
¼ z2 _ 2 þ f ¼ z2 z1 þ z4 þ f z_ 2 ¼ x€1 ¼ x_ 1 x1 þ x z_ 3 ¼ x_ 2 ¼ z4 z_ 4 ¼ x€2 ¼ x_ 1 x_ 2 x2 þ x3 ¼ z2 z4 z3 þ z5 z_ 5 ¼ x_ 3 ¼ z6 z_ 6 ¼ x€3 ¼ x_ 3 x3 þ x2 ¼ z6 z5 þ z3 The output is z . Hence, y ¼ z5 . In vector-matrix form, 5
0 1
z_
2 66 ¼6 664
0 0 0 0
1 1
0 0
0 1
0 0
0 0
0 1 0
0 1 0
1 1 0
0 1 0
0 0 1
1
0
0
0 1
3 23 77 66 77 77 þ 66 77 ð Þ 75 64 75
0 f t ; y 0 0
z
1 1
¼ ½0
0 0
0 1 0 z
0
3.3
First derive the state equations for the transfer function without zeros.
ð Þ ¼ 1 ð Þ s2 þ 7 s þ 9
X s R s
Cross multiplying yields s2
þ 7 s þ 9 X ðsÞ ¼ RðsÞ
Taking the inverse Laplace transform assuming zero initial conditions, we get x€
þ 7 x_ þ 9 x ¼ r
Defining the state variables as,
¼ x x2 ¼ x_ x1
Hence, x_ 1
¼ x2 x_ 2 ¼ x€ ¼ 7 x_ 9 x þ r ¼ 9 x1 7 x2 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 2 x2
Putting all equation in vector-matrix form yields, x_
"
0
1
# "#
¼ xþ 9 7 c ¼ ½ 1 2 x
0 1
r
9
10
Solutions to Skill-Assessment Exercises 3.4
The state equation is converted to a transfer function using
ð Þ ¼ CðsI AÞ1B
ð1Þ
G s
where A
4 4
1:5
¼
0
2 ; and C 0
¼ ;B
0:625 :
¼ ½ 1:5
Evaluating sI
ð AÞ yields 4 1:5 4 s
Þ¼ þ s
ð sI A
Taking the inverse we obtain
ð sI AÞ1 ¼
s2
1 4 s
þ þ6
1:5 sþ4
s
4
Substituting all expressions into Eq. (1) yields
ð Þ ¼ s2 3þ s 4þ s 5þ 6
G s 3.5
Writing the differential equation we obtain d2 x dt 2
ð1Þ
þ 2 x2 ¼ 10 þ d f ðt Þ
Letting x
¼ x þ d x and substituting into Eq. (1) yields d2 ð x þ d xÞ þ 2ðx þ d xÞ2 ¼ 10 þ d f ðt Þ o
o
ð2Þ
o
dt 2
Now, linearize x 2. 2
ð x þ d xÞ x2 o
d x2 dx
¼
o
from which
d x
xo
¼ 2 x d x o
ð x þ d xÞ2 ¼ x2 þ 2 x d x o
ð3Þ
o
o
Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives us the linearized intermediate differential equation, d2 d x dt 2
ð4Þ
þ 4 x d x ¼ 2 x2 þ 10 þ d f ðt Þ o
o
¼ 2 x2; 10 ¼ 2 x2 from
The force of the spring at equilibrium is 10 N. Thus, since F which
¼
xo
p
ffiffi
5
o
Chapter 4 Solutions to Skill-Assessment Exercises
Substituting this value of xo into Eq. (4) gives us the final linearized differential equation. d2 d x dt 2
p þ 4 5 d x ¼ d f ðt Þ
ffiffi
Selecting the state variables,
¼ d x x2 ¼ d_ x x1
Writing the state and output equations x_ 1
¼ x2 p x_ 2 ¼€d x ¼ 4 5x1 þ d f ðt Þ y ¼ x1
ffiffi
Converting to vector-matrix form yields the final result as x_ y
0 1 x 4 5 0
0 d f t 1
ffiffi þ ð Þ ¼ p ¼ ½ 1 0 x
CHAPTER 4 4.1
For a step input
ðs þ 4Þðs þ 6Þ A B C D E ð Þ ¼ sðs þ 110 ¼ þ þ þ þ Þðs þ 7Þðs þ 8Þðs þ 10Þ s s þ 1 s þ 7 s þ 8 s þ 10
C s
Taking the inverse Laplace transform,
ð Þ ¼ A þ Be þ Ce7 þ De8 þ Ee10 t
c t
t
t
t
4.2
¼ 50; T ¼ 1a ¼ 501 ¼ 0:02s; T ¼ 4a ¼ 504 ¼ 0:08 s; and 2:2 2:2 ¼ 50 ¼ 0:044 s. T ¼ a Since a
c
s
r
4.3
a. Since poles are at 6 j 19:08; c t A Be6t cos 19:08t f . b. Since poles are at 78:54 and 11:46; c t A Be78:54t Ce11:4t . c. Since poles are double on the real axis at 15 c t A Be15t Cte 15t : d. Since poles are at j 25; c t A B cos 25t f .
ðÞ¼ þ ð þ Þ ðÞ¼ þ þ ð Þ ¼ þ þ ð Þ ¼ þ ð þ Þ
4.4
a. b. c. d.
¼ p 400 ¼ 20 and 2zv ¼ 12; p v ¼ 900 ¼ 30 and 2zv ¼ 90; p v ¼ 225 ¼ 15 and 2zv ¼ 30; p v ¼ 625 ¼ 25 and 2zv ¼ 0; vn n n n
ffi ffiffiffi ffiffi ffiffi ffi ffiffiffi ffiffi ffiffi
¼ 0:3 and system is underdamped. z ¼ 1:5 and system is overdamped. z ¼ 1 and system is critically damped. z ¼ 0 and system is undamped.
n
;z
n
;
n
;
n
;
11
12
Solutions to Skill-Assessment Exercises 4.5
¼ p 361 ¼ 19 and 2zv ¼ 16; z ¼ 0:421: p 4 ¼ ¼ 0:182 s. Now, T ¼ 0:5 s and T ¼ zv v 1 z 2 From Figure 4.16, v T ¼ 1:4998. Therefore, T ¼ 0:079 s. p zp 2 Finally, %os ¼ e 1 z 100 ¼ 23:3%
ffiffi ffi ffi
vn
s
p
n
n
n
;
n
r
p ffiffi ffi ffi ffi ffi ffi
r
ffiffi
4.6
a. The second-order approximation is valid, since the dominant poles have a real part of 2 and the higher-order pole is at 15, i.e. more than five-times further. b. The second-order approximation is not valid, since the dominant poles have a real part of 1 and the higher-order pole is at 4, i.e. not more than five-times further.
4.7
1:5918 0:3023 ð Þ ¼ s1 þ s0:8942 þ 20 s þ 10 s þ 6:5.
a. Expanding G ( s) by partial fractions yields G s
But 0:3023 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. 1 0:9782 1:9078 0:0704 b. Expanding G ( s) by partial fractions yields G s . s s 20 s 10 s 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.
ð Þ ¼ þ þ þ þ
4.8
SeeFigure4.31inthetextbookfortheSimulinkblockdiagramandtheoutputresponses. 4.9
a. Since
sI
ðÞ¼
BU s
"
A 0
1= s
s
¼ #
3
ð þ 1Þ
2 ; ðsI AÞ1 ¼ 1 s þ 5 s2 þ 5 s þ 6 3 sþ5
2
s
:
Also,
.
ð Þ ¼ ðsI AÞ1½xð0Þ þ BUðsÞ ¼ ð s þ 1Þðs þ1 2Þðs þ 3Þ 2 s2 þ 7 s þ 7 5 s2 þ 2 s 4 Y ð s Þ ¼ ½ ð s Þ ¼ X . The output is 1 3 ð s þ 1Þðs þ 2Þðs þ 3Þ ¼ s2 4 s 6 12 17:5 s0:5 þ the inverse Laplace transform yields yðt Þ ¼ þ 1 s þ 22 s þ 3. Taking 0:5e 12e þ 17:5e 3 . b. The eigenvalues are given by the roots of j sI Aj ¼ s2 þ 5 s þ 6, or 2 and 3. The state vector is X s
"
#
t
t
t
4.10
a. Since sI
Þ¼
s
2 ; ðsI AÞ1 ¼ 1 s þ 5 s2 þ 5 s þ 4 2 sþ5
2
. Taking the s 2 Laplace transform of each term, the state transition matrix is given by 4 t 1 4t 2 t 2 4t e e e e 3 3 3 3 : F t 2 t 2 4t 1 t 4 4t e e e e 3 3 3 3
ð A
2 ð Þ ¼ 64
þ
þ
3 75
Chapter 5 Solutions to Skill-Assessment Exercises
2 ð Þ ¼ 64 þ 2 ð Þ¼ ð Þ ð Þ ¼ 64 R ð Þ ¼2 ð Þ ð Þ þ ð Þ 3 75 ð Þ ¼ 64 þ þ
4 ðt t Þ 1 4ð t t Þ 2 ðt t Þ 2 4ð t t Þ e e e e 3 3 3 3 b. Since F t t 2 ð t t Þ 2 4ð t t Þ 1 ð t t Þ 4 4ðt t Þ e e e e 3 3 3 3 2 t t 2 2t 4t e e e e 0 3 3 ; F t t Bu t : Bu t 1 t t 4 2t 4t e2t e e e e 3 3 t t Thus, x t F t x 0 0 F t 10 t 4 4t e e2t e 3 3 BU t dt 5 t 8 4t e e2t e 3 3
þ
þ
3 75
3 75
and
ð Þ ¼ ½ 2 1 x ¼ 5e e2
c. y t
t
t
CHAPTER 5 5.1
1 Combine the parallel blocks in the forward path. Then, push to the left past the s
pickoff point. s
R(s)
+
– s2 +
1 s
C (s)
1 s
–
s
Combine the parallel feedback paths and get 2 s. Then, apply the feedback formula, s3 1 simplify, and get, T s . 2 s4 s2 2 s
ðÞ¼
5.2
þ
þ þ
ð Þ ¼ 1 þ GGððssÞÞHðsÞ ¼ s2 þ as16 þ 16, where 16 a and GðsÞ ¼ and H ðsÞ ¼ 1. Thus, v ¼ 4 and 2zv ¼ a, from which z ¼ . sðs þ aÞ 8 % ln 100 ¼ 0:69. Since, z ¼ a8 ; a ¼ 5:52. But, for 5% overshoot, z ¼ % p2 þ ln2 100 Find the closed-loop transfer function, T s
n
s ffiffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi
n
13
14
Solutions to Skill-Assessment Exercises 5.3
Label nodes.
R(s) +
–
s –
+
s
N 1 (s)
N 2 (s)
N 3 (s)
1 s
N 5 (s)
1 s
N 4 (s)
+
N 6 (s)
s
N 7 (s)
Draw nodes. N 1(s)
R(s)
N 2(s)
N 3(s)
N 5(s)
N 4(s)
C (s)
N 6(s)
N 7(s)
Connect nodes and label subsystems. −1
R(s)
1 N 1 (s)
s
N 2 (s)
s
N 3 (s)
1
1 −1
1 N 4 (s) s
C (s)
1 N 5 (s)
1 s
N 7 (s)
N 6 (s)
s
Eliminate unnecessary nodes. –1
R(s)
1
s
s
1 s –s
1 s
C (s)
C (s)
Chapter 5 Solutions to Skill-Assessment Exercises 5.4
Forward-path gains are G 1G2G3 and G 1G3. Loop gains are G1 G2 H 1 ; G2 H 2; and G3 H 3. G1 G2 G3 H 1 H 3 and Nontouching loops are G1 G2 H 1 G3 H 3 G2 H 2 G3 H 3 G2 G3 H 2 H 3 . Also, D 1 G1 G2 H 1 G2 H 2 G3 H 3 G1 G2 G3 H 1 H 3 G2 G3 H 2 H 3 : Finally, D1 1 and D2 1.
½ ½ ¼ ½ ½ ¼ ¼ þ þ þ þ ¼ ¼ CðsÞ ¼ Substituting these values into T ðsÞ ¼ RðsÞ
P k
þ
T k Dk D
yields
ð Þ ¼ ½1 þ G2 ðsÞH 2ðsÞGþ1ðGsÞ1Gðs3ÞðGsÞ2½1ðsþÞH G1ð2sðÞsÞ½1 þ G3ðsÞH 3 ðsÞ
T s 5.5
The state equations are, x_ 1
¼ 2 x1 þ x2 x_ 2 ¼ 3 x2 þ x3 x_ 3 ¼ 3 x1 4 x2 5 x3 þ r y ¼ x2 Drawing the signal-flow diagram from the state equations yields 1
r
1
1 s
x 3
1 s
1
x 2
1 s
1
–3
–5
x 1
y
–2
–4 –3
5.6
ðs þ 5Þ we draw the signal-flow graph in controller canonical form ð Þ ¼ s100 2 þ 5 s þ 6
From G s
and add the feedback.
100
r
1
1 s
1 s
x 1
x 2
500 y
–5 –6
–1
15
16
Solutions to Skill-Assessment Exercises
Writing the state equations from the signal-flow diagram, we obtain x
¼
"
105
506
1
y
0
¼ ½ 100
# "# þ
x
1 0
r
500 x
5.7
From the transformation equations, 3 P1 ¼
2 4
1
Taking the inverse, P
0:4
¼
0:2 0:3
0:1
Now,
" ¼ "
# " #" #" # " # " #
3
# "
2 1 3 0:4 0:2 ¼ P1 AP 1 4 4 6 0:1 0:3 3 3 2 1 ¼ P1 B ¼ 11 1 4 3 0:4 0:2 ¼ ½ 0:8 1:4 CP ¼ ½ 1 4 0:1 0:3
6:5 9:5
8:5 11:5
#
Therefore,
"
# " #
6:5
8:5 3 z_ ¼ zþ 11 9:5 11:5 y ¼ ½ 0:8 1:4 z
u
5.8
First find the eigenvalues. l
0
0
l
1 4
3 6
j¼ ¼
jlI A
l
1 3 ¼ l2 þ 5l þ 6 lþ6 4
From which the eigenvalues are 2 and 3. Now use A xi l xi for each eigenvalue, l . Thus,
¼
For l
1
3
4 6
x1
x1
¼ x2
l
¼ 2, 3 x1 4 x1
þ 3 x2 ¼ 0 4 x2 ¼ 0
x2
Chapter 6 Solutions to Skill-Assessment Exercises
Thus x 1 For l
¼ x2 ¼ 3
Thus x 1
4 x1 4 x1
þ 3 x2 ¼ 0 3 x2 ¼ 0
¼ x2 and x1 ¼ 0:75 x2; from which we let 0:707 0:6 P¼ 0:707 0:8
Taking the inverse yields 5:6577 4:2433 P1 ¼
5
5
Hence, D
¼ P1AP ¼
P1 B CP
¼
"
"
5:6577 5
# " #" #" # " # ¼ # 4:2433 5
5:6577 4:2433 5
¼ ½1 4
"
1
5 0:707
0:707
4
0:707
6
1
18:39
3
20
0:6
0:8
3
¼ ½ 2:121
2:6
0:707
0:6 0:8
# " # ¼
2
0
0
3
Finally, z_
¼
" # " # 2
0
0
z
þ
3 y ¼ ½ 2:121
18:39
20 2:6 z
u
CHAPTER 6 6.1
Make a Routh table. s7
3
6
7
2
6
9
4
8
6
5
4.666666667
4.333333333
0
0
4
4.35714286
8
6
0
s s s
3
12.90163934
6.426229508
0
0
2
10.17026684
6
0
0
1
1.18515742
0
0
0
0
0
0
s s s
s0
6
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 6 s4 0 s2 6. Taking the derivative yields
þ
þ
17
18
Solutions to Skill-Assessment Exercises
24 s3 þ 0 s. 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
s5
1
s4
6
0
24
1 1
6
6
0
0
0
0
e
6
0
0
s1
144=e
0
0
0
s0
6
0
0
0
3
s
2
s
6 0
0 ROZ
There is one sign change below the even polynomial. Thus the even polynomial (4 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 halfplane poles, and 2 imaginary poles. th
6.3
ð Þ ¼ sðsKþðs2þÞðs20þÞ 3Þ ; T ðsÞ ¼ 1 þGðGsÞðsÞ ¼ s3 þ 5 s2 þKððs6þþ20K ÞÞ s þ 20K
Since G s
Form the Routh table. s3
1
2
s
30
1
s
ð6 þ K Þ
5
20K
15K
5 20K
s0
From the s1 row, K < 2. From the s0 row, K > 0. Thus, for stability, 0 < K < 2. 6.4
First find
2 j ¼ 64
j sI A
0 0 0 s 0 0 0 s
s
3 2 75 64
2 1 1 7 3 4
¼ s3 4 s2 33 s þ 51
3 ð Þ 75 ¼ 1 1 5
s
2 1 3
1 1 ðs 7Þ 1 4 ðs þ 5Þ
Now form the Routh table. s3
1
s2
4
1
s
0
s
33 51
20:25 51
There are two sign changes. Thus, there are two rhp poles and one lhp pole.
Chapter 7 Solutions to Skill-Assessment Exercises CHAPTER 7 7.1
a. First check stability. s2 þ 500 s þ 6000 10ðs þ 30Þðs þ 20Þ ð Þ ¼ 1 þGðGsÞðsÞ ¼ s3 þ1070 ¼ 2 s þ 1375 s þ 6000 ð s þ 26:03Þðs þ 37:89Þðs þ 6:085Þ
T s
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 15 15u t : 0 e step 1 lim G s 1
ðÞ
ð Þ¼ þ 1 ¼ þ1 ¼ ð Þ !0 15 15 ð Þ ¼ lim sG 1 ¼ ðsÞ 10 2030 ¼ 2:1875 s
15tu t :
ðÞ
eramp
!0
s
15t 2 u t : e parabola
ðÞ
ð Þ¼ 1
25 35 15 30 2 0 lim s G s
30 ¼ ¼ 1 ; since L 15t 2 ¼ 3 s ðÞ
!0
s
b. First check stability.
ð Þ 10 s2 þ 500 s þ 6000 T ðsÞ ¼ þ ð Þ ¼ s5 þ 110 s4 þ 3875 s3 þ 4:37e04 s2 þ 500 s þ 6000 Þðs þ 20Þ ¼ ð s þ 50:01Þðs þ 35Þ10ðs ðþs þ2530 Þðs2 7:189e 04 s þ 0:1372Þ G s 1 Gs
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 G s 1000 s 8 T s 1 Gs 1000 s s 9 s 7
ðs þ 8Þ ð Þ ¼ þ ð Þð Þ ¼ ð þ Þð þ Þð þþ Þ ð þ 8Þ ¼ s2 þ1000 1016 s þ 8063
and is of Type 0. Therefore,
1000 8 ¼ lim GðsÞ ¼ ¼ 127; K ¼ lim sGðsÞ ¼ 0; !0 !0 7 9
K p
s
¼ lim s2 GðsÞ ¼ 0 !0
and K a b.
v
s
s
e step
ð Þ ¼ 1 þ lim1 GðsÞ ¼ 1 þ1127 ¼ 7:8e 03 1 !0
s
eramp
1 1 ð Þ ¼ lim sG 1 ¼ ðsÞ 0 ¼ 1
!0
s
e parabola
1 ð Þ ¼ lim s12GðsÞ ¼ 10 ¼ 1 !0
s
19
20
Solutions to Skill-Assessment Exercises 7.3
System is stable for positive K . System is Type 0. Therefore, for a step input 1 12K e step 0:1. Solving for K p yields K p 9 lim G s ; from s!0 1 K p 14 18 which we obtain K 189.
1 ð Þ¼ þ
¼ ¼
¼ ¼
ðÞ¼
7.4
ð Þ ¼ 1000, and G 2ðsÞ ¼ ðð ss þþ 24ÞÞ,
System is stable. Since G 1 s eD
ð Þ¼ 1
1
1
lim
!0 G2 ðsÞ
s
þ lim G!10 ð sÞ s
¼ 2 þ 11000 ¼ 9:98e 04
7.5
ð Þ ¼ s þ1 1 1 ¼ s þ s1
System is stable. Create a unity-feedback system, where H e s The system is as follows: +
R(s)
E a(s)
C (s)
100 s+4
– –
−s s+1
Thus,
ð Þ ¼ 1 þ GGðSðsÞÞH ðsÞ ¼
Ge s
e
100 s 4 100 s s 1 s
ðþ Þ
1
ð þ Þð þ 4Þ
ðs þ 1Þ ¼ S100 2 95 s þ 4
Hence, the system is Type 0. Evaluating K p yields 100 K p 25 4
¼
¼
The steady-state error is given by e step
ð Þ ¼ 1 þ1K ¼ 1 þ1 25 ¼ 3:846e 02 1 p
7.6
ð Þ ¼ s2Kþðs2 sþþ7Þ10 ; eð 1Þ ¼ 1 þ1K ¼
Since G s
p
Calculating the sensitivity, we get
¼ K e@ @ Ke ¼
Se:K
K
þ
¼ 10 þ107K .
10 7
ð ðþ Þ Þ ¼
10 10 7K
þ
1 7K 1 10
10
7K 2
7K 10 7K
þ
Chapter 8 Solutions to Skill-Assessment Exercises 7.7
Given A
¼
0 3
1 ;B 6
0 ;C 1
¼
¼ ½ 1 1 ; RðsÞ ¼ 1s :
Using the final value theorem, e step
2 3 " # " # h i 5 ð Þ¼ 1 ð Þ ð Þ ¼ 4 ½ þ "þ # 3 2 " # 6 7 þ þ ¼ 7 ¼ 6 ½ þ þ ¼ 4 5 þ þ " #"# ð Þ¼ þ 1 ¼ ½ " # 2 3 "# ¼ þ½ ¼ þ ½ 4 5 ¼ lim sR s 1
C sI
!0
s
A
6
s
lim 1
1 1
!0
s
1 B
s2
lim 1
1 1
!0
s
1
s
3 s
6
2 3
2 3
1 0 1
1
0
3 s s 6 s 3
s2 lim s!0 s2
1
5 s 6 s
Using input substitution, step
CA1 B
1
1
1 1
6
1
1
3
6
1 0 1
1
3
1 1
0
0
0
3
1
1
1 3 0
1 1
2 3
CHAPTER 8 8.1
ð5 þ j9Þð3 þ j9Þ ð 7 þ j9Þ ¼ ðð7 7þþ j9 jÞ9ðþ72þÞð j97þþ3 jÞ9ðþ74þÞ0:0339 ¼ j9 þ 6Þ ð7 þ j9Þð 4 þ j9Þð 1 þ j9Þ 66 j72Þ ¼ 0:0339 j0:0899 ¼ 0:096 < 110:7 ¼ ðð944 j378Þ
a. F
b. The arrangement of vectors is shown as follows: jw
(–7+ j9)
s-plane
M 1
–7
X –6
M 2
M 3
–5
–4
M 4
X –3
M 5
–2
–1
X 0
s
21
22
Solutions to Skill-Assessment Exercises
From the diagram,
ð3 þ j9Þð5 þ j9Þ ¼ ð66 j72Þ 4 ð 7 þ j9Þ ¼ M M 1M 2M ¼ ð1 þ j9Þð 4 þ j9Þð7 þ j9Þ ð944 j378Þ 3 M 5
F
¼ 0:0339 j0:0899 ¼ 0:096 <; 110:7 8.2
a. First draw the vectors. jw
X
j3
j2 s-plane j1
–2
–3
–1
s
0
– j1
– j2
– j3
X
From the diagram,
P
angles ¼ 180 tan1
¼
3 1
tan1
3 1
180
108:43 þ 108:43 ¼ 180:
b. Since the angle is 180 , the point is on the root locus. P pole lengths
¼ P zero lengths ¼
c. K
p ffiffi ffi ffiþffi ffi ffi ffi p ffi ffi ffi þffi ffi ffi ffi ffi 12
12
32
1
32
¼ 10
8.3
First, find the asymptotes.
P ¼
poles # poles
s a
P
zeros ¼ ð2 4 6Þ ð0Þ ¼ 4 # zeros 30
¼ ð2k þ3 1Þp ¼ p3 ; p;
u a
5p 3
Chapter 8 Solutions to Skill-Assessment Exercises
Next draw root locus following the rules for sketching. 5 4 3 2 s 1 i x A g 0 a m I
–1 –2 –3 –4 –5 –8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
Real Axis
8.4
a. jw
j3
X
s-plane
s
O
0
–2
2
X
– j3
b. Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function. G s K s 2 T s 1 Gs 2K 13 s2 K 4 s
ð Þ ¼ þ ð Þð Þ ¼ þ ð ðÞ þþ ðÞ þ Þ
Using the denominator of T (s), make a Routh table. s2
1
s1
40 2K þ 13
2K
0
K
0
s
þ 13
0
We get a row of zeros for K 4. From the s 2 row with K 4; s2 21 0. From which we evaluate the imaginary axis crossing at 21. c. From part (b), K 4. d. Searching for the minimum gain to the left of 2 on the real axis yields 7 at a gain of 18. Thus the break-in point is at 7.
¼
p
¼
ffiffi ffi
¼
þ ¼
23
24
Solutions to Skill-Assessment Exercises
e. First, draw vectors to a point e close to the complex pole. jw s-plane j3
X
s –2
0
2
– j3
X
At the point e close to the complex pole, the angles must add up to zero. Hence, angle fromzero – angle frompole in 4th quadrant – angle from pole in 1st quadrant 3 180 , or tan1 90 u 180. Solving for the angle of departure, 4 u 233:1.
¼
¼
¼
8.5
a. z = 0.5 X
jw j4 s-plane
–3
X
0
– j4
o
o
2
4
s
Chapter 8 Solutions to Skill-Assessment Exercises
j4:06. b. Search along the imaginary axis and find the 180 point at s c. For the result in part (b), K 1. d. Searching between 2 and 4 on the real axis forthe minimum gain yieldsthe break-in at s 2:89. e. Searching along z 0:5 for the 180 point we find s 2:42 j4:18. f. For the result in part (e), K 0:108. g. Using the result from part ( c) and the root locus, K < 1.
¼
¼
¼
¼
¼
¼
þ
8.6
a. jw
z = 0.591 s-plane
X
X
X
–6
–4
–2
s 0
b. Searching along the z 0:591 (10% overshoot) line for the 180 point yields 2:028 j2:768 with K 45:55. p p 4 4 c. T s 1:97 s; T p 1:13 s; v n T r 1:8346 from the Re 2:028 Im 2:768 rise-time chart and graph in Chapter 4. Since vn is the radial distance to the pole,
þ ¼ j j ¼
¼ ¼
¼
¼ j j ¼
p ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ¼ þ ¼ 2:0282
vn
¼ 2K 46 ¼
K p
¼
¼
2:7682 3:431. Thus, T r 45:55 0:949. Thus, 48
¼ 0:53 s; since the system is Type 0,
¼
ð Þ ¼ 1 þ1K ¼ 0:51: 1 d. Searching the real axis to the left of 6 for the point whose gain is 45.55, we find 7:94. Comparing this value to the real part of the dominant pole,2:028, we find e step
p
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 p i as the root locus variable. Thus, 100 ð Þ ¼ 1 þGðGsÞðsÞ ¼ s2 þ p100 ¼ ¼ s þ 100 ð s2 þ 100Þ þ p s
T s
i
i
100 s2 100
þ p s 1þ 2 s þ 100 i
25
26
Solutions to Skill-Assessment Exercises
ð Þ ð Þ ¼ s2 þp s100. The following shows the root locus.
Hence, KG s H s
i
jw
s-plane
X j10
s
O
0
X – j10
8.8
Following the rules for plotting the root locus of positive-feedback systems, we obtain the following root locus: jw s-plane
o
X
X
X
–4
–3
–2
–1
0
s
8.9
ð Þ ¼ s2 þ ðKK ðsþþ21Þ sÞ þ K . Differentiating the
The closed-loop transfer function is T s denominator with respect to K yields
þ ðK þ 2Þ @ @ K s þ ðs þ 1Þ ¼ ð2 s þ K þ 2Þ @ @ K s þ ðs þ 1Þ ¼ 0 @ s @ s ðs þ 1Þ . Thus, S ¼ K @ s ¼ Kðs þ 1Þ : ¼ Solving for , we get @ K @ K ð2 s þ K þ 2Þ s @ K sð2 s þ K þ 2Þ 10ðs þ 1Þ. Substituting K ¼ 20 yields S ¼ sðs þ 11Þ Now find the closed-loop poles when K ¼ 20. From the denominator of T ðsÞ ; s1;2 ¼ 21:05; 0:95, when K ¼ 20. For the pole at 21:05, 10ð 21:05 þ 1Þ 0:05 ¼ 0:9975: DK ¼ 12:05 D s ¼ sðS Þ 21:05ð 21:05 þ 11Þ K 2 s
@ s @ K
s:K
s:K
s:K
Chapter 9 Solutions to Skill-Assessment Exercises
For the pole at
0:95,
D s
DK
¼ sðS Þ K ¼ 0:95 s:K
10 0:95
ð 0:95 þ 1Þ ð 0:95 þ 11Þ
0:05
¼ 0:0025:
CHAPTER 9 9.1
a. Searching along the 15% overshoot line, we find the point on the root locus at 3:5 j5:8 at a gain of K 45:84. Thus, for the uncompensated system, K v lim sG s K =7 45:84=7 6:55.
þ ¼ !0 ð Þ ¼
¼
¼ ¼ ð Þ ¼ 1=K ¼ 0:1527. 1 Hence, e b. Compensator zero should be 20 x further to the left than the compensator pole. ðs þ 0:2Þ . Arbitrarily select G ðsÞ ¼ ð s þ 0:01Þ c. Insert compensator and search along the 15% overshoot line and find the root locus at 3:4 þ j5:63 with a gain, K ¼ 44:64. Thus, for the compensated system, 44:64ð0:2Þ 1 ¼ 1 ð Þ ¼ ¼ 0:0078. 127:5 and e K ¼ ð7Þð0:01Þ K e 0:1527 ¼ ¼ 19:58 d. 0:0078 e s
ramp uncompensated
v
c
v
ramp compensated
v
ramp uncompensated ramp compensated
9.2
a. Searching along the 15% overshoot line, we find the point on the root locus at 3:5 j5:8 at a gain of K 45:84. Thus, for the uncompensated system,
þ
¼
4 ¼ jRe4 j ¼ 3:5 ¼ 1:143 s:
T s
b. The real part of the design point must be three times larger than the unj 3 5:8 compensated pole’s real part. Thus the design point is 3 3:5 10:5 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 180 130:8 49:2 . Using the following diagram,
ð Þþ ð Þ ¼
þ
¼
jw
j17.4 s-plane
49.2° – pc
s –10.5
17:4 tan49:2 , from which, p c 25:52. Adding this pole, we find P c 10:5 the gain at the design point to be K 476:3. A higher-order closed-loop pole is found to be at 11:54. This pole may not be close enough to the closed-loop zero at 10. Thus, we should simulate the system to be sure the design requirements have been met. we find
¼
¼
¼
27
28
Solutions to Skill-Assessment Exercises 9.3
a. Searching along the 20% overshoot line, we find the point on the root locus at 3:5 6:83 at a gain of K 58:9. Thus, for the uncompensated system,
þ
¼
4 4 ¼ jRe ¼ j 3:5 ¼ 1:143 s:
T s
b. For the uncompensated system, K v
¼ lim sGðsÞ ¼ K =7 ¼ 58:9=7 ¼ 8:41. Hence, !0 s
ð Þ ¼ 1=K ¼ 0:1189. 1 c. In order to decrease the settling time by a factor of 2, the design point is twice the uncompensated value, or 7 þ j 13:66. Adding the angles from the plant’s poles and the compensator’s zero at 3 to the design point, we obtain 100:8 . Thus, the compensator pole must contribute 180 100:8 ¼ 79:2. Using the following eramp
uncompensated
v
diagram,
jw j13.66
s-plane
79.2° – pc
s
–7
13:66 tan79:2 , from which, pc 9:61. Adding this pole, we find the P c 7 gain at the design point to be K 204:9. Evaluating K v for the lead-compensated system: we find
¼
¼
¼
¼ lim sGðsÞ G ¼ K ð3Þ=½ð7Þð9:61Þ ¼ ð204:9Þð3Þ=½ð7Þð9:61Þ ¼ 9:138: !0
K v
lead
s
K v for the uncompensated system was 8.41. For a 10 x improvement in steady-state error, K v must be 8:41 10 84:1. Since lead compensation gave us K v 9:138,
ð
Þð Þ ¼
¼
we need an improvement of 84:1=9:138 9:2. Thus, the lag compensator zero should be 9.2 x further to the left than the compensator pole. Arbitrarily select s 0:092 . Gc s s 0:01 Using all plant and compensator poles, we find the gain at the design point to be K 205:4. Summarizing the forward path with plant, compensator, and gain yields
ð Þ ¼ ðð þþ
¼
Þ Þ
¼
ðs þ 3Þðs þ 0:092Þ : ð Þ ¼ s205:4 ðs þ 7Þð9:61Þðs þ 0:01Þ
Ge s
Higher-order poles are found at 0:928 and 2:6. It would be advisable to simulate the system to see if there is indeed pole-zero cancellation.
Chapter 9 Solutions to Skill-Assessment Exercises 9.4
The configuration for the system is shown in the figure below.
R(s)
+
+
K
s(s+ 7)(s +10)
–
–
C (s)
1
K f s
Minor-Loop Design:
ð Þ ð Þ ¼ ð s þ 7ÞK ðs þ 10Þ. Using the following diagram, we f
For the minor loop, G s H s
find that the minor-loop root locus intersects the 0.7 damping ratio line at 8:5 j 8:67. The imaginary part was found as follows: u cos1 z 45:57 . Hence, Im tan45:57 , from which Im 8:67. 8:5
þ ¼
¼
¼
¼
jw
z = 0.7 (-8.5 + j8.67)
Im
X −10 −8.5
s-plane
X –7
q
s
The gain, K f , is found from the vector lengths as
p ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi p ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ¼ þ þ ¼
K f
1:52
8:672
1:52
8:672
77:42
Major-Loop Design: Using the closed-loop poles of the minor loop, we have an equivalent forward-path transfer function of
ð Þ ¼ sðs þ 8:5 þ j8:67K Þðs þ 8:5 j8:67Þ ¼ sðs2 þ 17 sK þ 147:4Þ :
Ge s
Using the three poles of Ge s as open-loop poles to plot a root locus, we search along z 0:5 and find that the root locus intersects this damping ratio line at 4:34 j7:51 at a gain, K 626:3.
¼
ðÞ
¼
þ
29
30
Solutions to Skill-Assessment Exercises 9.5
a. An active PID controller must be used. We use the circuit shown in the following figure: Z 2(s)
V i (s)
I 2(s)
Z 1(s)
V 1(s)
–
V o(s)
I a(s)
I 1(s)
+
where the impedances are shown below as follows: C 1 R2
C 2
R1 Z 1(s)
Z 2(s)
Matching the given transfer function with the transfer function of the PID controller yields 2 ð Þ ¼ ðs þ 0:1 sÞðs þ 5Þ ¼ s þ 5:1 s s þ 0:5 ¼ s þ 5:1 þ 0:5 8
Gc s
2 ¼ 64
1 R2 R1
þ
C 1 C 2
þ
R2 C 1 s
þ R1 sC 2
Equating coefficients
1
¼ 0:5 R2 C 1 ¼ 1 R2 C 1 þ ¼ 5:1 R1 C 2
ð1Þ ð2Þ ð3Þ
R1 C 2
3 75
In Eq. (2) we arbitrarily let C 1 105 . Thus, R2 105 . Using these values along with Eqs. (1) and (3) we find C 2 100mF and R 1 20 kV. b. The lag-lead compensator can be implemented with the following passive network, since the ratio of the lead pole-to-zero is the inverse of the ratio of the lag pole-to-zero:
¼ ¼
¼ ¼
R1 + vi (t )
+ C 1
R2 vo (t ) C 2
–
–
Chapter 10 Solutions to Skill-Assessment Exercises
Matchingthegiventransferfunctionwiththetransferfunctionofthepassivelag-lead compensator yields s 0:1 s 2 s 0:1 s 2 Gc s 2 s 0:01 s 20 s 20:01 s 0:2
ð Þ ¼ ð ð þþ ÞÞðð þþ Þ Þ ¼ ð þþ Þð þþ Þ 1 1 s þ sþ R1 C 1 R2 C 2 ¼ 1 1 1 1 þ þ s2 þ s þ R1 C 1 R2 C 2 R2 C 1 R1 R2 C 1 C 2
Equating coefficients
1 R1 C 1
1 R2 C 2
1
¼ 0:1
ð1Þ
¼ 0:1
ð2Þ
1
R1 C 1
1
þ R2C 2 þ R2C 1
¼
20:01
ð3Þ
Substituting Eqs. (1) and (2) in Eq. (3) yields 1 17:91 R2 C 1
ð4Þ
¼
Arbitrarily letting C 1 100 mF in Eq. (1) yields R1 Substituting C 1 100 mF into Eq. (4) yields R 2 Substituting R2 558 kV into Eq. (2) yields C 2
¼ ¼
¼ 100 kV. ¼ 558 kV. ¼ 900 mF .
¼
CHAPTER 10 10.1
a.
ð Þ ¼ ð s þ 2Þ1ðs þ 4Þ ; GðjvÞ ¼ ð8 þ v21Þ þ j6v
Gs
q ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ð Þ¼ ð Þ þð Þ p ffiffi ð Þ ¼ ffiffi ð Þ ¼ þ p
For v < For v < b.
v2
8
M v
8, f v
tan1
8, f v
p
6v
8
v2
tan1
2
6v
:
6v
8
2
v2
:
Bode Diagrams 0 –20 ) –40 B d ( –60 e d u –80 t i n g –100 a M 0 ; ) g e –50 d ( e s a –100 h P
–150
–200 10–1
100 101 Frequency (rad/sec)
102
31
32
Solutions to Skill-Assessment Exercises
c. Nyquist Diagrams
0.08
0.06 0.04 0.02
s i x A y r 0 a n i g a m –0.02 I
–0.04 –0.06
–0.08
–0.05
0
0.05
0.1
0.15
0.2
Real Axis
10.2
Asymptotic –20 dB/dec –40
–40 dB/dec Actual
–60 M g o –80 l 0 2
–20 dB/dec –40 dB/dec
–100 –120 1
0.1
100
10
1000
Frequency (rad/s)
–45o /dec ) s –50 e e r g e d ( –100 e s a h–150 P
–90o /dec –45o /dec –90o /dec –45o /dec
–45o /dec
Asymptotic –200
0.1
1
10 Frequency (rad/s)
Actual
100
1000
Chapter 10 Solutions to Skill-Assessment Exercises 10.3
The frequency response is 1/8 at an angle of zero degrees at v 0. Each pole rotates 90 in going from v 0 to v . Thus, the resultant rotates 180 while its magnitude goes to zero. The result is shown below.
¼
¼
¼1
Im
w = 0
w = ∞ 0
1
Re
8
10.4
a. The frequency response is 1/48 at an angle of zero degrees at v 0. Each pole rotates 90 in going from v 0 to v . Thus, the resultant rotates 270 while its magnitude goes to zero. The result is shown below.
¼
¼
¼1
Im
w = 6.63 w = ∞ – 1 480
w = 0 1 48
0
Re
ð Þ ¼ ð s þ 2Þðs þ1 4Þðs þ 6Þ ¼ s3 þ 12 s2 þ1 44 s þ 48 and 48 12v2 j 44v v3 simplifying, we obtain GðjvÞ ¼ 6 . The Nyquist diagram v þ 56v4 þ 784v2 þ 2304 crosses the real axis when the imaginary part of G ðjvÞ is zero. Thus, the Nyquist p diagram crosses the real axis at v2 ¼ 44; or v ¼ 44 ¼ 6:63 rad=s. At this fre1 quency G ðjvÞ ¼ . Thus, the system is stable for K < 480. 480
b. Substituting jv into G s
ffiffi ffi
33
10.5
If K
¼ 100, the Nyquist diagram will intersect the real axis at 100=480. Thus, ¼ 20log 480 ¼ 13:62 dB. From Skill-Assessment Exercise Solution 10.4, the 100
GM
180 frequency is 6.63 rad/s. 10.6
a.
–80 –100 –120 –140 –160 –180
1
10
100
1000
Chapter 10 Solutions to Skill-Assessment Exercises 10.8
j1350 v2 1350 v 160 6750000 101250v2 For both parts find that G jv . v6 2925v4 1072500v2 25000000 27 For a range of values for v, superimpose G jv on the a. M and N circles, and on the b. Nichols chart. a.
ð Þ ¼
þ ð Þ
þ
þ
þ
Im 3 G-plane Φ = 20°
2
M = 1.3
25°
M = 1.0 30° 1.4 40°
1.5 1
M = 0.7
50°
1.6 70°
1.8
0.6
2.0 0.4
0.5
Re
–70° –1 –50° –40° –30° –25° –2
–20°
–3
–4
–3
–2
1
–1
2
b.
Nichols Charts
0
) B d (
n i a G p o o L n e p O
0 dB
0.25 dB 0.5 dB 1 dB 3 dB 6 dB
–1 dB –3 dB –6 dB –12 dB –20 dB –40 dB
–50
–60 dB –80 dB –100 dB
–100
–120 dB –140 dB
–150
–160 dB –180 dB
–200
–200 dB –220 dB
–350
–300
–250
–200 –150 Open-Loop Phase (deg)
–100
–50
0
–240 dB
35
36
Solutions to Skill-Assessment Exercises
Plotting the closed-loop frequency response from a. or b. yields the following plot: 0 –20 –40
M g o l 0 2
–60 –80
–100 –120 1
10
100
1000
Frequency (rad/s) 0 –50 ) s –100 e e r g –150 e d ( e s –200 a h P
–250 –300
1
10
100
1000
Frequency (rad/s)
10.9
The open-loop frequency response is shown in the following figure:
Bode Diagrams 40 20 0 ) B –20 d ( e d u t i –40 n g a M ; ) g e d ( –100 e s a h P –120
–140 –160
10–1
100
101 Frequency (rad/sec)
102
Chapter 10 Solutions to Skill-Assessment Exercises
The open-loop frequency response is 7 at v 14:5 rad=s. Thus, the estimated bandwidth is vWB 14:5rad=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 180 151:98 28:02. This phase margin corresponds to
¼
¼
¼
p ffiffi ffi ffi ¼ q ffiðffi ffi ffiffi ffi ffi ffi ffi Þffi ffiþffi ffip ffi ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffiþffi ffi ffi ¼ q ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi p ffi p ffiffi ffiffi ffi ffi ffi ð Þ þ þ ¼
z ¼ 0:25: Therefore; %OS ¼ e
¼ v 4 z
T s
¼
T p
1
2z 2
4z 4
zp=
1 z 2
100
4z 2
2
4z 4
4z 2
44:4%
1:64 s and
BW
p
vBW
10.10
1
z 2
1
2z 2
2
0:33 s
The initial slope is 40 dB/dec. Therefore, the system is Type 2. The initial slope intersects 0 dB at v 9:5 rad=s. Thus, K a 9:52 90:25 and K p K v .
¼
¼
¼
¼ ¼ 1
10.11
ð Þ ¼ jvðjv10þ 1Þ ¼ vð 10 , from which the zero dB frev þ jÞ p 10 ¼ quency is found as follows: M ¼ p 1. Solving for v; v v2 þ 1 ¼ 10, or v v2 þ 1 after squaring both sides and rearranging, v4 þ v2 100 ¼ 0. Solving for the roots, v2 ¼ 10:51; 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 ffðv þ jÞ ¼ ffð3:08 þ jÞ ¼ 162. Therefore the phase margin is 180 162 ¼ 18 .
a. Without delay, G jv
ffiffi ffi ffi ffi ffi ffi ffi
ffiffi ffi ffi ffi ffi ffi ffi
b. With a delay of 0.1 s,
¼ ffðv þ jÞ vT ¼ ffð3:08 þ jÞ ð3:08Þð0:1Þð180=piÞ ¼ 162 17:65 ¼ 179:65 Therefore the phase margin is 180 179:65 ¼ 0:35. Thus, the system is table. f
c. With a delay of 3 s, f
¼ ffð v þ jÞ vT ¼ ffð3:08 þ jÞ ð3:08Þð3Þð180=piÞ ¼ 162 529:41 ¼ 691:41 ¼ 28:59 deg:
Therefore the phase margin is 28:59 unstable.
180 ¼ 151:41 deg. Thus, the system is
10.12
Drawing judicially selected slopes on the magnitude and phase plot as shown below yields a first estimate.
37
38
Solutions to Skill-Assessment Exercises Experimental 20 10 0 ) B d ( –10 n i a G –20 –30 –40
) g e d ( e s a h P
–45 –50 –55 –60 –65 –70 –75 –80 –85 –90 –95
1
2
3
4 5 6 7 8 10
20 30 40 50 70 100
200 300
500
1000
Frequency (rad/sec)
We see an initial slope on the magnitude plot of 20 dB/dec. We also see a final 20 dB/dec slope with a break frequency around 21 rad/s. Thus, an initial estimate is 1 . Subtracting G1 s from the original frequency response yields the G1 s s s 21 frequency response shown below.
ðÞ¼ ð þ Þ
ðÞ
Experimental Minus 1/s(s+21) 90 80 ) B 70 d ( n i a G 60
50 40
100 80 ) g 60 e d ( e s a h 40 P
20 0
1
2
3
4 5 6 7 8 10
20
30 40 50
Frequency (rad/sec)
70
100
200 300 500
1000
Chapter 11 Solutions to Skill-Assessment Exercises
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 20log 5:7K , or K 27:8. Thus, we estimate G2 s 27:8 s 7 . Thus, 27:8 s 5:7 . It is interesting to note that the original problem G s G1 s G2 s s s 21 30 s 5 was developed from G s . s s 20
¼ ð Þ ¼ ð Þ ¼ ð Þ ð Þ ¼ ð ðþþ Þ Þ ð Þ ¼ ð ðþþ ÞÞ
ðÞ¼
ðþ Þ
CHAPTER 11 11.1
The Bode plot for K
¼ 1 is shown below. Bode Diagrams
–60 –80 –100 –120 ) –140 B d ( e –160 d u t i –180 n g a M ; ) g e d ( –100 e s a h P
–150
–200
–250 10–1
100
101
102
103
Frequency (rad/sec)
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ffi ¼ ¼ s þ log
A 20% overshoot requiresz
% 100
0:456. This damping ratio implies a % 2 p2 log 100 phase margin of 48.10, which is obtained when the phase angle 1800 48:10 131:9 . This phase angle occurs at v 27:6 rad=s. The magnitude at this frequency 1 is 5:15 106. Since the magnitude must be unity K 194; 200. 5:15 106
¼
¼
¼
11.2
To meet the steady-state error requirement, K gain is shown below.
¼
þ
¼
¼ 1; 942; 000. The Bode plot for this
39
40
Solutions to Skill-Assessment Exercises Bode Diagrams 60 40 20 0 ) B –20 d ( e d –40 u t i n g a M ; ) g e d ( –100 e s a h P
–150
–200
–250 10–1
100
101
102
103
Frequency (rad/sec)
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ffi ¼ ¼ s þ log
A 20% overshoot requires z
% 100
0:456. This damping ratio % 2 p2 log 100 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 180 58:1 129:9 . 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 23:2 dB. Insert a break at 0:1 20:4 2:04 rad=s. At this frequency, draw 23:2 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 s 2:04 , where the gain is chosen to yield 0 dB at low frequencies, Gc s K c s 0:141 or K c 0:141=2:04 0:0691. In summary,
þ
¼
ð
Þ¼
ð Þ ¼ ðð þþ ÞÞ ¼ ¼ ðs þ 2:04Þ and GðsÞ ¼ 1;942;000 G ðsÞ ¼ 0:0691 ð s þ 0:141Þ sðs þ 50Þðs þ 120Þ c
11.3
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ¼ ¼ s þ q ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi p ffi ¼ ð Þþ þ ¼ log
A 20% overshoot requires z
is then calculated as v BW
4
T s z
% 100
p2
log2
1
2z 2
0:456. The required bandwidth
% 100
4z 4
4z 2
2
57:9 rad/s. In order
¼ 50 ¼ ð50ÞK ð120Þ, we calculate
to meet the steady-state error requirement of K v
¼ 300; 000. The uncompensated Bode plot for this gain is shown below.
K
Chapter 11 Solutions to Skill-Assessment Exercises Bode Plot for K = 300000 40 20 0 –20 ) B –40 d ( e d –60 u t i n g a M ; ) g e d ( –100 e s a h P
–150
–200
–250 10–1
100
101
102
103
Frequency (rad/sec)
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 144:8. Therefore, the uncompensated system’s phase margin is 180 144:8 35:2 . The required phase margin based on the required damping 2z ratio is FM tan1 48:1. Adding a 10 correction factor, the 2z 2 1 4z 4 required phase margin is 58:1. Hence, the compensator must contribute 1 b 1 sinfmax fmax 58:1 35:2 22:9 . Using fmax sin1 , b 0:44. 1 b 1 sinfmax 1 The compensator’s peak magnitude is calculated as M max 1:51. Now find
þ
¼
¼
¼
q ffiffi ffi ffi ffi ffi ffiþffi ffip ffi ffiffi ffi ffiþffi ffi ffi ffi ffi ffi ¼
¼
þ
¼
¼ þ ¼ p b ¼
ffiffi
¼
the frequency at which the uncompensated system has a magnitude 1=M max, or 3:58 dB. From the Bode plot, this magnitude occurs at vmax 50 rad=s. The 1 1 compensator’s zero is at zc . vmax Therefore, zc 33:2.
¼
¼ T p b ¼ z 1 The compensator’s pole is at P ¼ ¼ ¼ 75:4. The compensator gain is bT b ¼ T
c
ffiffi
c
chosen to yield unity gain at dc. Hence,
¼ 75:4=33:2 ¼ 2:27.
K c
ð Þ sðs þ300;000 . 50Þðs þ 120Þ
G s
Þ, ð Þ ¼ 2:27 ðð ss þþ 33:2 75:4Þ
Summarizing, Gc s
and
41
42
Solutions to Skill-Assessment Exercises 11.4
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ¼ ¼ s þ q ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffi p ffi ffiffi ffiffi ffi ffi ffi ð Þ þ þ ¼ ¼ p log
A 10% overshoot requires z
log2
p2 p
is then calculated as vBW
T p
% 100
z 2
1
0:591. The required bandwidth
% 100
1
2z 2
4z 4
4z 2
In order to meet the steady-state error requirement of K v calculate K
2
7:53 rad= s:
¼ 10 ¼ ð8ÞK ð30Þ, we
¼ 2400. The uncompensated Bode plot for this gain is shown below. Bode Diagrams
40 20 0 –20 –40 ) –60 B d ( –80 e d u –100 t i n g a M ; ) g e d ( –100 e s a h P
–150
–200
–250 10–1
100
101
102
103
Frequency (rad/sec)
Let us select a new phase-margin frequency at 0:8vBW 6:02 rad=s. The required phase margin based on the required damping ratio is FM tan1 2z 58:6 . Adding a 5 correction factor, the required phase 2z 2 1 4z 4 margin is 63:6. At 6.02 rad/s, the new phase-margin frequency, the phase angle is–which represents a phase margin of 180 138:3 41:7. Thus, the lead compensator must contribute fmax 63:6 41:7 21:9. 1 b 1 sinfmax Using fmax sin1 ,b 0:456. 1 b 1 sinfmax We now design the lag compensator by first choosing its higher break frequency one decade below the new phase-margin frequency, that is, zlag 0:602 rad=s. The lag compensator’s pole is plag bzlag 0:275. Finally, the lag compensator’s gain is K lag b 0:456.
¼
q ffiffi ffi ffi ffi ffi ffiþffi ffip ffi ffiffi ffi ffi ffiþffi ffi ffi ffi ffi ¼ ¼
¼ ¼
¼ ¼ ¼ þ ¼
þ
¼
¼
¼
¼
¼
Chapter 12 Solutions to Skill-Assessment Exercises
Now we design the lead compensator. The lead zero is the product of the new phase margin frequency and
p b, or z ¼ 0:8v p b ¼ 4:07. Also, p ¼ z b
ffiffi
lead
BW
¼ 8:93. Finally, K ¼ 1b ¼ 2:19. Summarizing, lead
lead
ffiffi
lead
Þ ; G ðsÞ ¼ 2:19 ðs þ 4:07Þ ; ¼ ðsÞ ¼ 0:456 ðð ss þþ 0:602 ð s þ 8:93Þ 0:275Þ
Glag
and k
lead
¼ 2400:
CHAPTER 12 12.1
We first find the desired characteristic equation. A 5% overshoot requires % log p 100 z 0:69. Also, vn 14:47 rad= s. Thus, the char2 T z 1 p 2 % p2 log 100
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ¼ ¼ s þ
¼
p ffiffi ffiffi ffi ffi ffi ¼
acteristic equation is s2 2zvn s v2n s2 19:97 s 209:4. Adding a pole at 10 to cancel the zero at 10 yields the desired characteristic equation, 2 19:97 s 209:4 s 10 s s3 29:97 s2 409:1 s 2094. The compensated system matrix in phase-variable form is
A
þ
þ
BK
2 ¼ 64 ð
þ þ ¼ þ ð þ Þ ¼ þ þ
0 0
1 0
0 1
3 75 Þ
þ
þ
. The characteristic equation for this
Þ ð36 þ k2Þ ð15 þ k3 system is j sI ð BKÞj ¼ s3 þ ð15 þ k3 Þ s2 þ ð36 þ k2 Þ s þ ðk1 Þ. Equating coeffik1 A
cients of this equation with the coefficients of the desired characteristic equation yields the gains as K
¼ ½ k1
k2
k3
¼ ½ 2094
373:1 14:97 :
12.2
2 ¼ 64
3 75
2 1 1 The controllability matrix is CM B AB A2 B 1 4 9 . Since 1 1 16 CM 80, CM is full rank, that is, rank 3. We conclude that the system is controllable.
¼
j j¼
12.3
First check controllability. The controllability matrix is CMz B AB A2 B 0 0 1 0 1 17 . Since CMz 1, CMz is full rank, that is, rank 3. We conclude that 1 9 81 the system is controllable. We now find the desired characteristic equation. A 20% % log 4 100 overshoot requires z 0:456. Also, vn 4:386 rad=s. T z s 2 % p2 log 100
2 64
3 75
¼
j
j¼
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ¼ ¼ s þ
¼
¼
¼
43
44
Solutions to Skill-Assessment Exercises
Thus, the characteristic equation is s2 2zvn s v2n s2 4 s 19:24. Adding a pole at 6 to cancel the zero at 6 yields the resulting desired characteristic equation,
þ
s2
þ ¼ þ þ
þ 4 s þ 19:24 ðs þ 6Þ ¼ s3 þ 10 s2 þ 43:24 s þ 115:45: ðs þ 6Þ sþ6 ¼ Since GðsÞ ¼ ð s þ 7Þðs þ 8Þðs þ 9Þ s3 þ 24 s2 þ 191 s þ 504, we can write the phase-
variable
representation
2 ¼ 64 3 75 þ Þ
as
0 0
Ap
1 0
3 75
0 1
23 ¼ 64 75
; Bp
0 0 ; Cp
¼
504 191 24 1 6 1 0 . The compensated system matrix in phase-variable form is A p Bp Kp 0 1 0 0 0 1 . The characteristic equation for this 504 k1 191 k2 24 k3 3 system is sI 24 k3 s2 191 k2 s 504 k1 . Equat s Ap Bp Kp ing coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as Kp 388:55 147:76 14 . We k1 k2 k3 now develop the transformation matrix to transform back to the z -system.
½
2 64 ð
þ Þ ð þ Þ ð j j¼ þð þ Þ þð þ Þ þð ¼ ½ ¼ ½
h ¼
CMz
CMp
Bz
h ¼
A2z B z
Az Bz
Bp
2 i ¼ 64
A2p Bp
Ap Bp
Therefore,
P
2 ¼ 64
1 Mz CM x
¼C
0 0 1
0 1 9
1 17 81
32 7564
¼ ½ 40:23
p
62:24
2 i ¼ 64
0 1 9
14 .
1 17 81
0 0 1
0 1 24
3 2 75 ¼ 64 2 64
191 24 1 24 1 0 1 0 0
¼ K P1 ¼ ½ 388:55 147:76
Hence, K z
0 0 1
14
3 75
¼
þ Þ
and
3 75
1 24 : 385
1 0 0 7 1 0 56 15 1
3 75
1 7
0 1
0 0
49
15
1
3 75
12.4
2 ð Þ ¼ 64 ð ð
3 75
þ Þ 1 0 þ Þ 0 1 e . The characteristic For the given system e ¼ ðA LC e 504 þ l 3 Þ 0 0 polynomial is given by j½ sI ðA LCÞj ¼ s3 þ ð24 þ l 1 Þ s2 þ ð191 þ l 2 Þ s þ ð504 þ l 3 Þ. Now we find the desired characteristic equation. The dominant poles from Skill-Assessment Exercise 12.3 come from s2 þ 4 s þ 19:24 . Factoring yields ð2 þ j3:9Þ and ð2 j3:9Þ. Increasing these poles by a factor of 10 and adding a x_
x
24 l 1 191 l 2
x
third pole 10 times the real part of the dominant second-order poles yields the
Chapter 12 Solutions to Skill-Assessment Exercises
desired characteristic polynomial, s 20 j39 s 20 j39 s 200 s3 240 s2 9921 s 384200. Equating coefficients of the desired characteristic equation to the 216 system’s characteristic equation yields L 9730 .
þ
ð þ þ Þð þ Þð þ Þ ¼ þ
þ
2 3 75 ¼ 64 2 3 2 ¼ 64 75 ¼ 64 383696
12.5
C
The
observability
matrix
is OM
64 674
CA
CA2
A2
2 ¼ 64
6
8
80 78 848
814
3 75
,
where
25 28 32 7 4 11 . The matrix is of full rank, that is, rank 3, since 77 95 94 1576. Therefore the system is observable.
jO j ¼ M
3 75
4
12.6
The system is represented in cascade form by the following state and output equations: z_
2 ¼ 64
7 0 0
y
3 23 75 þ 64 75
1 8
0
¼ ½1
0 1 z
0 0 u
9
1
0 0 z
2 3 2 75 ¼ 64 ¼ 64
1 7 49
Cz
The observability matrix is OMz
Cz Az
Cz A2z
2 ¼ 64
where A2z
49 0 0
15 64 0
1
17 81
3 75
3 75
0 0 1 0 , 15 1
ð Þ ¼ ð s þ 7Þðs þ1 8Þðs þ 9Þ
. Since G s
¼ s3 þ 24 s2 þ1191 s þ 504, we can write the observable canonical form as 24 1 0 0 x_ ¼ 191 0 1 x þ 0 u 504 0 0 1
2 64
y
¼ ½1
3 23 75 64 75
0 0 x
2 3 2 75 ¼ 64 ¼ 64
The observability matrix for this form is OMx
Cx
1
Cx Ax
24 385
Cx A2x
0
0
3 75
1 0 , 24 1
45
46
Solutions to Skill-Assessment Exercises
where
2 ¼ 64
385
A2x
24 191
4080 12096
1
3 75
0 : 0
504
We next find the desired characteristic equation. A 10% overshoot requires % log 4 100 z 0:591. Also, vn 67:66 rad=s. Thus, the characteristic z T s % 2 p2 log 100
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ¼ ¼ s þ
¼
¼
equation is s 2 2zvn s v2n s2 80 s 4578:42. Adding a pole at 400, or 10 times the real part of the dominant second-order poles, yields the resulting desired characteristic equation, s2 80 s 4578:42 s 400 s3 480 s2 36580 s 1:831 x106 . For Ax Lx Cx e x the system represented in observable canonical form ex_ 24 l 1 1 0 191 l 2 0 1 ex . The characteristic polynomial is given by 504 l 3 0 0 sI s3 Ax Lx Cx 24 l 1 s2 191 l 2 s 504 l 3 . Equating coefficients of the desired characteristic equation to the system’s characteristic equation 456 yields L x 36; 389 . 1; 830; 496 Now, develop the transformation matrix between the observer canonical and cascade forms.
þ
2 ð 64 ð ð
þ Þ þ Þ þ Þ j½ ð
þ ¼ þ
þ
þ
ð þ
þ
3 75
¼ O1 O Mz
2 6 ¼ 64 2 6 ¼ 64 2 6 ¼ 64
Mx
1
0
7
1
49
15 0
0
7
1
0
Finally,
2 ¼ 64
1 17 81
0 0 1 0 9 1
32 7564
15 1
þ
¼ ð
1
0
0
0
24
1
0
1
385
24
1
0 0
17
1 0
9
456 36; 389 1; 830; 496
1
0
0
24
1
0
385
1
81
1
32 77 66 54 32 7766 54 3 77 5 0
1
56
x
þ
Þ ¼
3 75
P
¼ PL
Þ¼ þ
Þj ¼ þ ð þ Þ þ ð þ Þ þ ð þ Þ
2 ¼ 64
Lz
24 1
3 77 5
3 77 5
1
3 2 75 ¼ 64
456 28; 637 1; 539; 931
3 2 75 64
3 75
456 28; 640 . 1; 540; 000
Chapter 12 Solutions to Skill-Assessment Exercises 12.7
We first find the desired characteristic equation. A 10% overshoot requires % log 100 0:591 z % 2 p2 log 100
ffiffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffi ffi ffi ffi ffiffi ¼ ¼ s þ
¼ p 2 ¼ 1:948 rad=s. Thus, the characteristic equation is s2þ T 1 z 2 2zv s þ v ¼ s2 þ 2:3 s þ 3:79. Adding a pole at 4, which corresponds to the original system’s zero location, yields the resulting desired characteristic equation, s2 þ 2:3 s þ 3:79 ðs þ 4Þ ¼ s3 þ 6:3 s2 þ 13 s þ 15:16. ðA BKÞ BK x þ 0 r ; and y ¼ ½ C 0 x , x_ ¼ Now, x_ C x x 0 1 Also, vn
p ffiffi ffi ffi ffi ffi ffi p
n
n
e
N
N
N
where A
BK
¼ ½ ¼ ¼ 0 7
1 9
0 1
k1
0
0 7
k2
1 9
0
0
k1
k2
1
ð7 þ k1Þ ð9 þ k2Þ C ¼ ½4 1
¼ ¼ 0 k 1 e
Bke
0
ke
Thus,
2 3 2 64 75 ¼ 64 ð þ x_1
0
x_2
7
x_N
1 k1
4
0
Þ ð9 þ k2Þ 1
ke
0
32 3 7564 75 þ x1
0
x2
1
xN
x1
r ; y
¼ ½4
1 0
A
sI
Þ
BK
C
s
7
k1
4
BK e
0
¼ 264
0 0 0 s 0
s
0 0
1 0 s þ ð9 þ k2 Þ k
e
1
s
3 75 ¼
s
3 2 75 64 ð þ
s3
0
7
x2
1 k1
4
0
Þ ð9 þ k2Þ 1
ke
0
e
þ 6:3 s2 þ 13 s þ 15:16 ¼ s3 þ ð9 þ k2 Þ s2 þ ð7 þ k1 þ k Þ s þ 4k e
Solving for the k ’s, K
3 75
þ ð9 þ k2 Þ s2 þ ð7 þ k1 þ k Þ s þ 4k
Equating this polynomial to the desired characteristic equation, s3
.
xN
Finding the characteristic equation of this system yields
ð 2 ¼ 64 ð þ Þ
2 3 64 75
¼ ½ 2:21 2:7 and k ¼ 3:79: e
e
e
47
48
Solutions to Skill-Assessment Exercises CHAPTER 13 13.1
f ðt Þ ¼ sinðvkT Þ; f ðt Þ ¼
1
1
P
sin vkT d t
Þ ð kT Þ; e e
ð
k 0
¼
1
jvkT
jvkT
ekTs
X X ðÞ¼ ð Þ ¼ X ¼ P ¼ ðÞ¼ ¼ ð ð Þ ¼ ¼
F
s
sin vkT ekTs
k 0
1 1 eT ðs jvÞ 2 j k¼0
But,
2 j
k 0
¼
1 k x k¼0
¼ k eT ðsþ jvÞ
k
1
x1
1
Thus, F
s
1 2 j 1
eTs
1
1
1
eT ðs jvÞ
eT ðsþ jvÞ
1
sin vT eTs 2cos vT
ð Þ þ e2
eTs e jvT eTs e jvT eTs e jvT eTs e jvT e2Ts
1 2 j 1
Ts
1
z1 sin vT 2z1 cos vT
ð Þ ð Þ þ z2
Þþ
13.2
zðz þ 1Þðz þ 2Þ ð Þ ¼ ðz 0:5 Þðz 0:7Þðz 0:9Þ
F z
ð Þ¼
F z z
ð þ 1Þðz þ 2Þ ðz 0:5Þðz 0:7Þðz 0:9Þ z z
¼ 46:875 z z0:5 114:75 z 10:7 þ 68:875 z z0:9 ð Þ ¼ 46:875 z z0:5 114:75 z z0:7 þ 68:875 z z0:9 ;
F z
k
k
ð Þ ¼ 46:875ð0:5Þ 114:75ð0:7Þ þ 68:875ð0:9Þ
f kT
k
13.3
ð Þ ¼ 1 e
Since G s
8
Ts
ðþÞ ¼ 4
s s
,
z1
8
1 z A þ B ¼ z 1 z 2 þ 2 1 G z z sðs þ 4Þ z s s þ 4 z s s þ 4 2 2 Let G2 ðsÞ ¼ þ . Therefore, g 2 ðt Þ ¼ 2 2e4 , or g 2 ðkT Þ ¼ 2 2e4 . s s þ 4 2z 1 e4 2z 2z ¼ ðz 1Þðz e4 Þ. Hence, G 2 ðzÞ ¼ z 1 z e4 ð Þ¼
z
t
T
kT
T
T
:
Chapter 13 Solutions to Skill-Assessment Exercises
2 1 e4 1 G2 ðzÞ ¼ Therefore, G ðzÞ ¼ z ðz e4 Þ . 1 1:264 For T ¼ s, GðzÞ ¼ . z 0:3679 4 T
z
T
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. R(s)
+ –
G1(s)G2(s)
C (s)
H (s)G1(s)G2(s)
1 G2 ðzÞ ð Þ ¼ 1 þGHG . 1 G2 ðzÞ
Hence, T z 13.5
ð Þ ¼ s 20 þ 5.
ð Þ ¼ G sðsÞ ¼ sðs20þ 5Þ ¼ s4 s þ4 5. Taking the inverse Laplace transform and letting t ¼ kT , g 2 ðkT Þ ¼ 4 4e5 . Taking the z-transform 4z 1 e5 4z 4z ¼ ðz 1Þðz e5 Þ. yields G 2 ðzÞ ¼ z 1 z e5 4 1 e5 z1 G2 ðzÞ ¼ ðz e5 Þ . Now, GðzÞ ¼ z 4 1 e5 GðzÞ ¼ Finally, T ðzÞ ¼ . 1 þ GðzÞ z 5e5 þ 4 The pole of the closed-loop system is at 5e5 4. Substituting values of T , we find Let G s
Let G2 s
kT
T
T
T
T
T
T
T
T
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 s þ 11 into DðzÞ ¼ z3 z2 0:5z þ 0:3, we obtain DðsÞ ¼ s3 8 s2
27 s 6. The Routh table for this polynomial is shown below. s3
1
s2
8 27:75 6
1
s
s0
27 6
0
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.
49
50
Solutions to Skill-Assessment Exercises 13.7
Defining G( s) as G 1( s) in cascade with a zero-order-hold,
eTs
ð Þ ¼ 20 1
G s
ðs þ 3Þ ¼ 20 1 e sðs þ 4Þðs þ 5Þ
Taking the z-transform yields
Ts
3=20
ð Þ þ ð Þ ð Þ ¼ z1
ð Þ ¼ 20 1
G z
3=20 z z 1
1=4 z
2=5 z
e4T
z
z
e5T
1=4 s 4
2=5 : s 5
s
þð þ Þð þ Þ
3
þ z5ðz e14 Þ z8ðz e15 Þ : T
T
1 ¼ 0:1 second, K ¼ lim GðzÞ ¼ 3, and K v ¼ lim ðz 1ÞGðzÞ ¼ 0, and !1 T !1 1 2 K a ¼ 2 lim ðz 1Þ GðzÞ ¼ 0. Checking for stability, we find that the system is T !1 GðzÞ 1:5z 1:109 ¼ stable for T ¼ 0:1 second, since T ðzÞ ¼ has poles 2 1 þ GðzÞ z þ 0:222z 0:703 inside the unit circle at 0:957 and þ 0:735. Again, checking for stability, we find that GðzÞ ¼ the system is unstable for T ¼ 0:5 second, since T ðzÞ ¼ 1 þ GðzÞ 3:02z 0:6383 has poles inside and outside the unit circle at þ 0:208 and 2 z þ 2:802z 0:6272 3:01, respectively. Hence for T
p
z
z
z
13.8
Draw the root locus superimposed over the z 0:5 curve shown below. Searching along a 54:3 line, which intersects the root locus and the z 0:5 curve, we find the point 0:587 54:3 0:348 j 0:468 and K 0:31.
¼
ff
¼ ð
þ
Þ
¼
¼
z-Plane Root Locus
1.5
1
(0.348 + j 0.468) K = 0.31
0.5 s i x A g 0 a m I
54.3°
–0.5
–1
–1.5
–3
–2.5
–2
–1.5
–1
–0.5 Real Axis
0
0.5
1
1.5
2
13.9
Let
ðs þ 25:3Þ ¼ 342720ðs þ 25:3Þ : ð Þ ¼ GðsÞG ðsÞ ¼ sðs þ 36100ÞðsK þ 100Þ 2:38 ð s þ 60:2Þ sðs þ 36Þðs þ 100Þðs þ 60:2Þ
Ge s
c
Chapter 13 Solutions to Skill-Assessment Exercises
The following shows the frequency response of G e jv .
ð Þ
Bode Diagrams 40 20 0 –20 ) B –40 d ( e d u –60 t i n g a M ; ) g e d ( –100 e s a h P
–150
–200
–250 10–1
100
101
102
103
Frequency (rad/sec)
We find that the zero dB frequency, vFM , for Ge jv is 39 rad/s. Using Astrom’s guideline the value of T should be in the range, 0:15=vFM 0:0038 second to 0:5=vFM 0:0128 second. Let us use T 0:001 second. Now find the Tustin 2z 1 transformation for the compensator. Substituting s into Gc s T z 1 2:38 s 25:3 with T 0:001 second yields s 60:2
ð Þ
¼
¼
ðþ Þ ðþ Þ
¼ ¼ ðð ÞÞ
ðÞ¼
¼
0:975Þ ð Þ ¼ 2:34 ððzz 0:9416 Þ:
Gc z
13.10
z2 3761z þ 1861 ð Þ ¼ XEððzzÞÞ ¼ 1899 . Cross-multiply and obtain z2 1:908z þ z2 1:908z þ 0:9075 0:9075 X ðzÞ ¼ 1899z2 3761z þ 1861 EðzÞ. Solve for the highest power of z operating on the output, X (z), and obtain z2 X ðzÞ ¼ 1899z2 3761z þ 1861 EðzÞ ð 1:908z þ 0:9075Þ X ðzÞ. Solving for X (z) on the left-hand side yields
Gc z
51
52
Solutions to Skill-Assessment Exercises
ð Þ ¼ 1899 3761z1 þ 1861z2 EðzÞ 1:908z1 þ 0:9075z2 X ðzÞ. Finally,
X z
we implement this last equation with the following flow chart:
x *(t )
e*(t )
1899 Delay 0.1 second
Delay 0.1 second +
e*(t -0.1)
–3761
+
–
+ +
–1.9.08
– Delay 0.1 second
Delay 0.1 second e*(t -0.2)
x *(t -0.1)
1861
x *(t -0.2)
0.9075
Copyright # 2011 by John Wiley & Sons, Inc. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008 or online at http://www.wiley.com/go/permissions. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. ISBN 13 978-0470-54756-4 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 1
