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Solutions to Skill-Assessment
Exercises
CHAPTER 2
2.1
The Laplace transform of t is
F ðsÞ ¼
1
ð s þ 5Þ 2
1
using Table 2.1, Item 3. Using Table 2.2, Item 4,
s2
.
2.2
Expanding F(s) by partial fractions yields:
A
B
C
D
þ
F ðsÞ ¼ þ
þ
s s þ 2 ð s þ 3Þ 2 ð s þ 3Þ
where,
5
A¼
¼
2 ðs þ 2Þðs þ 3Þ S!0 9
10
C¼
B¼
2 s ð s þ 3Þ
10
¼ 5
S!2
10 10
40
2 dF ðsÞ
;
and
D
¼
ð
s
þ
3
Þ
¼
¼
sðs þ 2Þ S!3
3
ds s!3
9
Taking the inverse Laplace transform yields,
5
10
40
f ðtÞ ¼ 5e2t þ te3t þ e3t
9
3
9
2.3
Taking the Laplace transform of the differential equation assuming zero initial
conditions yields:
s3 CðsÞ þ 3s2 CðsÞ þ 7sCðsÞ þ 5CðsÞ ¼ s2 RðsÞ þ 4sRðsÞ þ 3RðsÞ
Collecting terms,
3
s þ 3s2 þ 7s þ 5 CðsÞ ¼ s2 þ 4s þ 3 RðsÞ
Thus,
C ðsÞ
s2 þ 4s þ 3
¼ 3
RðsÞ s þ 3s2 þ 7s þ 5
1
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2.4
G ðsÞ ¼
C ðsÞ
2s þ 1
¼
RðsÞ s2 þ 6s þ 2
Cross multiplying yields,
d2 c
dc
dr
þ 6 þ 2c ¼ 2 þ r
dt2
dt
dt
2.5
CðsÞ ¼ RðsÞGðsÞ ¼
1
s
1
A
B
C
¼
¼ þ
þ
s2 ðs þ 4Þðs þ 8Þ sðs þ 4Þðs þ 8Þ
s ð s þ 4 Þ ð s þ 8Þ
where
1
1
A¼
¼
ðs þ 4Þðs þ 8ÞS!0 32
1 1
1 1
B¼
¼ ; and C ¼
¼
sðs þ 8ÞS!4
16
sðs þ 4ÞS!8 32
Thus,
c ðt Þ ¼
1
1
1
e4t þ e8t
32 16
32
2.6
Mesh Analysis
Transforming the network yields,
s
1
V(s)
I9(s)
+
_
V1(s)
1
+
s
I1(s)
s
V2(s)
_
I2(s)
Now, writing the mesh equations,
ðs þ 1ÞI 1 ðsÞ sI 2 ðsÞ I 3 ðsÞ ¼ V ðsÞ
sI 1 ðsÞ þ ð2s þ 1ÞI 2 ðsÞ I 3 ðsÞ ¼ 0
I 1 ðsÞ I 2 ðsÞ þ ðs þ 2ÞI 3 ðsÞ ¼ 0
Solving the mesh equations for I2(s),
ð s þ 1Þ V ð s Þ
1 s
0
1 2
1
0
ð s þ 2Þ s þ 2s þ 1 V ðsÞ
¼
I 2 ðsÞ ¼ sðs2 þ 5s þ 2Þ
s
1 ð s þ 1Þ
s
ð2s þ 1Þ
1 1
1
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Chapter 2 Solutions to Skill-Assessment Exercises
But, VL ðsÞ ¼ sI 2 ðsÞ
Hence,
2
s þ 2s þ 1 V ðsÞ
VL ðsÞ ¼
ðs2 þ 5s þ 2Þ
or
VL ðsÞ s2 þ 2s þ 1
¼ 2
V ðsÞ
s þ 5s þ 2
Nodal Analysis
Writing the nodal equations,
1
þ 2 V 1 ðsÞ VL ðsÞ ¼ V ðsÞ
s
2
1
þ 1 VL ðsÞ ¼ V ðsÞ
V 1 ðsÞ þ
s
s
Solving for VL(s),
1
þ2
V ðsÞ s
1
2
1
V ðsÞ s þ 2s þ 1 V ðsÞ
s
¼
VL ðsÞ ¼ 1
ðs2 þ 5s þ 2Þ
þ
2
1
s
2
1
þ
1
s
or
VL ðsÞ s2 þ 2s þ 1
¼ 2
V ðsÞ
s þ 5s þ 2
2.7
Inverting
G ðsÞ ¼ Z2 ðsÞ 100000
¼ s
¼ Z 1 ðsÞ
105 =s
Noninverting
G ðsÞ ¼
½ Z 1 ðsÞ þ Z ðsÞ
¼
Z1 ðsÞ
105
þ 105
s
!
105
s
!
¼sþ1
2.8
Writing the equations of motion,
2
s þ 3s þ 1 X1 ðsÞ ð3s þ 1ÞX2 ðsÞ ¼ F ðsÞ
ð3s þ 1ÞX1 ðsÞ þ s2 þ 4s þ 1 X2 ðsÞ ¼ 0
3
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Solutions to Skill-Assessment Exercises
Solving for X2(s),
2
s þ 3s þ 1 F ðsÞ ð3s þ 1Þ
ð3s þ 1ÞF ðsÞ
0 ¼
X 2 ðsÞ ¼ sðs3 þ 7s2 þ 5s þ 1Þ
2
s
þ
3s
þ
1
ð
3s
þ
1
Þ
2
ð3s þ 1Þ
s þ 4s þ 1 Hence,
X 2 ðsÞ
ð3s þ 1Þ
¼ 3
F ðsÞ
sðs þ 7s2 þ 5s þ 1Þ
2.9
Writing the equations of motion,
2
s þ s þ 1 u1 ðsÞ ðs þ 1Þu2 ðsÞ ¼ T ðsÞ
ðs þ 1Þu1 ðsÞ þ ð2s þ 2Þu2 ðsÞ ¼ 0
where u1 ðsÞ is the angular displacement of the inertia.
Solving for u2 ðsÞ,
2
s þ s þ 1 T ðsÞ ð s þ 1Þ
ðs þ 1ÞF ðsÞ
0 ¼
u 2 ðsÞ ¼ 2
s þ s þ 1 ðs þ 1Þ 2s3 þ 3s2 þ 2s þ 1
ðs þ 1Þ
ð2s þ 2Þ From which, after simplification,
u 2 ðsÞ ¼
2s2
1
þsþ1
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,
2
s þ s u1 ðsÞ sua ðsÞ ¼ T ðsÞ
su1 ðsÞ þ ðs þ 1Þua ðsÞ ¼ 0
where u1 ðsÞ is the angular displacement of the 1-kg inertia.
Solving for ua ðsÞ,
2
s þ s T ðsÞ s
sT ðsÞ
0 ¼ 3
u a ðsÞ ¼ 2
s þs
s
þ
s2 þ s
s s
ðs þ 1Þ
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Chapter 2 Solutions to Skill-Assessment Exercises
From which,
ua ðsÞ
1
¼
T ðsÞ s2 þ s þ 1
1
But, u2 ðsÞ ¼ ua ðsÞ:
2
Thus,
u2 ðsÞ
1=2
¼ 2
T ðsÞ s þ s þ 1
2.11
First find the mechanical constants.
1 1 2
1
Jm ¼ Ja þ JL
¼ 1 þ 400
¼2
5 4
400
1 1 2
1
¼ 5 þ 800
Dm ¼ Da þ DL
¼7
5 4
400
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,
Kt T stall 200
¼2
¼
¼
100
Ra
Ea
Ea
100
Kb ¼
¼4
¼
25
vnoload
Substituting all values into the motor transfer function,
KT
u m ðsÞ
1
R
a J m
¼ ¼ 1
KT Kb
15
Ea ðsÞ
s sþ
Dm þ
s sþ
Jm
2
Ra
where um ðsÞ is the angular displacement of the armature.
1
Now uL ðsÞ ¼ um ðsÞ. Thus,
20
u L ðsÞ
1=20
¼ 15
Ea ðsÞ
s sþ
2
2.12
Letting
u1 ðsÞ ¼ v1 ðsÞ=s
u2 ðsÞ ¼ v2 ðsÞ=s
5
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Solutions to Skill-Assessment Exercises
in Eqs. 2.127, we obtain
K
K
v1 ðsÞ v2 ðsÞ ¼ T ðsÞ
J 1 s þ D1 þ
s
s
K
K
v 2 ðsÞ
v 1 ðsÞ þ J 2 s þ D 2 þ
s
s
From these equations we can draw both series and parallel analogs by considering
these to be mesh or nodal equations, respectively.
J1
T(t)
D1
J2
1
K
–
+
w1(t)
D2
w2(t)
Series analog
1
K
w1(t)
J1
T(t)
w2(t)
1
D1
J2
1
D2
Parallel analog
2.13
Writing the nodal equation,
C
dv
þ ir 2 ¼ iðtÞ
dt
But,
C¼1
v ¼ vo þ dv
ir ¼ evr ¼ ev ¼ evo þdv
Substituting these relationships into the differential equation,
dðvo þ dvÞ
þ evo þdv 2 ¼ iðtÞ
dt
We now linearize ev.
The general form is
f ð vÞ f ð vo Þ df dv
dv vo
Substituting the function, f ðvÞ ¼ ev , with v ¼ vo þ dv yields,
dev dv
evo þdv evo dv vo
ð1Þ
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Chapter 3 Solutions to Skill-Assessment Exercises
Solving for evo þdv,
e
vo þdv
dev ¼e þ
dv ¼ evo þ evo dv
dv vo
vo
Substituting into Eq. (1)
ddv
þ evo þ evo dv 2 ¼ iðtÞ
dt
ð2Þ
Setting iðtÞ 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
þ 2dv ¼ iðtÞ
dt
Taking the Laplace transform,
ðs þ 2ÞdvðsÞ ¼ I ðsÞ
Solving for the transfer function, we obtain
dvðsÞ
1
¼
I ðsÞ
sþ2
or
V ðsÞ
1
¼
about equilibrium:
I ðsÞ s þ 2
CHAPTER 3
3.1
Identifying appropriate variables on the circuit yields
C1
R
iR
iC1
v1(t) +
L
–
iL
+
C2
vo(t)
–
iC2
Writing the derivative relations
dvC1
¼ iC1
dt
diL
¼ vL
L
dt
dvC2
¼ iC2
C2
dt
C1
ð1Þ
7
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Using Kirchhoff’s current and voltage laws,
iC1 ¼ iL þ iR ¼ iL þ
1
ð vL vC 2 Þ
R
vL ¼ vC1 þ vi
1
iC2 ¼ iR ¼ ðvL vC2 Þ
R
Substituting these relationships into Eqs. (1) and simplifying yields the state
equations as
dvC1
1
1
1
1
¼
vC 1 þ
iL vC 2 þ
vi
RC1
C1
RC1
RC1
dt
diL
1
1
¼ vC 1 þ vi
L
L
dt
dvC2
1
1
1
¼
vC 1 vC 2
vi
RC2
RC2
RC2
dt
where the output equation is
vo ¼ vC 2
Putting the equations in vector-matrix form,
2
1
RC1
6
6
6
1
6
x_ ¼ 6 6
L
6
4
1
RC2
y ¼ ½0
0
1
C1
0
0
3
3
2
1
1
6 RC1 7
RC1 7
7
7
6
7
7
6
7
6 1 7
7 vi ð t Þ
0 7x þ 6
7
6 L 7
7
7
6
4 1 5
1 5
RC2
RC2
1 x
3.2
Writing the equations of motion
2
sX 2 ðsÞ
¼ F ðsÞ
s þ s þ 1 X 1 ðsÞ
2
sX 1 ðsÞ þ s þ s þ 1 X 2 ðsÞ
X 3 ðsÞ ¼ 0
2
X 2 ðsÞ þ s þ s þ 1 X 3 ðsÞ ¼ 0
Taking the inverse Laplace transform and simplifying,
x€1 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
Defining state variables, zi,
z1 ¼ x1 ; z2 ¼ x_ 1 ; z3 ¼ x2 ; z4 ¼ x_ 2 ; z5 ¼ x3 ; z6 ¼ x_ 3
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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
z_ 2 ¼ x€1
z_ 3 ¼ x_ 2
z_ 4 ¼ x€2
z_ 5 ¼ x_ 3
¼ x_ 1 x1 þ x_ 2 þ f ¼ z2 z1 þ z4 þ f
¼ z4
¼ x_ 1 x_ 2 x2 þ x3 ¼ z2 z4 z3 þ z5
¼ z6
z_ 6 ¼ x€3 ¼ x_ 3 x3 þ x2 ¼ z6 z5 þ z3
The output is
2
0
6 1
6
6
6 0
z_ ¼ 6
6 0
6
6
4 0
0
z5. Hence, y ¼ z5 . In vector-matrix form,
2 3
3
0
1
0
0
0
0
7
6
7
1 0
1
0
0 7
617
6 7
7
607
0
0
1
0
0 7
7z þ 6 7 f ðtÞ; y ¼ ½ 0
607
7
1 1 1 1
0 7
6 7
6 7
7
405
0
0
0
0
1 5
0
1
1
0
1
0 0
0
1
0 z
0
3.3
First derive the state equations for the transfer function without zeros.
X ðsÞ
1
¼ 2
RðsÞ s þ 7s þ 9
Cross multiplying yields
s2 þ 7s þ 9 X ðsÞ ¼ RðsÞ
Taking the inverse Laplace transform assuming zero initial conditions, we get
x€ þ 7x_ þ 9x ¼ r
Defining the state variables as,
x1 ¼ x
x2 ¼ x_
Hence,
x_ 1 ¼ x2
x_ 2 ¼ x€ ¼ 7x_ 9x þ r ¼ 9x1 7x2 þ r
Using the zeros of the transfer function, we find the output equation to be,
c ¼ 2x_ þ x ¼ x1 þ 2x2
Putting all equation in vector-matrix form yields,
"
#
" #
0
1
0
x_ ¼
xþ
r
9 7
1
c ¼ ½ 1 2 x
9
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3.4
The state equation is converted to a transfer function using
GðsÞ ¼ CðsI AÞ1 B
ð1Þ
where
A¼
4
1:5
4
0
2
;B ¼
0
; and C ¼ ½ 1:5
0:625 :
Evaluating ðsI AÞ yields
ðsI AÞ ¼
sþ4
1:5
4
s
Taking the inverse we obtain
ðsI AÞ1 ¼
s2
s 1:5
1
þ 4s þ 6 4 s þ 4
Substituting all expressions into Eq. (1) yields
GðsÞ ¼
3s þ 5
s2 þ 4s þ 6
3.5
Writing the differential equation we obtain
d2 x
þ 2x2 ¼ 10 þ df ðtÞ
dt2
ð1Þ
Letting x ¼ xo þ dx and substituting into Eq. (1) yields
d2 ðxo þ dxÞ
þ 2ðxo þ dxÞ2 ¼ 10 þ df ðtÞ
dt2
ð2Þ
Now, linearize x2.
2
ðxo þ dxÞ x2o
d x2 ¼
dx ¼ 2xo dx
dx xo
from which
ðxo þ dxÞ2 ¼ x2o þ 2xo dx
ð3Þ
Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives
us the linearized intermediate differential equation,
d2 dx
þ 4xo dx ¼ 2x2o þ 10 þ df ðtÞ
dt2
ð4Þ
The force of the spring at equilibrium is 10 N. Thus, since F ¼ 2x2 ; 10 ¼ 2x2o from
which
pffiffiffi
xo ¼ 5
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Chapter 4 Solutions to Skill-Assessment Exercises
Substituting this value of xo into Eq. (4) gives us the final linearized differential
equation.
pffiffiffi
d2 dx
þ 4 5 dx ¼ df ðtÞ
dt2
Selecting the state variables,
x1 ¼ dx
_
x2 ¼ dx
Writing the state and output equations
x_ 1 ¼ x2
pffiffiffi
x_ 2 ¼€dx ¼ 4 5x1 þ df ðtÞ
y ¼ x1
Converting to vector-matrix form yields the final result as
x_ ¼
0
1
0
pffiffiffi
xþ
df ðtÞ
4 5 0
1
y ¼ ½1
0 x
CHAPTER 4
4.1
For a step input
C ðsÞ ¼
10ðs þ 4Þðs þ 6Þ
A
B
C
D
E
¼ þ
þ
þ
þ
sðs þ 1Þðs þ 7Þðs þ 8Þðs þ 10Þ
s s þ 1 s þ 7 s þ 8 s þ 10
Taking the inverse Laplace transform,
cðtÞ ¼ A þ Bet þ Ce7t þ De8t þ Ee10t
4.2
Since a ¼ 50; T c ¼
Tr ¼
1
1
4
4
¼
¼ 0:02s; T s ¼ ¼
¼ 0:08 s; and
a 50
a 50
2:2 2:2
¼
¼ 0:044 s.
a
50
4.3
a.
b.
c.
d.
Since
Since
Since
Since
4.4
a.
b.
c.
d.
vn
vn
vn
vn
poles
poles
poles
poles
are
are
are
are
at 6 j 19:08; cðtÞ ¼ A þ Be6t cosð19:08t þ fÞ.
at 78:54 and 11:46; cðtÞ ¼ A þ Be78:54t þ Ce11:4t .
double on the real axis at 15 cðtÞ ¼ A þ Be15t þ Cte15t :
at j 25; cðtÞ ¼ A þ B cosð25t þ fÞ.
pffiffiffiffiffiffiffiffi
¼ 400 ¼ 20 and 2zvn
pffiffiffiffiffiffiffiffi
¼ 900 ¼ 30 and 2zvn
pffiffiffiffiffiffiffiffi
¼ 225 ¼ 15 and 2zvn
pffiffiffiffiffiffiffiffi
¼ 625 ¼ 25 and 2zvn
¼ 12;
¼ 90;
¼ 30;
¼ 0;
;z 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.
11
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4.5
vn ¼
pffiffiffiffiffiffiffiffi
361 ¼ 19 and 2zvn ¼ 16; ;z ¼ 0:421:
4
p
¼ 0:5 s and T p ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:182 s.
zvn
vn 1 z 2
From Figure 4.16, vn T r ¼ 1:4998. Therefore, T r ¼ 0:079 s.
pzp
ffiffiffi
2
Finally, %os ¼ e 1 z 100 ¼ 23:3%
Now, T s ¼
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 0:8942 1:5918 0:3023
a. Expanding G(s) by partial fractions yields GðsÞ ¼ þ
.
s s þ 20 s þ 10 s þ 6:5
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 GðsÞ ¼ þ
.
s s þ 20 s þ 10 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
See Figure 4.31 in the textbook for the Simulink block diagram and the output responses.
4.9
s 2
sþ5
1
; ðsI AÞ1 ¼ 2
sI A ¼
s þ 5s þ 6 3
"
#3 s þ 5
0
BUðsÞ ¼
.
1=ðs þ 1Þ
a. Since
2
:
s
Also,
1
The state vector is XðsÞ ¼ ðsI AÞ1 ½xð0Þ þ BUðsÞ ¼
ð
s
þ
1
Þ
ð
s
þ
2Þ ð s þ 3Þ
" #
2 s2 þ 7s þ 7
5s2 þ 2s 4
¼
.
The
output
is
Y
ð
s
Þ
¼
½
1
3
X
ð
s
Þ
¼
ðs þ 1Þðs þ 2Þðs þ 3Þ
s2 4s 6
0:5
12
17:5
þ
. Taking the inverse Laplace transform yields yðtÞ ¼
sþ1 sþ2 sþ3
t
2t
0:5e 12e þ 17:5e3t .
b. The eigenvalues are given by the roots of jsI Aj ¼ s2 þ 5s þ 6, or 2 and 3.
4.10
s 2
sþ5 2
1
; ðsI AÞ1 ¼ 2
. Taking the
s þ 5s þ 4 2 s
2 sþ5
Laplace transform of each term, the state transition matrix is given by
2
3
4 t 1 4t
2 t 2 4t
e e
6 3e 3e
7
3
3
7:
FðtÞ ¼ 6
4 2
2 4t
1 t 4 4t 5
t
e þ e
e þ e
3
3
3
3
a. Since ðsI AÞ ¼
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Chapter 5 Solutions to Skill-Assessment Exercises
2
3
4 ðttÞ 1 4ðttÞ
2 ðttÞ 2 4ðttÞ
e
e
e
6 3e
7
3
3
3
7 and
b. Since Fðt tÞ ¼ 6
4 2
2 4ðttÞ
1 ðttÞ 4 4ðttÞ 5
ðttÞ
e
þ e
e
þ e
3
3 2
3
3 3
2 t t 2 2t 4t
6 3e e 3e e
7
0
7:
BuðtÞ ¼ 2t ; Fðt tÞBuðtÞ ¼ 6
4
5
1
4
e
t t
2t 4t
e e þ e e
3
3
Rt
Thus, xðtÞ ¼ FðtÞxð0Þ þ 0 Fðt tÞ
2
3
10 t
4 4t
2t
6 3 e e 3e
7
7
BUðtÞdt ¼ 6
4 5
5
8
t
2t
4t
e þe þ e
3
3
c. yðtÞ ¼ ½ 2 1 x ¼ 5et e2t
CHAPTER 5
5.1
Combine the parallel blocks in the forward path. Then, push
1
to the left past the
s
pickoff point.
s
R(s)
+
–
s2 +
1
s
1
s
C(s)
–
s
Combine the parallel feedback paths and get 2s. Then, apply the feedback formula,
s3 þ 1
.
simplify, and get, T ðsÞ ¼ 4
2s þ s2 þ 2s
5.2
G ðsÞ
16
¼ 2
, where
1 þ GðsÞH ðsÞ s þ as þ 16
16
a
and H ðsÞ ¼ 1. Thus, vn ¼ 4 and 2zvn ¼ a, from which z ¼ .
and GðsÞ ¼
s ð s þ aÞ
8
%
ln
a
100
But, for 5% overshoot, z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 0:69. Since, z ¼ 8 ; a ¼ 5:52.
%
p2 þ ln2
100
Find the closed-loop transfer function, T ðsÞ ¼
13
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Solutions to Skill-Assessment Exercises
5.3
Label nodes.
R(s) +
–
s
–
+
s
N1 (s)
N2 (s)
N3 (s)
1
s
N5 (s)
1
s
N4 (s)
+
N6 (s)
s
N7 (s)
Draw nodes.
R(s)
N1(s)
N3(s)
N2(s)
N5(s)
N4(s)
C (s)
N6(s)
N7(s)
Connect nodes and label subsystems.
−1
R(s) 1 N1 (s)
s
N2 (s)
s
N3 (s)
1
1
−1
1
N4 (s) s
C(s)
1
N5 (s)
1
s
N7 (s)
N6 (s)
s
Eliminate unnecessary nodes.
–1
R(s)
1
s
s
1
s
–s
1
s
C(s)
C(s)
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Chapter 5 Solutions to Skill-Assessment Exercises
5.4
Forward-path gains are G1G2G3 and G1G3.
Loop gains are G1 G2 H 1 ; G2 H 2; and G3 H 3.
Nontouching loops are ½G1 G2 H 1 ½G3 H 3 ¼ G1 G2 G3 H 1 H 3 and
½G2 H 2 ½G3 H 3 ¼ G2 G3 H 2 H 3 .
Also, D ¼ 1 þ G1 G2 H 1 þ G2 H 2 þ G3 H 3 þ G1 G2 G3 H 1 H 3 þ G2 G3 H 2 H 3 :
Finally, D1 ¼ 1 and D2 ¼ 1.
P
Substituting these values into T ðsÞ ¼
T ðsÞ ¼
CðsÞ
¼
R ðsÞ
k
T k Dk
D
yields
G1 ðsÞG3 ðsÞ½1 þ G2 ðsÞ
½1 þ G2 ðsÞH 2 ðsÞ þ G1 ðsÞG2 ðsÞH 1 ðsÞ½1 þ G3 ðsÞH 3 ðsÞ
5.5
The state equations are,
x_ 1 ¼ 2x1 þ x2
x_ 2 ¼ 3x2 þ x3
x_ 3 ¼ 3x1 4x2 5x3 þ r
y ¼ x2
Drawing the signal-flow diagram from the state equations yields
1
r
1
1
s
x3
1
s
1
x2
1
s
1
–3
–5
x1
y
–2
–4
–3
5.6
100ðs þ 5Þ
we draw the signal-flow graph in controller canonical form
s2 þ 5s þ 6
and add the feedback.
From GðsÞ ¼
100
r
1
1
s
1
s
x1
x2
500
y
–5
–6
–1
15
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Solutions to Skill-Assessment Exercises
Writing the state equations from the signal-flow diagram, we obtain
"
x¼
105
506
#
" #
1
xþ
r
1
0
0
y ¼ ½ 100 500 x
5.7
From the transformation equations,
P1 ¼
3 2
1 4
Taking the inverse,
0:2
0:3
0:4
0:1
P¼
Now,
"
1
P AP ¼
3
2
#"
1
3
#"
0:4
0:2
#
4 6 0:1 0:3
"
#" # "
#
3
2
1
3
P1 B ¼
¼
1 4 3
11
"
#
0:4 0:2
CP ¼ ½ 1 4 ¼ ½ 0:8 1:4 0:1 0:3
1
4
"
¼
6:5
8:5
9:5
11:5
#
Therefore,
"
z_ ¼
8:5
6:5
"
#
3
#
u
zþ
11
9:5 11:5
y ¼ ½ 0:8 1:4 z
5.8
First find the eigenvalues.
l 0
1
jlI Aj ¼ 0 l
4
l 1
¼
6 4
3
From which the eigenvalues are 2 and 3.
Now use Axi ¼ lxi for each eigenvalue, l.
Thus,
1
3
x1
4
6
x2
¼l
For l ¼ 2,
3x1 þ 3x2 ¼ 0
4x1 4x2 ¼ 0
x1
x2
3 ¼ l2 þ 5l þ 6
l þ 6
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Chapter 6 Solutions to Skill-Assessment Exercises
Thus x1 ¼ x2
For l ¼ 3
4x1 þ 3x2 ¼ 0
4x1 3x2 ¼ 0
Thus x1 ¼ x2 and x1 ¼ 0:75x2 ; from which we let
P¼
0:707
0:707
0:6
0:8
P1 ¼
5:6577
5
4:2433
5
Taking the inverse yields
Hence,
"
1
D ¼ P AP ¼
"
P1 B ¼
CP ¼ ½ 1
5:6577
5
#" #
4:2433 1
4
#"
5
5:6577
5
"
4:2433
5
0:707
0:6
0:707
0:8
3
#
¼
1
#"
3
4 6
"
#
18:39
0:707
0:6
0:707
0:8
"
#
¼
2
0
0
3
#
20
¼ ½ 2:121
2:6 Finally,
"
z_ ¼
2
0
#
"
18:39
zþ
20
0 3
y ¼ ½ 2:121 2:6 z
#
u
CHAPTER 6
6.1
Make a Routh table.
s7
3
6
7
2
s6
9
4
8
6
5
s
4.666666667
4.333333333
0
0
s4
4.35714286
8
6
0
3
s
12.90163934
6.426229508
0
0
s2
10.17026684
6
0
0
s1
1.18515742
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 6s4 þ 0s2 þ 6. Taking the derivative yields
17
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Solutions to Skill-Assessment Exercises
24s3 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
6
1
6
s
5
1
0
1
0
s
4
6
0
6
0
s3
24
0
0
0
2
e
6
0
0
s1
144=e
0
0
0
s0
6
0
0
0
s
ROZ
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 halfplane poles, and 2 imaginary poles.
6.3
Since GðsÞ ¼
Kðs þ 20Þ
G ðsÞ
K ðs þ 20Þ
; T ðsÞ ¼
¼
sðs þ 2Þðs þ 3Þ
1 þ GðsÞ s3 þ 5s2 þ ð6 þ KÞs þ 20K
Form the Routh table.
s3
1
ð6 þ K Þ
s2
5
30 15K
5
20K
20K
s
1
s0
From the s1 row, K < 2. From the s0 row, K > 0. Thus, for stability, 0 < K < 2.
6.4
First find
2
s
6
jsI Aj ¼ 4 0
0
0
s
0
0
3
2
2
7 6
05 4 1
s
3
3 ð s 2Þ
7 7 1 5 ¼ 1
4 5 3
1
1
1
ðs 7Þ
4
1 1 ð s þ 5Þ ¼ s3 4s2 33s þ 51
Now form the Routh table.
s3
1
33
s2
4
51
s
1
20:25
s
0
51
There are two sign changes. Thus, there are two rhp poles and one lhp pole.
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Chapter 7 Solutions to Skill-Assessment Exercises
CHAPTER 7
7.1
a. First check stability.
GðsÞ
10s2 þ 500s þ 6000
10ðs þ 30Þðs þ 20Þ
¼ 3
T ðsÞ ¼
¼
2
1 þ GðsÞ s þ 70s þ 1375s þ 6000 ðs þ 26:03Þðs þ 37:89Þðs þ 6:085Þ
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
¼0
15uðtÞ :
estep ð1Þ ¼
¼
1 þ lim GðsÞ 1 þ 1
s!0
15
15
¼ 2:1875
¼
lim sGðsÞ 10 20 30
s!0
25 35
15
30
30
¼ 1; since L 15t2 ¼ 3
15t2 uðtÞ : eparabola ð1Þ ¼
¼
2
0
s
lim s GðsÞ
15tuðtÞ :
eramp ð1Þ ¼
s!0
b. First check stability.
G ðsÞ
10s2 þ 500s þ 6000
¼ 5
T ðsÞ ¼
1 þ GðsÞ s þ 110s4 þ 3875s3 þ 4:37e04s2 þ 500s þ 6000
10ðs þ 30Þðs þ 20Þ
¼
ðs þ 50:01Þðs þ 35Þðs þ 25Þðs2 7:189e 04s þ 0:1372Þ
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
GðsÞ
1000ðs þ 8Þ
1000ðs þ 8Þ
T ðsÞ ¼
¼
¼ 2
1 þ GðsÞ ðs þ 9Þðs þ 7Þ þ 1000ðs þ 8Þ s þ 1016s þ 8063
and is of Type 0. Therefore,
K p ¼ lim GðsÞ ¼
s!0
1000 8
¼ 127; K v ¼ lim sGðsÞ ¼ 0;
s!0
7 9
and Ka ¼ lim s2 GðsÞ ¼ 0
s!0
b.
estep ð1Þ ¼
1
1
¼ 7:8e 03
¼
1 þ lim GðsÞ 1 þ 127
s!0
eramp ð1Þ ¼
1
1
¼ ¼1
lim sGðsÞ 0
s!0
eparabola ð1Þ ¼
1
1
¼ ¼1
lim s2 GðsÞ 0
s!0
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Solutions to Skill-Assessment Exercises
7.3
System is stable for positive K. System is Type 0. Therefore, for a step input
1
12K
estep ð1Þ ¼
¼ 0:1. Solving for Kp yields K p ¼ 9 ¼ lim GðsÞ ¼ ; from
s!0
1 þ Kp
14 18
which we obtain K ¼ 189.
7.4
ð s þ 2Þ
,
ð s þ 4Þ
1
¼ 9:98e 04
¼
2 þ 1000
System is stable. Since G1 ðsÞ ¼ 1000, and G2 ðsÞ ¼
e D ð 1Þ ¼ 1
1
þ lim G1 ðsÞ
lim
s!0 G2 ðsÞ
s!0
7.5
System is stable. Create a unity-feedback system, where H e ðsÞ ¼
The system is as follows:
+
R(s)
–
Ea(s)
100
s+4
1
s
1¼
sþ1
sþ1
C(s)
–
−s
s+1
Thus,
100
GðsÞ
100ðs þ 1Þ
ð s þ 4Þ
¼
Ge ðsÞ ¼
¼ 2
100s
1 þ GðSÞH e ðsÞ
S
95s þ 4
1
ð s þ 1Þ ð s þ 4Þ
Hence, the system is Type 0. Evaluating Kp yields
100
¼ 25
Kp ¼
4
The steady-state error is given by
estep ð1Þ ¼
1
1
¼ 3:846e 02
¼
1 þ K p 1 þ 25
7.6
Since GðsÞ ¼
K ð s þ 7Þ
1
; eð1Þ ¼
¼
1 þ Kp
þ 2s þ 10
s2
1
10
.
¼
7K 10 þ 7K
1þ
10
Calculating the sensitivity, we get
Se:K ¼
K @e
¼
e @K
K
ð10Þ7
7K
¼
2
10
10 þ 7K
ð10 þ 7K Þ
10 þ 7K
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Chapter 8 Solutions to Skill-Assessment Exercises
7.7
Given
A¼
0
1
3
6
; B¼
0
1
1
; C ¼ ½ 1 1 ; RðsÞ ¼ :
s
Using the final value theorem,
2
"
#1 " #3
h
i
0
s
1
5
estep ð1Þ ¼ lim sRðsÞ 1 CðsI AÞ1 B ¼ lim 41 ½ 1 1 s!0
s!0
1
3 sþ6
"
#
3
2
sþ6 1
" #7
6
2
0 7
6
3s s
7 ¼ lim s þ 5s þ 2 ¼ 2
6
¼ lim 61 ½ 1 1 2
2
s!0 4
s þ 6s þ 3 1 7
5 s!0 s þ 6s þ 3 3
Using input substitution,
step ð1Þ
"
1
¼ 1 þ CA B ¼ 1 ½ 1
"
¼ 1 þ ½1 1
6 1
3
0
1
0
1
3
6
#1 " #
0
1
#
" #
0
3
1
2
¼ 1 þ ½1
3
1
2
1 4 3 5 ¼
3
0
CHAPTER 8
8.1
ð5 þ j9Þð3 þ j9Þ
a. F ð7 þ j9Þ ¼ ð7 þ j9 þ 2Þð7 þ j9 þ 4Þ0:0339 ¼
ð7 þ j9Þð7 þ j9 þ 3Þð7 þ j9 þ 6Þ ð7 þ j9Þð4 þ j9Þð1 þ j9Þ
ð66 j72Þ
¼
¼ 0:0339 j0:0899 ¼ 0:096 < 110:7
ð944 j378Þ
b. The arrangement of vectors is shown as follows:
jw
(–7+j9)
s-plane
M1
–7
X
–6
M2
M3
–5
–4
M4
X
–3
M5
–2
–1
X
0
s
21
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Solutions to Skill-Assessment Exercises
From the diagram,
F ð7 þ j9Þ ¼
M2 M4
ð3 þ j9Þð5 þ j9Þ
ð66 j72Þ
¼
¼
M1 M3 M5 ð1 þ j9Þð4 þ j9Þð7 þ j9Þ ð944 j378Þ
¼ 0:0339 j0:0899 ¼ 0:096 <; 110:7
8.2
a. First draw the vectors.
jw
X
j3
j2
s-plane
j1
–3
–2
–1
s
0
–j1
–j2
X
–j3
From the diagram,
P
angles ¼ 180 tan1
3
3
tan1
¼ 180 108:43 þ 108:43 ¼ 180 :
1
1
b. Since the angle is 180, the point
is on the root locus.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
1 þ3
12 þ 32
P pole lengths
¼ 10
¼
c. K ¼
1
P zero lengths
8.3
First, find the asymptotes.
sa ¼
P
P
poles zeros ð2 4 6Þ ð0Þ
¼ 4
¼
30
# poles # zeros
ua ¼
ð2k þ 1Þp p
5p
¼ ; p;
3
3
3
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Chapter 8 Solutions to Skill-Assessment Exercises
Next draw root locus following the rules for sketching.
5
4
3
2
Imag Axis
E1SM
1
0
–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
–2
0
–j3
2
X
b. Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function.
G ðsÞ
K ð s þ 2Þ
¼
T ðsÞ ¼
1 þ GðsÞ s2 þ ðK 4Þs þ ð2K þ 13Þ
Using the denominator of T(s), make a Routh table.
s2
1
2K þ 13
s1
K 40
0
0
2K þ 13
0
s
We get a row of zeros for K ¼ 4. From the s2 row with K ¼ 4; s2 þ 21 ¼ 0. From
pffiffiffiffiffi
which we evaluate the imaginary axis crossing at 21.
c. From part (b), K ¼ 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.
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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
2
0
–j3
X
At the point e close to the complex pole, the angles must add up to zero. Hence,
th
st
angle from zero –angle
from pole in 4 quadrant – angle from pole in 1 quadrant ¼
3
90 u ¼ 180. Solving for the angle of departure,
180 , or tan1
4
u ¼ 233:1.
8.5
a.
z = 0.5
X
jw
j4
s-plane
–3
X
0
–j4
o
o
2
4
s
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Chapter 8 Solutions to Skill-Assessment Exercises
b. Search along the imaginary axis and find the 180 point at s ¼ j4:06.
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 2:42 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.
jw
z = 0.591
s-plane
X
–6
X
–4
X
–2
s
0
b. Searching along the z ¼ 0:591 (10% overshoot) line for the 180 point yields
2:028 þ j2:768 with K ¼ 45:55.
4
4
p
p
¼
¼ 1:97 s; T p ¼
¼
¼ 1:13 s; vn T r ¼ 1:8346 from the
c. T s ¼
jRej 2:028
jImj 2:768
rise-time chart and graph in Chapter 4. Since vn is the radial distance to the pole,
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
vn ¼ 2:0282 þ 2:7682 ¼ 3:431. Thus, T r ¼ 0:53 s; since the system is Type 0,
K
45:55
¼ 0:949. Thus,
Kp ¼ ¼
2 46
48
estep ð1Þ ¼
1
¼ 0:51:
1 þ Kp
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
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,
100
2
G ðsÞ
100
100
¼
¼
¼ s þ 100
T ðsÞ ¼
1 þ GðsÞ s2 þ pi s þ 100 ðs2 þ 100Þ þ pi s 1 þ pi s
s2 þ 100
25
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Hence, KGðsÞH ðsÞ ¼
pi s
. The following shows the root locus.
s2 þ 100
jw
s-plane
X j10
O
s
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
–4
X
–3
X
–2
X
–1
0
s
8.9
The closed-loop transfer function is T ðsÞ ¼
K ð s þ 1Þ
. Differentiating the
s2 þ ðK þ 2Þs þ K
denominator with respect to K yields
2s
@s
@s
@s
þ ð K þ 2Þ
þ ðs þ 1Þ ¼ ð2s þ K þ 2Þ
þ ð s þ 1Þ ¼ 0
@K
@K
@K
@s
@s
ðs þ 1Þ
K @s
K ðs þ 1Þ
, we get
¼
. Thus, Ss:K ¼
¼
:
@K
@K ð2s þ K þ 2Þ
s @K sð2s þ K þ 2Þ
10ðs þ 1Þ
Substituting K ¼ 20 yields Ss:K 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,
DK
10ð21:05 þ 1Þ
Ds ¼ sðSs:K Þ
¼ 12:05
0:05 ¼ 0:9975:
K
21:05ð21:05 þ 11Þ
Solving for
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Chapter 9 Solutions to Skill-Assessment Exercises
For the pole at 0:95,
Ds ¼ sðSs:K Þ
DK
10ð0:95 þ 1Þ
¼ 0:95
0:05 ¼ 0:0025:
K
0:95ð0:95 þ 11Þ
CHAPTER 9
9.1
a. Searching along the 15% overshoot line, we find the point on the root locus at
3:5 þ j5:8 at a gain of K ¼ 45:84. Thus, for the uncompensated system,
Kv ¼ lim sGðsÞ ¼ K=7 ¼ 45:84=7 ¼ 6:55.
s!0
Hence, eramp uncompensated ð1Þ ¼ 1=Kv ¼ 0:1527.
b. Compensator zero should be 20x further to the left than the compensator pole.
ðs þ 0:2Þ
.
Arbitrarily select Gc ðsÞ ¼
ðs 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 127:5 and eramp compensated ð1Þ ¼
Kv ¼
¼ 0:0078.
ð7Þð0:01Þ
Kv
eramp uncompensated 0:1527
¼ 19:58
d.
¼
0:0078
eramp compensated
9.2
a. Searching along the 15% overshoot line, we find the point on the root locus at
3:5 þ j5:8 at a gain of K ¼ 45:84. Thus, for the uncompensated system,
Ts ¼
4
4
¼
¼ 1:143 s:
jRej 3:5
b. The real part of the design point must be three times larger than the uncompensated pole’s real part. Thus the design point is 3ð3:5Þþ j 3ð5:8Þ ¼
10:5 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
180 130:8 ¼ 49:2 . Using the following diagram,
jw
j17.4
s-plane
49.2°
–pc
s
–10.5
17:4
¼ tan 49:2 , from which, pc ¼ 25:52. Adding this pole, we find
Pc 10:5
the gain at the design point to be K ¼ 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
27
E1SM
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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 þ 6:83 at a gain of K ¼ 58:9. Thus, for the uncompensated system,
Ts ¼
4
4
¼
¼ 1:143 s:
jRej 3:5
b. For the uncompensated system, K v ¼ lim sGðsÞ ¼ K=7 ¼ 58:9=7 ¼ 8:41. Hence,
s!0
eramp uncompensated ð1Þ 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 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
diagram,
jw
j13.66
s-plane
79.2°
–pc
s
–7
13:66
¼ tan79:2 , from which, pc ¼ 9:61. Adding this pole, we find the
Pc 7
gain at the design point to be K ¼ 204:9.
Evaluating Kv for the lead-compensated system:
we find
K v ¼ lim sGðsÞGlead ¼ K ð3Þ=½ð7Þð9:61Þ ¼ ð204:9Þð3Þ=½ð7Þð9:61Þ ¼ 9:138:
s!0
K v for the uncompensated system was 8.41. For a 10x improvement in steady-state
error, Kv must be ð8:41Þð10Þ ¼ 84:1. Since lead compensation gave us K v 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
ðs þ 0:092Þ
.
G c ðsÞ ¼
ðs þ 0:01Þ
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
G e ðsÞ ¼
205:4ðs þ 3Þðs þ 0:092Þ
:
sðs þ 7Þð9:61Þðs þ 0:01Þ
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.
E1SM
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Chapter 9 Solutions to Skill-Assessment Exercises
9.4
The configuration for the system is shown in the figure below.
R(s)
+
+
K
1
s(s+ 7)(s +10)
–
–
C(s)
Kf s
Minor-Loop Design:
Kf
. Using the following diagram, we
ðs þ 7Þðs þ 10Þ
find that the minor-loop root locus intersects the 0.7 damping ratio line at
8:5 þ j 8:67. The imaginary part was found as follows: u ¼ cos1 z ¼ 45:57 . Hence,
Im
¼ tan45:57 , from which Im ¼ 8:67.
8:5
For the minor loop, GðsÞH ðsÞ ¼
jw
z = 0.7
(-8.5 + j8.67)
Im
X
−10
−8.5
s-plane
X
–7
q
s
The gain, Kf , is found from the vector lengths as
Kf ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1:52 þ 8:672 1:52 þ 8:672 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 ðsÞ ¼
K
K
¼
:
sðs þ 8:5 þ j8:67Þðs þ 8:5 j8:67Þ sðs2 þ 17s þ 147:4Þ
Using the three poles of Ge ðsÞ 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 4:34 þ
j7:51 at a gain, K ¼ 626:3.
29
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Solutions to Skill-Assessment Exercises
9.5
a. An active PID controller must be used. We use the circuit shown in the following
figure:
Z2(s)
Vi (s)
I2(s)
Z1(s)
V1(s)
–
Vo(s)
Ia(s)
I1(s)
+
where the impedances are shown below as follows:
C1
C2
R2
R1
Z1(s)
Z2(s)
Matching the given transfer function with the transfer function of the PID
controller yields
ðs þ 0:1Þðs þ 5Þ s2 þ 5:1s þ 0:5
0:5
¼
¼ s þ 5:1 þ
s
s
8
2
3
1
6 R2 C 1
R C 7
6
¼ 4
þ
þ R2 C 1 s þ 1 2 7
R1 C 2
s 5
G c ðsÞ ¼
Equating coefficients
1
¼ 0:5
R1 C 2
ð1Þ
R2 C 1 ¼ 1
R2 C 1
þ
¼ 5:1
R1 C 2
ð2Þ
ð3Þ
In Eq. (2) we arbitrarily let C1 ¼ 105 . 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 network, since the ratio of the lead pole-to-zero is the inverse of the ratio of the lag
pole-to-zero:
R1
+
vi (t)
+
C1
R2
vo (t)
C2
–
–
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Chapter 10 Solutions to Skill-Assessment Exercises
Matching the given transfer function with the transfer function of the passive lag-lead
compensator yields
ðs þ 0:1Þðs þ 2Þ
ðs þ 0:1Þðs þ 2Þ
¼
G c ðsÞ ¼
ðs þ 0:01Þðs þ 20Þ s2 þ 20:01s þ 0:2
1
1
sþ
sþ
R1 C 1
R2 C2
¼
1
1
1
1
2
s þ
þ
þ
sþ
R1 C 1 R2 C 2 R2 C 1
R1 R2 C1 C2
Equating coefficients
1
¼ 0:1
R1 C 1
1
¼ 0:1
R2 C 2
1
1
1
þ
þ
R1 C1 R2 C2 R2 C1
ð1Þ
ð2Þ
¼ 20:01
ð3Þ
Substituting Eqs. (1) and (2) in Eq. (3) yields
1
¼ 17:91
R2 C 1
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.
1
1
; GðjvÞ ¼
2
ðs þ 2Þðs þ 4Þ
ð8 þ v Þ þ j6v
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
MðvÞ ¼ ð8 v2 Þ þ ð6vÞ2
6v
1
fðvÞ ¼ tan
:
8 v2
6v
fðvÞ ¼ p þ tan1
:
8 v2
GðsÞ ¼
For v <
pffiffiffi
8,
For v <
pffiffiffi
8,
b.
Bode Diagrams
0
–20
Phase (deg); Magnitude (dB)
E1SM
–40
–60
–80
–100
0
–50
–100
–150
–200
10–1
100
101
Frequency (rad/sec)
102
31
Page 32
Solutions to Skill-Assessment Exercises
c.
Nyquist Diagrams
0.08
0.06
0.04
0.02
Imaginary Axis
32
9:29:20
0
–0.02
–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
–20 dB/dec
–80
–40 dB/dec
–100
–120
1
0.1
1000
100
10
Frequency (rad/s)
–45o/dec
Phase (degrees)
11/11/2010
20 log M
E1SM
–50
–90o/dec
–45o/dec
–90o/dec
–100
–45o/dec
–150
–45o/dec
Asymptotic
–200
0.1
1
10
Frequency (rad/s)
Actual
100
1000
E1SM
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Chapter 10 Solutions to Skill-Assessment Exercises
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 180 while its
magnitude goes to zero. The result is shown below.
Im
w=∞
0
w=0
1
8
Re
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 270 while
its magnitude goes to zero. The result is shown below.
Im
w = 6.63 w = ∞
– 1
480
0
w=0
1
48
Re
1
1
¼ 3
and
2
ðs þ 2Þðs þ2 4Þðs þ 6Þ s 3 þ 12s þ 44s þ 48
48 12v j 44v v
. The Nyquist diagram
simplifying, we obtain GðjvÞ ¼ 6
v þ 56v4 þ 784v2 þ 2304
crosses the real axis when the imaginary part of GðjvÞ is zero. Thus, the Nyquist
pffiffiffiffiffi
diagram crosses the real axis at v2 ¼ 44; or v ¼ 44 ¼ 6:63 rad=s. At this fre1
. Thus, the system is stable for K < 480.
quency GðjvÞ ¼ 480
b. Substituting jv into GðsÞ ¼
33
34
9:29:22
Page 34
Solutions to Skill-Assessment Exercises
10.5
If K ¼ 100, the Nyquist diagram will intersect the real axis at 100=480. Thus,
480
¼ 13:62 dB. From Skill-Assessment Exercise Solution 10.4, the
GM ¼ 20 log
100
180 frequency is 6.63 rad/s.
10.6
a.
–60
–80
–100
20 log M
11/11/2010
–120
–140
–160
–180
1
10
100
1000
100
1000
Frequency (rad/s)
0
–50
Phase (degrees)
E1SM
–100
–150
–200
–250
–300
1
10
Frequency (rad/s)
b. The phase angle is 180 at a frequency of 36.74 rad/s. At this frequency the gain is
99:67 dB. Therefore, 20 logK ¼ 99:67, or K ¼ 96; 270. We conclude that the
system is stable for K < 96; 270.
c. For K ¼ 10; 000, the magnitude plot is moved up 20log10; 000 ¼ 80 dB. Therefore,
the gain margin is 99:67 80 ¼ 19:67 dB. The 180 frequency is 36.7 rad/s. The
gain curve crosses 0 dB at v ¼ 7:74 rad=s, where the phase is 87:1. We calculate
the phase margin to be 180 87:1 ¼ 92:9.
10.7
lnð%=100Þ
Using z ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, we find z ¼ 0:456, which corresponds to 20% overp2 þ ln2 ð%100Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4
2
4
2
ð1 2z Þ þ 4z 4z þ 2 ¼ 5:79 rad=s.
shoot. Using T s ¼ 2; vBW ¼
T sz
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Chapter 10 Solutions to Skill-Assessment Exercises
10.8
160 6750000 101250v2 þ j1350 v2 1350 v
For both parts find that GðjvÞ ¼
.
v6 þ 2925v4 þ 1072500v2 þ 25000000
27
For a range of values for v, superimpose GðjvÞ on the a. M and N circles, and on the
b. Nichols chart.
a.
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
–3
–4
–2
1
–1
2
b.
Nichols Charts
0 dB
0.25 dB
0.5 dB
1 dB
3 dB
6 dB
0
Open-Loop Gain (dB)
E1SM
–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
Page 36
Solutions to Skill-Assessment Exercises
Plotting the closed-loop frequency response from a. or b. yields the following plot:
0
–20
–40
20 log M
36
9:29:22
–60
–80
–100
–120
1
10
100
1000
Frequency (rad/s)
0
–50
Phase (degrees)
11/11/2010
–100
–150
–200
–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
Phase (deg); Magnitude (dB)
E1SM
–20
–40
–100
–120
–140
–160
10–1
100
101
Frequency (rad/sec)
102
E1SM
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Chapter 10 Solutions to Skill-Assessment Exercises
The open-loop frequency response is 7 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 180 151:98 ¼ 28:02. This phase margin corresponds to
z ¼ 0:25: Therefore; %OS ¼ e
Ts ¼
Tp ¼
4
vBW z
pffiffiffiffiffiffiffi2ffi
zp=
1z
100 ¼ 44:4%
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
ð1 2z Þ þ 4z4 4z2 þ 2 ¼ 1:64 s and
p
pffiffiffiffiffiffiffiffiffiffiffiffiffi
vBW 1 z2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð1 2z2 Þ þ 4z4 4z2 þ 2 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, K a ¼ 9:52 ¼ 90:25 and K p ¼ K v ¼ 1.
10.11
10
10
¼
, from which the zero dB frejvðjv þ 1Þ vðv þ jÞ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
10
quency is found as follows: M ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 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 .
b. With a delay of 0.1 s,
a. Without delay, GðjvÞ ¼
f 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.
c. With a delay of 3 s,
f ¼ ffðv 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 28:59 180 ¼ 151:41 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.
37
Page 38
Solutions to Skill-Assessment Exercises
Experimental
20
Gain(dB)
10
0
–10
–20
–30
–40
Phase(deg)
38
9:29:24
–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 ðsÞ from the original frequency response yields the
G 1 ðsÞ ¼
sðs þ 21Þ
frequency response shown below.
Experimental Minus 1/s(s+21)
90
80
Gain(dB)
11/11/2010
70
60
50
40
100
80
Phase(deg)
E1SM
60
40
20
0
1
2
3
4 5 6 7 8 10
20
30 40 50 70 100
Frequency (rad/sec)
200 300
500
1000
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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 ¼ 20logð5:7K Þ, or K ¼ 27:8. Thus, we estimate G2 ðsÞ ¼ 27:8ðs þ 7Þ. Thus,
27:8ðs þ 5:7Þ
. It is interesting to note that the original problem
GðsÞ ¼ G1 ðsÞG2 ðsÞ ¼
sðs þ 21Þ
30ðs þ 5Þ
was developed from GðsÞ ¼
.
sðs þ 20Þ
CHAPTER 11
11.1
The Bode plot for K ¼ 1 is shown below.
Bode Diagrams
–60
–80
–100
–120
Phase (deg); Magnitude (dB)
E1SM
–140
–160
–180
–100
–150
–200
–250
10–1
100
101
Frequency (rad/sec)
102
103
%
log
100
A 20% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 0:456. This damping ratio implies a
%
p2 þ log2
100
phase margin of 48.10, which is obtained when the phase angle ¼ 1800 þ 48:10
¼ 131:9 . This phase angle occurs at v ¼ 27:6 rad=s. The magnitude at this frequency
1
is 5:15 106. Since the magnitude must be unity K ¼
¼ 194; 200.
5:15 106
11.2
To meet the steady-state error requirement, K ¼ 1; 942; 000. The Bode plot for this
gain is shown below.
39
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9:29:25
Page 40
Solutions to Skill-Assessment Exercises
Bode Diagrams
60
40
20
0
Phase (deg); Magnitude (dB)
E1SM
–20
–40
–100
–150
–200
–250
10–1
100
101
102
103
Frequency (rad/sec)
%
log
100
A 20% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi ¼ 0:456. This damping ratio
%
2
p2 þ 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 þ 58:1 ¼ 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 0:1ð20:4Þ ¼ 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
ðs þ 2:04Þ
, where the gain is chosen to yield 0 dB at low frequencies,
G c ðsÞ ¼ K c
ðs þ 0:141Þ
or K c ¼ 0:141=2:04 ¼ 0:0691. In summary,
Gc ðsÞ ¼ 0:0691
11.3
ðs þ 2:04Þ
1;942;000
and GðsÞ ¼
ðs þ 0:141Þ
sðs þ 50Þðs þ 120Þ
%
log
100
A 20% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi ¼ 0:456. The required bandwidth
%
2
p2 þ log
100
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4
ð1 2z2 Þ þ 4z4 4z2 þ 2 ¼ 57:9 rad/s. In order
is then calculated as vBW ¼
T sz
K
, we calculate
to meet the steady-state error requirement of K v ¼ 50 ¼
ð50Þð120Þ
K ¼ 300; 000. The uncompensated Bode plot for this gain is shown below.
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Chapter 11 Solutions to Skill-Assessment Exercises
Bode Plot for K = 300000
40
20
0
–20
Phase (deg); Magnitude (dB)
E1SM
–40
–60
–100
–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 þ 144:8 ¼ 35:2 . The required phase margin based on the required damping
2z
ratio is FM ¼ tan1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 48:1 . Adding a 10 correction factor, the
2
4
2z þ 1 þ 4z
required phase margin is 58:1. Hence, the compensator must contribute
1b
1 sinfmax
, b¼
fmax ¼ 58:1 35:2 ¼ 22:9 . Using fmax ¼ sin1
¼ 0:44.
1þb
1 þ sinfmax
1
The compensator’s peak magnitude is calculated as Mmax ¼ pffiffiffi ¼ 1:51. Now find
b
the frequency at which the uncompensated system has a magnitude 1=Mmax, or
3:58 dB. From the Bode plot, this magnitude occurs at vmax ¼ 50 rad=s. The
1
1
compensator’s zero is at zc ¼ . vmax ¼ pffiffiffi Therefore, zc ¼ 33:2.
T
T b
1
zc
¼ ¼ 75:4. The compensator gain is
The compensator’s pole is at Pc ¼
bT
b
chosen to yield unity gain at dc.
ðs þ 33:2Þ
, and
Hence, K c ¼ 75:4=33:2 ¼ 2:27. Summarizing, Gc ðsÞ ¼ 2:27
ðs þ 75:4Þ
300;000
GðsÞ
.
sðs þ 50Þðs þ 120Þ
41
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Solutions to Skill-Assessment Exercises
11.4
%
log
100
A 10% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi ¼ 0:591. The required bandwidth
%
2
p2 þ log
100
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
is then calculated as vBW ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 2z2 Þ þ 4z4 4z2 þ 2 ¼ 7:53 rad=s:
T p 1 z2
K
, we
In order to meet the steady-state error requirement of K v ¼ 10 ¼
ð8Þð30Þ
calculate K ¼ 2400. The uncompensated Bode plot for this gain is shown below.
Bode Diagrams
40
20
0
–20
–40
Phase (deg); Magnitude (dB)
E1SM
–60
–80
–100
–100
–150
–200
–250
10–1
100
101
Frequency (rad/sec)
102
103
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 tan1
2z
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 58:6 . Adding a 5 correction factor, the required phase
2z2 þ 1 þ 4z4
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 ¼ 41:7. Thus, the lead compensator must contribute fmax ¼ 63:6 41:7 ¼ 21:9.
1b
1 sinfmax
,b¼
Using fmax ¼ sin1
¼ 0:456.
1þb
1 þ 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 ¼ 0:602 rad=s. The lag
compensator’s pole is plag ¼ bzlag ¼ 0:275. Finally, the lag compensator’s gain is
K lag ¼ b ¼ 0:456.
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Chapter 12 Solutions to Skill-Assessment Exercises
Now we design the lead compensator. The lead zero is the product of the new
pffiffiffi
pffiffiffi
zlead
phase margin frequency and b, or zlead ¼ 0:8vBW b ¼ 4:07. Also, plead ¼
b
1
¼ 8:93. Finally, K lead ¼ ¼ 2:19. Summarizing,
b
Glag ¼ ðsÞ ¼ 0:456
ðs þ 0:602Þ
ðs þ 4:07Þ
; Glead ðsÞ ¼ 2:19
;
ðs þ 0:275Þ
ðs þ 8:93Þ
and k ¼ 2400:
CHAPTER 12
12.1
We first find
the
desired characteristic equation. A 5% overshoot requires
%
log
p
100
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
s
z¼
ffi ¼ 0:69. Also, vn ¼ pffiffiffiffiffiffiffiffiffiffiffiffi2ffi ¼ 14:47 rad=s. Thus, the charTp 1 z
%
p2 þ log2
100
acteristic equation is s2 þ 2zvn s þ v2n ¼ s2 þ 19:97s þ 209:4. Adding a pole at 10 to
cancel
the zero at 10 yields the desired characteristic equation,
2
s þ 19:97s þ 209:4 ðs þ 10Þ ¼ s3 þ 29:97s2 þ 409:1s þ 2094. The compensated system matrix in phase-variable form is
2
6
A BK ¼ 4
0
1
0
0
0
1
3
7
5. The characteristic equation for this
ðk1 Þ ð36 þ k2 Þ ð15 þ k3 Þ
system is jsI ðA BKÞj ¼ s3 þ ð15 þ k3 Þs2 þ ð36 þ k2 Þs þ ðk1 Þ. Equating coefficients of this equation with the coefficients of the desired characteristic equation
yields the gains as
K ¼ ½ k1
12.2
k2
k3 ¼ ½ 2094 373:1
14:97 :
2
3
2
1
1
6
7
The controllability matrix is CM ¼ B AB A2 B ¼ 4 1
4 9 5. Since
1 1 16
jCM j 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 ¼
2
3
0
0
1
6
7
1 17 5. Since jCMz j ¼ 1, CMz is full rank, that is, rank 3. We conclude that
40
1 9
81
the system is controllable. We now
find
the desired characteristic equation. A 20%
%
log
4
100
overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi ¼ 0:456. Also, vn ¼ zT ¼ 4:386 rad=s.
s
%
p2 þ log2
100
43
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Solutions to Skill-Assessment Exercises
Thus, the characteristic equation is s2 þ 2zvn s þ v2n ¼ s2 þ 4s þ 19:24. Adding a pole
at 6 to cancel the zero at 6 yields the resulting desired characteristic equation,
2
s þ 4s þ 19:24 ðs þ 6Þ ¼ s3 þ 10s2 þ 43:24s þ 115:45:
ð s þ 6Þ
sþ6
¼ 3
, we can write the phase2
ðs þ 7Þðs þ 8Þðs þ 9Þ s þ 24s
2 þ 191s þ 504
3
2 3
0
1
0
0
6
7
6 7
variable
representation
as
Ap ¼ 4 0
0
1 5 ; Bp ¼ 4 0 5; Cp ¼
504 191 24
1
½ 6 1 0 . The compensated system matrix in phase-variable form is Ap Bp Kp ¼
2
3
0
1
0
6
7
0
0
1
4
5. The characteristic equation for this
Since GðsÞ ¼
ð504 þ k1 Þ ð191 þ k2 Þ ð24 þ k3 Þ
system is jsI Ap Bp Kp j ¼ s3 þ ð24 þ k3 Þs2 þ ð191 þ k2 Þs þ ð504 þ k1 Þ. Equating coefficients of this equation with the coefficients of the desired characteristic
equation yields the gains as Kp ¼ ½ k1 k2 k3 ¼ ½ 388:55 147:76 14 . We
now develop the transformation matrix to transform back to the z-system.
2
3
0
0
1
h
i
6
7
CMz ¼ Bz Az Bz A2z B z ¼ 4 0
1 17 5 and
9
1
h
CMp ¼ Bp
Ap Bp
A2p Bp
i
2
0
6
¼ 40
81
3
1
7
24 5:
0
1
1 24
385
Therefore,
2
P ¼ CMz C1
Mx
0
6
¼ 40
0
1
1 9
32
1
191 24
76
17 54 24
1
81
1
0
3 2
1
1
7 6
05 ¼ 4 7
0
1
3
0
7
05
0
15
1
56
2
Hence, Kz ¼ Kp P1 ¼ ½ 388:55 147:76
1
0
6
14 4 7
12.4
62:24
15 1
14 .
2
ð24 þ l1 Þ
3
7
05
1
49
¼ ½ 40:23
0
1
0
3
6
7
For the given system ex_ ¼ ðA LCÞex ¼ 4 ð191 þ l2 Þ 0 1 5ex . The characteristic
ð504 þ l3 Þ 0 0
polynomial is given by j½sI ðA LCÞj ¼ s3 þ ð24 þ l1 Þs2 þ ð191 þ l2 Þs þ
ð504 þ l3 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 . Factoring yields
ð2 þ j3:9Þ and ð2 j3:9Þ. 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
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Chapter 12 Solutions to Skill-Assessment Exercises
desired characteristic polynomial, ðs þ 20 þ j39Þðs þ 20 j39Þðs þ 200Þ ¼ s3 þ 240s2
þ 9921s þ 384200. Equating coefficients of the
2 desired3characteristic equation to the
216
6
7
system’s characteristic equation yields L ¼ 4 9730 5.
383696
12.5
The
2
observability
2
25
28
matrix
32
is
3
C
3
2
4
6
7 6
OM ¼ 4 CA 5 ¼ 4 64
674
CA2
6
8
3
7
80 78 5,
848 814
where
6
7
A2 ¼ 4 7 4 11 5. The matrix is of full rank, that is, rank 3, since
77 95
94
jOM j ¼ 1576. Therefore the system is observable.
12.6
The system is represented in cascade form by the following state and output
equations:
2
3
2 3
7
1
0
0
6
7
6 7
z_ ¼ 4 0 8
1 5z þ 4 0 5u
0
0 9
1
The observability matrix is OMz
2
49
6
where A2z ¼ 4 0
0
¼
s3
þ
24s2
15
64
0
1
y ¼ ½ 1 0 0 z
2
3 2
Cz
1
6
7 6
¼ 4 Cz Az 5 ¼ 4 7
Cz A2z
49
0
1
3
0
7
0 5,
15 1
3
7
17 5. Since GðsÞ ¼
81
1
ð s þ 7Þ ð s þ 8Þ ð s þ 9Þ
1
, we can write the observable canonical form as
þ 191s þ 504
2
3
2 3
24 1 0
0
6
7
6 7
x_ ¼ 4 191 0 1 5x þ 4 0 5u
504 0 0
1
y ¼ ½1
0
0 x
2
The observability matrix for this form is OMx
3
2
1
0
6
7 6
1
¼ 4 Cx Ax 5 ¼ 4 24
2
Cx Ax
385 24
Cx
3
0
7
0 5,
1
45
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Solutions to Skill-Assessment Exercises
where
2
24
385
6
A2x ¼ 4 4080 191
12096
504
1
3
7
0 5:
0
We next find
the
desired characteristic equation. A 10% overshoot requires
%
log
4
100
z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi ¼ 0:591. Also, vn ¼ zT ¼ 67:66 rad=s. Thus, the characteristic
s
%
p2 þ log2
100
equation is s2 þ 2zvn s þ v2n ¼ s2 þ 80s þ 4578:42. Adding a pole at 400, 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 ðs þ 400Þ ¼ s3 þ 480s2 þ 36580s þ 1:831x106 . For
the system represented
in observable canonical form ex_ ¼ ðAx Lx Cx Þ ex ¼
2
3
ð24 þ l1 Þ 1 0
6
7
The
characteristic
polynomial
is
given
by
4 ð191 þ l2 Þ 0 1 5ex .
ð504 þ l3 Þ 0 0
j½sI ðAx Lx Cx Þj ¼ s3 þ ð24 þ l1 Þs2 þ ð191 þ l2 Þs þ ð504 þ l3 Þ. Equating coefficients of the2 desired characteristic
equation to the system’s characteristic equation
3
456
6
7
yields Lx ¼ 4 36; 389 5.
1; 830; 496
Now, develop the transformation matrix between the observer canonical and
cascade forms.
2
1
6
6
P ¼ O1
O
¼
6 7
Mz Mx
4
49
2
1
6
6
¼6 7
4
56
2
1
6
6
¼ 6 17
4
81
0
0
31 2
1
0
0
3
7
7 6
7
7 6
1 07
0 7 6 24
5
5 4
385 24 1
15 1
32
3
0 0
1
0 0
76
7
76
7
1 0 76 24
1 07
54
5
385 24 1
15 1
3
0 0
7
7
1 07
5
9 1
1
Finally,
2
1
6
Lz ¼ PLx ¼ 4 17
0
1
81
9
32
3 2
3 2
3
0
456
456
456
76
7 6
7 6
7
0 54 36; 389 5 ¼ 4 28; 637 5 4 28; 640 5.
1
1; 830; 496
1; 539; 931
1; 540; 000
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Chapter 12 Solutions to Skill-Assessment Exercises
12.7
We first find the desired characteristic equation. A 10% overshoot requires
%
log
100
z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi ¼ 0:591
%
p2 þ log2
100
p
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1:948 rad=s. Thus, the characteristic equation is s2 þ
T p 1 z2
2
2zvn s þ vn ¼ s2 þ 2:3s þ 3:79. Adding a pole at 4, which corresponds to the
system’szero location, yields the resulting desired characteristic equation,
original
2
s þ 2:3s þ 3:79 ðs þ 4Þ ¼ s3 þ 6:3s2 þ 13s þ 15:16.
ðA BKÞ BK e
0
x
x
x_
Now,
¼
þ
r; and y ¼ ½ C 0 ,
_xN
1
xN
C
0
xN
Also, vn ¼
where
A BK ¼
¼
0
7
1
0
½ k1
9
1
0
ð7 þ k1 Þ
k2 ¼
0
7
1
0
9
k1
0
k2
1
ð9 þ k2 Þ
C ¼ ½4 1
0
0
ke ¼
1
ke
Bke ¼
Thus,
2
3 2
0
x_1
6
7 6
4 x_2 5 ¼ 4 ð7 þ k1 Þ
4
x_N
1
0
32
3
2
0
76 7
ke 5 4 x 2 5 þ
r; y ¼ ½ 4
1
0
xN
ð9 þ k 2 Þ
1
Finding the characteristic equation of
2
s
sI ðA BKÞ BK e ¼ 6
40
C
0 0
2
s
6
¼ 4 ð 7 þ k1 Þ
4
x1
this system yields
3 2
0 0
0
7 6
s 0 5 4 ð7 þ k1 Þ
0
s
4
1
x1
3
6 7
0 4 x2 5 .
xN
3
1
0 7
ð9 þ k2 Þ ke 5
1
0 3
7
s þ ð9 þ k2 Þ ke 5 ¼ s3 þ ð9 þ k2 Þs2 þ ð7 þ k1 þ ke Þs þ 4ke
1
s 1
0
Equating this polynomial to the desired characteristic equation,
s3 þ 6:3s2 þ 13s þ 15:16 ¼ s3 þ ð9 þ k2 Þs2 þ ð7 þ k1 þ ke Þs þ 4ke
Solving for the k’s,
K ¼ ½ 2:21
2:7 and ke ¼ 3:79:
47
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Solutions to Skill-Assessment Exercises
CHAPTER 13
13.1
f ðtÞ ¼ sinðvkT Þ; f ðtÞ ¼
1
P
sinðvkT Þdðt kT Þ;
jvkT
1
1
X
X
e
ejvkT ekTs
F ðsÞ ¼
sinðvkT ÞekTs ¼
2j
k¼0
k¼0
1
k
1 X T ðsjvÞ k
¼
e
eT ðsþjvÞ
2j k¼0
k¼0
But,
1
P
xk ¼
k¼0
1
1 x1
Thus,
1
1
1
1
eTs ejvT eTs ejvT
¼
2j 1 eT ðsjvÞ 1 eT ðsþjvÞ
2j 1 ðeTs ejvT eTs ejvT Þ þ e2Ts
sinðvT Þ
z1 sinðvT Þ
¼ eTs
¼
Ts
2Ts
1 e 2cosðvT Þ þ e
1 2z1 cosðvT Þ þ z2
F ðsÞ ¼
13.2
F ð zÞ ¼
zðz þ 1Þðz þ 2Þ
ðz 0:5Þðz 0:7Þðz 0:9Þ
F ð zÞ
zðz þ 1Þðz þ 2Þ
¼
z
ðz 0:5Þðz 0:7Þðz 0:9Þ
¼ 46:875
F ðzÞ ¼ 46:875
z
1
z
114:75
þ 68:875
z 0:5
z 0:7
z 0:9
z
z
z
114:75
þ 68:875
;
z 0:5
z 0:7
z 0:9
f ðkT Þ ¼ 46:875ð0:5Þk 114:75ð0:7Þk þ 68:875ð0:9Þk
13.3
Since GðsÞ ¼ 1 eTs
8
,
s ð s þ 4Þ
8
z1 A
B
z1 2
2
1
z
þ
z þ
G ð zÞ ¼ 1 z z
¼
¼
:
s ð s þ 4Þ
z
s sþ4
z
s sþ4
2
2
þ
. Therefore, g2 ðtÞ ¼ 2 2e4t , or g2 ðkT Þ ¼ 2 2e4kT .
s sþ4
2z 1 e4T
2z
2z
.
¼
Hence, G2 ðzÞ ¼
z 1 z e4T ðz 1Þðz e4T Þ
Let G2 ðsÞ ¼
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Chapter 13 Solutions to Skill-Assessment Exercises
2 1 e4T
z1
G2 ðzÞ ¼
.
z
ðz e4T Þ
1
1:264
.
For T ¼ s, GðzÞ ¼
4
z 0:3679
Therefore, GðzÞ ¼
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)
Hence, T ðzÞ ¼
G1 G2 ðzÞ
.
1 þ HG1 G2 ðzÞ
13.5
20
GðsÞ
20
4
4
. Let G2 ðsÞ ¼
¼
¼ . Taking the inverse
sþ5
s
s ð s þ 5Þ s s þ 5
Laplace transform and letting t ¼ kT, g2 ðkT Þ ¼ 4 4e5kT . Taking the z-transform
4z 1 e5T
4z
4z
.
¼
yields G2 ðzÞ ¼
z 1 z e5T ðz 1Þðz e5T Þ
4 1 e5T
z1
G2 ðzÞ ¼
.
Now, GðzÞ ¼
z
ðz e5T Þ
4 1 e5T
G ð zÞ
¼
.
Finally, T ðzÞ ¼
1 þ GðzÞ z 5e5T þ 4
The pole of the closed-loop system is at 5e5T 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.
Let GðsÞ ¼
13.6
sþ1
into DðzÞ ¼ z3 z2 0:5z þ 0:3, we obtain DðsÞ ¼ s3 8s2
s1
27s 6. The Routh table for this polynomial is shown below.
Substituting z ¼
s3
s
2
s1
s0
1
27
8
6
27:75
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
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Solutions to Skill-Assessment Exercises
13.7
Defining G(s) as G1(s) in cascade with a zero-order-hold,
GðsÞ ¼ 20 1 eTs
3=20
ð s þ 3Þ
1=4
2=5
þ
¼ 20 1 eTs
:
sðs þ 4Þðs þ 5Þ
s
ðs þ 4Þ ðs þ 5Þ
Taking the z-transform yields
ð3=20Þz
ð1=4Þz
ð2=5Þz
5ð z 1Þ 8ð z 1Þ
þ
GðzÞ ¼ 20 1 z1
:
¼3þ
4T
5T
z1
ze
ze
z e4T z e5T
Hence for T ¼ 0:1 second, K p ¼ lim GðzÞ ¼ 3, and K v ¼
z!1
1
lim ðz 1ÞGðzÞ ¼ 0, and
T z!1
1
lim ðz 1Þ2 GðzÞ ¼ 0. Checking for stability, we find that the system is
T 2 z!1
GðzÞ
1:5z 1:109
¼
stable for T ¼ 0:1 second, since T ðzÞ ¼
has poles
1 þ GðzÞ z2 þ 0:222z 0:703
inside the unit circle at 0:957 and þ 0:735. Again, checking for stability, we find that
G ð zÞ
¼
the system is unstable for T ¼ 0:5 second, since T ðzÞ ¼
1 þ G ð zÞ
3:02z 0:6383
has poles inside and outside the unit circle at þ0:208 and
2
z þ 2:802z 0:6272
3:01, respectively.
Ka ¼
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 ð0:348 þ j 0:468Þ and K ¼ 0:31.
z-Plane Root Locus
1.5
1
(0.348 + j0.468)
K = 0.31
0.5
Imag Axis
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54.3°
0
–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
Ge ðsÞ ¼ GðsÞGc ðsÞ ¼
100K
2:38ðs þ 25:3Þ
342720ðs þ 25:3Þ
¼
:
sðs þ 36Þðs þ 100Þ ðs þ 60:2Þ
sðs þ 36Þðs þ 100Þðs þ 60:2Þ
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Chapter 13 Solutions to Skill-Assessment Exercises
The following shows the frequency response of Ge ðjvÞ.
Bode Diagrams
40
20
0
–20
Phase (deg); Magnitude (dB)
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–40
–60
–100
–150
–200
–250
10–1
100
101
Frequency (rad/sec)
102
103
We find that the zero dB frequency, vFM , for Ge ðjvÞ 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
2ð z 1Þ
into Gc ðsÞ ¼
transformation for the compensator. Substituting s ¼
T ðz 1Þ
2:38ðs þ 25:3Þ
with T ¼ 0:001 second yields
ðs þ 60:2Þ
Gc ðzÞ ¼ 2:34
ðz 0:975Þ
:
ðz 0:9416Þ
13.10
X ðzÞ 1899z2 3761z þ 1861
¼ 2
. Cross-multiply and obtain z2 1:908z þ
EðzÞ z 1:908z þ 0:9075
0:9075X 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 ð1:908z þ 0:9075ÞX ðzÞ. Solving for X(z) on the left-hand side yields
Gc ðzÞ ¼
51
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Solutions to Skill-Assessment Exercises
X ðzÞ ¼ 1899 3761z1 þ 1861z2 EðzÞ 1:908z1 þ 0:9075z2 X ðzÞ. Finally,
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
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