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ECE 421, Spring 2006
Solutions to ‘Textbook Problems’ for HW Assignment #10
The solutions to the additional problems appear after in a separate
f
le.
For these HW problems, if there is a choice of the desired dominant closedloop pole
s
1
,
it can be
done based on the equations for the standard secondorder system. If
s
1
is chosen correctly based on the
transient performance speci
f
cations, and the design procedure is applied correctly, then veri
f
cation that the
speci
f
cations are actually satis
f
ed is not necessary.
Problem #B711
The closedloop poles of the given plant model with
G
c
(
s
)=1
are located at
s
=
−
2
±
j
2
√
3
.
Thus, the
desired dominant closedloop pole used for design is
s
1
=
−
2+
j
2
√
3
.
Since that is already a closedloop
pole,

G
p
(
s
1
)

=1
and
6
G
p
(
s
1
)=
−
180
◦
.
Therefore, no additional phase shift needs to be added, only the
steadystate error needs to be considered.
For the given system, the velocity error constant is
K
v
−
actual
= lim
s
→
0
∙
s
·
16
s
(
s
+4)
¸
=4
(1)
The speci
f
ed value for
K
v
is
K
v
=20
,
the units are sec
−
1
.
A special lag compensator can be designed for
the system with the parameter
α
g
being
α
g
=
z
cg
p
cg
=
K
v
−
spec
K
v
−
actual
=
e
ss
−
actual
e
ss
−
spec
=
20
4
=5
(2)
In order to have the compensator’s pole and zero separated by a factor of
5
and still have

G
c
(
s
1
)

≈
1
and
6
G
c
(
s
1
)
≈
0
◦
,
they must be located must closer to
s
=0
than the point
s
1
.
The compensator zero is
normally placed to the right of the real part of
s
1
by a factor of
100
,
and the compensator pole is placed to
the right of the zero by the factor
α
g
.
The special lag compensator is
G
c
(
s
)=
(
s
+0
.
02)
(
s
+0
.
004)
(3)
The gain of the special lag compensator is always
1
.
Problem #B712
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 Spring '08
 Cook,G

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