ECE 421, Spring 2006
Solutions to ‘Textbook Problems’ for HW Assignment #10
The solutions to the additional problems appear after in a separate
fi
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
fi
cations, and the design procedure is applied correctly, then veri
fi
cation that the
speci
fi
cations are actually satis
fi
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
fi
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
.
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 Spring '08
 Cook,G
 GC, special lag compensator

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