16.30/31,
Fall
2010
—
Recitation
#
2
September
22,
2010
In
this
recitation,
we
will
consider
two
problems
from
Chapter
8
of
the
Van
de
Vegte
book.
R
G
(
s
)
C
G
c
(
s
)
E

+
Figure
1:
The
standard
block
diagram
for
a
unitfeedback
loop.
Problem
1
(VDV
8.14)
If
Figure
1
models
a
radar
tracking
system
with
G
= 1
/
[(0
.
1
s
+
1)
s
],
design
series
compen
sation
to
meet
the
following
specifications:
1.
The
steadystate
error
following
ramp
inputs
must
not
exceed
2%.
2.
The
error
in
response
to
sinusoidal
inputs
up
to
5
rad/sec
should
not
exceed
about
5%.
3.
The
crossover
frequency
should
be
about
50
rad/sec
to
meet
bandwidth
requirements
while
limiting
the
response
to
highfrequency
noise.
4.
The
ratio
of
the
break
frequencies
of
G
c
should
not
exceed
5
to
limit
noise
effects.
5.
The
phase
margin
should
be
about
50
◦
.
Solution
—
The
very
first
thing
we
should
do
is
check
closedloop
stability.
A
quick
root
locus
sketch
will
reveal
that
the
closedloop
system
is
stable
for
all
positive
gains,
i.e.,
it
is
stable
for
G
c
(
s
) =
K
,
for
K >
0.
For
simplicity,
we
can
choose
K
=
1
to
get
a
baseline
stabilizing
controller.
Re(
s
)
Im(
s
)
Figure
2:
Root
locus
sketch
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Just
for
practice,
let
us
figure
out
where
the
closedloop
poles
would
be
for
K
=
1,
us
ing
rootlocus
techniques.
First
of
all,
write
the
transfer
function
in
rootlocus
form,
i.e.,
1
G
(
s
)
=
10
s
(
s
+10)
.
Applying
the
magnitude
condition,
for
G
c
=
K
=
1,
we
know
that
the
closedloop
poles
will
be
at
locations
that
satisfy

s
·
s
+
10

=
10.
Let
us
check
whether
the
closedloop
poles
would
be
on
the
part
of
the
root
locus
lying
on
the
real
axis.
In
such
a
case,
s
would
be
a
negative
real
number,
and
the
magnitude
condition
would
read
−
s
(
s
+10)
=
10
K
,
i.e.,
s
2
+
10
s
+
10
=
0,
which
has
the
two
roots
s
=
−
5
±
√
25
−
10
=
{−
8
.
8730
,
−
1
.
1270
}
.
Let
us
sketch
the
Bode
plot
of
such
a
system
(recall
we
chose
G
c
=
1
for
the
time
being,
for
convenience).
The
straightline
approximation
is
shown
in
Figure
3
.
Since
we
have
an
integrator,
the
magnitude
Bode
plot
starts
with
a
negative
“unit”
slope
(i.e.,
20dB/decade).
It
will
cross
the
line
ω
= 1
at
K
=
1.
The
effect
of
the
pole
at
s
=
10
is
to
decrease
the
slope
to
40
dB/decade,
starting
at
the
break
frequency
ω
=
10.
The
phase
will
start
at
90
degrees
for
low
frequencies
(because
of
the
integrator),
be
equal
to
135
degrees
at
the
pole
break
frequency
ω
=
10,
and
will
ultimately
approach
180
degrees
at
high
frequencies
(because
of
the
additional
pole).
Since
the
phase
never
drops
below
180
degrees,
the
phase
margin
will
always
be
positive—but
will
be
small
if
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '04
 EricFeron

Click to edit the document details