MAS
239
U
as
=
5
-4
=
'
✗ 4)
Poles
Clt)
Problem
b)
If
-
-
Gist
(s)
✗
=
40 ✗ (t )
+
=
2s
2
K,
+
041
2 V61
+
duct
+
2
V41
dt
e-
real
poles
"t
e-
yz
+
10£
2
1s
at
63,2%
.
T
constant
C final
=
step Amplitude
Cs
40
different
Ko
=
(25+2) UH
+
dxltl
V41
dt
2
:
=
✗ (s )
14
+
step
Time
14s
+
dt
a)
-
-1
=
26+1 )
=
(5+145+40)
✗ G)
d
Zeroes
-10
,
(5+414+10)
(5)
(s)
✗
ZV
Ole Andre
15.02.21
(5+4) (5+10)
Poles
✗
C)
.
2 ( Stl )
G)
(5)
by
assignment
I
Problem
✗
Mandatory
-
=
K
Ts
,
1,02
300
=
2
=
+
=
1,02s
=
150
0,025 -11
-
l
150
,
=
steady
0,02
state
Seconds
300
rad /s
Tuntland
Problem 3
There
as
(
is
961
(
-
2) (
.
5- y
input
-4+-52
Which
)
Stl
-12)
Is
stable
is
this
also
?
Wn
k
5+25
=
s
at
poles
system
=
step
Unit
,
complex
(5+1)
s
0,5
=
under damped
a
=
G (5)
final
two
are
This
b)
(
-1
correct
( final
"
Wns
.
K=
U
complex
poles
Wn
?
0,5
=
I
=
0,5
10
C. (5)
Wn
C)
5+85+20
=
22+-1412
=
4
=
Under
Problem
It
The
seeing
The
over
=
4,47
0,89
=
2hr5
dg
2T5
swn
g5
=
damped
,
There
are
two
at
-4+-52
4
would
be
derivative
the
Integral
time
pace
needed
in
control
of
control
case
as
decreases
overshoot
and
settling
time
by
change
removes
steady
state
error
by
sunning
up
error