G (s)
=
where
L ( s )
E (s)
=
K G (s)
a
p
=
K K ( Bs
a
i
+K
)
o
(
o = 0.12 s ( s + 0.0325 ) s + 2.5675 s + 6667
(
2
= s 0.12 s + 0.312 s + 80.05 s + 26
G ( s) =
3
(
2
)
43.33( s + 500)
s s + 2.6 s + 667.12 s + 216.67
3
2
(b) B ode Diagram:
Gain crossover
1 0-37 Forward-path Transfer Function:
G ( s) =
K s K K Ki N
1
s J t L a s + ( Ra J t + L a Bt + K 1 K 2 J t ) s + R a Bt + K 1 K 2 Bt + K b K i + K K Ki Kt
1
2
1.5 10 K
7
G ( s) =
R amp Error Constant:
Thus
(
s s + 3408.33s + 1,204,000 + 1.5 10 KK t
K
2
S tate Diagram:
Transfer Functions:
m ( s)
=
Ea ( s )
L ( s)
=
E a (s )
m ( s)
=
Ea ( s )
L ( s)
E a (s )
=
(
)
Ki s + BL s + KL / Ra
2
J m J L s + ( K e J L + BL J L + BL J m ) s + ( J m K L + J L K L + K e BL ) s + K L K e
2
2
K i ( B Ls + K L ) / R a
0.0008
0.0009
0.0010
53.83
52.68
51.38
43.72
41.81
40.09
1.125
1.140
1.162
T he phase margin is at a maximum of 54.69 deg when T = 0.0006. The performance worsens
if the value of a is less than 1000.
1 0-24 (a) B ode Plot.
The attributes of the frequency
1 0-9 (a) Forward-path Transfer Function:
G ( s) =
100 K P +
KI
s
s + 10 s + 100
2
For
K v = 10,
K v = lim sG ( s ) = lim s
s 0
s 0
100 ( K P s + K I )
(
s s + 10 s + 100
2
)
= K I = 10
Thus the forward-path transfer function becomes
G ( s) =
10 (1 + 0.1K
Maximum overshoot = 20.8%
(b)
S elect a relatively large value for K
The closed-loop zero at s
D
= K P
and a small value for K
/K
D
P
dynamics are governed by the other closed-loop poles. Let K
The following results show that the value of K
Kp
For BW
40 r
6.5500E+00
(b) T ime Responses: x ( 0 ) =
0 .1
0
0
0
x ( 0 ) =
0 .1
0
0
0 , the initial position of
With the initial states
'
'
x 1 or y1 is preturbed downward
x 3 = y 2 is
from its stable equilibrium position. The steel ball is initially pulled toward th
Comparing these results with the part 1, the final values are approximately the same but the shape of responses is
closed to the first order system behavior. Then the system time constant is obviously different and it can be
identified from open loop resp
7
8
9
10
I-6 (a)
x( k
20.781250
32.171875
49.257813
74.886719
17
18
19
20
+ 2 ) x ( k + 1) + 0 .1 x ( k ) = u s ( k )
x (0)
=
1311.681641
1968.522461
2953.783691
4431.675781
=0.
x (1 )
Taking the z-transform on both
sides,
2
z X (z)
z
X ( z) =
2
x(0)
zx
(c) L ( s ) =
(
K ( s + 1)
s s + 3s + 1
3
P
)
=1
P
=
2
For stability, Z = 0. 11 = ( 0.5 P + P 1 ) 180 = 450
o
For K > 0, 11 = 90
o
For K < 0, 11 = +90
o
o
The system is unstable.
.
when K > 1.
11 = 270
o
when K < 1.
The Nyquist plot of L ( j ) crosses the
0.5 Hz
1 Hz
327
2Hz
5Hz
328
10Hz
50Hz
329
As frequency increases, the phase shift of the input and output also increase. Also, the amplitude of the
output starts to decrease when the frequency increases above 0.5Hz.
11-17
As proportional gain increases, t
c. -10 V:
13. Study of the effect of viscous friction:
303
As seen in above figure, two different values for B are selected, zero and 0.0075. We could change the
final speed by 50% in open loop system. The same values selected for closed loop speed contro
11-3
a)
b)
c)
d)
50rad/sec
0.0795 seconds
2.5A. The current
When Jm is increased by a factor of 2, it takes 0.159 seconds to reach 63% of its steady state
speed, which is exactly twice the original time period . This means that the time constant has
been