Unformatted text preview: Example Prob. 6.50 on page 456: K
:
s1 + s=51 + s=20
Design a compensator such that
KGs = ess 0:01 for unit ramp input. P.M.= 45o 3o.
ess 1=250 for sinusoidal input with ! 0:2. Noise be attenuated by a factor of 100 for ! 100. Analysis:
ess
to 1=250 for sinusoidal input with ! 0:2 is equivalent
j1 + DGj!j 250 = jDGj!j 251 = 48dB; for all ! 0:2. Noise be attenuated by a factor of 100 for ! 100 is equivalent to that
jDGj!j 0:01
j1 + DGj!j
that is guaranteed by jDGj!j 0:01 = ,40dB:
1 Design:
,50dB We choose K = 20 such that KGs attenuated to
for ! 100. It gives the Bode plot of Gain dB 50 0 −50 −100 −1
10 0 1 10 10 2 10 Frequency (rad/sec) Phase deg 0 −90 −180 −270
−1 10 0 1 10 10 2 10 Frequency (rad/sec) K = 20 implies that Kv = 20. Hence we need boost the low
frequency gain by a factor of 5. Moreover it will also provide
the necessary gain to achieve speci cation for ! 0:2. 2 We thus take = 5 for lag compensator. Since !c 8:6,
zlag = 1; plag = 0:2 are chosen that gives s+1
:
s + 0:2
We make Bode plot for KDlagG as follows:
Dlags = 100 Gain dB 50
0
−50
−100 −1
10 0 1 10 10 2 10 Phase deg Frequency (rad/sec) −180 −270
−1 10 0 1 10 10
Frequency (rad/sec) It is seen that P.M. = 0. We need design lead. 3 2 10 By P:M: = 45o, we take
1 , sin 45o
= 0:1716:
=
1 + sin 45o
However ,20 log10 = 15:3dB. Thus high frequency gain
will add 15:3dB that will violate the ,40dB requirement. We
thus need iterate our design. Iterative Design I:
We choose a smaller K : K = 10 that gives Bode plot of
KGs: Gain dB 50 0 −50 −100 −1
10 0 1 10 10 2 10 Frequency (rad/sec) Phase deg 0 −90 −180 −270
−1 10 0 1 10 10
Frequency (rad/sec) 4 2 10 We need take = 10 for lag compensator in order to get
Kv = 100 as required. Since !c 6, we choose zlag = 1 and
plag = 0:1. Hence we obtain
s+1
Dlags =
:
s + 0:1
We now make Bode plots for KDlagsGs as follows:
100 Gain dB 50
0
−50
−100 −1
10 0 1 10 10 2 10 Phase deg Frequency (rad/sec) −180 −270
−1 10 0 1 10 10
Frequency (rad/sec) We note that at ! = 0:2, jKDlagsGsjs=0:2j = 227 250: 5 2 10 To get gain 250 at ! = 0:2, we set lag compensator as
s + 1:1
Dlags =
s + 0:1
that gives Kv = 110 100, and jKDlagsGsjs=0:2j 250:
100 Gain dB 50
0
−50
−100 −1
10 0 1 10 10 2 10 Phase deg Frequency (rad/sec) −180 −270
−1 10 0 1 10 10 2 10 Frequency (rad/sec) It is seen that the high frequency gain is much smaller than
,40dB. 6 We now design lead compensator. We choose !c 2!c = 14.
Then at s = 14j ,
0 6 KDlagG = ,196o: Hence P.M. = ,16o. The requirement of P.M. = 45o gives
m = 45 + 16 = 61o: The is now given by
We now have that 1 , sin 61o
= 0:067:
=
1 + sin 61o p = 54; zlead = !c=
0 plag = p !c = 3:623:
0 Hence, the lead compensator is
1 s + 3:623
Dleads =
:
0:067 s + 54 7 We nally have
Ds = KDleadsDlagsGs; K = 10;
s + 1:1
1 s + 3:623
; Dlags =
:
0:067 s + 54
s + 0:1
The Bode plots of DsGs are as follows:
Dleads = Gain dB 100 50 0 −50 −1
10 0 1 10 10 2 10 Frequency (rad/sec) Phase deg −120
−150
−180
−210
−240
−1 10 0 1 10 10
Frequency (rad/sec) 8 2 10 Noted that low frequency gain is satis ed: it is more than 48dB.
For P.M. and high frequency gain, we magnify the Bode plot as
follows:
40 Gain dB 20
0
−20
−40 0
10 1 10
Frequency (rad/sec) 2 10 Phase deg −120
−150
−180
−210
−240
0 10 1 10
Frequency (rad/sec) 2 10 P.M. 45o. But the gain at ! = 100 is 1:29=100 = ,37:8dB.
We thus reduce value for lead compensator:
= 0:067 1:29 = 0:0864 =
p
p
zlead = 14 0:0864 = 4:116; plead = 14= 0:0864 = 47:6: 9 We now replace the original lead by
1 s + 4:116
:
Dleads =
0:0864 s + 47:6
The Bode plots are given as follows. It can be seen that the
P.M. and ,40dB for ! 100 are satis ed.
40 Gain dB 20
0
−20
−40 0
10 1 10
Frequency (rad/sec) 2 10 Phase deg 120
−150
−180
−210
−240
0 10 1 10
Frequency (rad/sec) 10 2 10 ...
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This note was uploaded on 01/23/2012 for the course EE 4580 taught by Professor Gu during the Fall '10 term at LSU.
 Fall '10
 GU

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