2008090111364000001(ech10 Design)

η ψ
2
Ⅱ Several types of compensator (a) Cascade compensator
R(s)
compensator
process
C (s )
Gc (s )
H (s )
G (s )
(b) Feedback compensator
R(s)
G (s )
C (s )
Gc (s )
therefore set
1 = 0 . 15 T
20loga = 20db
Gc ( s ) =
01 = a
1 + 6.66 s 1 + 66.6s
22
The characteristic of phase-lag compensator (1)to increase phase margin. (2) to enhance state-steady accuracy.
Steady state error: Open-loop gain: K
ess
1
Typical frequency domain indices: Cutoff frequency: ω c Phase margin: φ pm Gain margin: GM Peak value :
Mp
The resonant frequency: ω r bandwidth: ω B Typical complex domain indices:
G (s) H ( s) = K s ( s + 2)
It is required that the velocity constant equal to 20. The phaseMargin φ pm ≥ 450 answer .
1 + 6.66 s Gc ( s ) = 1 + 66.6s
01 = a
−1
3. Evaluate a form Eq. φ m = sin
a −1 a +1
4. Evaluate 10loga and determine the frequency ω m uncompensated magnitude curve is equal to –10log a. 5.Evaluate T from Eq. ω m =
K = 20 2
so
K=40
20 GH ( jω ) = jω (0.5 jω + 1)
10
(2) Calculate phase-margin of uncompensated system determine cutoff frequency
40
ω c1 (ω c1 )2 + 4
ω c1 = 6.2rad / s
H (s )
3
(c) Input compensator(Prefilter)
R(s)
C (s )
Gc (s )
G (s )
H (s )
(d) Compound compensator
R(s)
G p (s )
C (s )
Gc (s )
G (s )
H (s )
4
Ⅲ Approaches to system design (a). Frequency Method (b). Root Locus Method (c). Equivalent Structure and Equivalent Transfer Function Method
ω c' = 44rad / s
We set therefore
10 log a = 6
1 T a
= 44
a = 4 T = 0.01136
φ pm = 49.80
The loop transfer function of compensated system is
100(0.04544s + 1) Gc ( s )G ( s ) = s (0.01136 s + 1)(0.1s + 1)
(a > 1)
R1
vi
R2
v0
R1 + R2 a= R2
R1 R2 T= C R1 + R2
6
The phase-lead compensation transfer function can be written
1 + aTs Gc ( s ) = 1 + Ts
(a > 1)
Bode diagram of the phase-lead network is
17
(b) phase-lag compensation R1
R2
C
vi
v0
Vo ( s ) R2Cs + 1 Gc ( s ) = = Vi ( s ) ( R1 + R2 )Cs + 1
1 + Ts 1 ( s + z ) = Gc ( s ) = 1 + aTs a ( s + p )
a= R1 + R2 R2
ϕ (ω c ) = −900 − tg −1 0.5ω c − tg −1 0.065ω c + tg −1 0.21ω c = −134.90
φ pm = 180 + ϕ (ω c ) = 45.1
0
0
13
Bode diagram
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Bode diagram
15
Bode diagram
16
It is required that the velocity constant equal to 20. The phaseMargin φ pm ≥ 450 Answer: (1) determine value of K
K v = lim s ⋅ G ( s ) H ( s ) =
s →0
.
1 T a
,where the
6. Check the result,and repeat the steps if necessary.
9
Example : A feedback control system has a loop transfer function
G (s) H ( s) = K s ( s + 2)
(5) Evaluate T
ω
m
=
1 T a
= 8 .5
12
T=0.065 (6) Check the rsult Then compensated loop transfer function is
G c ( s ) GH ( s ) = 20 ( 0 . 21 s + 1) s ( 0 . 5 s + 1)( 0 . 065 s + 1)
dB
+ 20
π
2
0
0
(aT )
−1
ω m T −1 φm
10 lg a
20 lg a
ω
7
where
ω
m
=
1 T
−1
a
φ m = sin
a −1 a +1
20 lg Gc ( jω m ) = 10 lg a
phase- lead design
1 + aTs Gc ( s ) = 1 + Ts
(a > 1)
10.2 Cascade Compensation design using the bode diagram The compensator Gc (s ) can be written
Gc ( s ) =
K ∏ ( s + zi )
M
∏ (s + p )
i i =1
5
i =1 N
Consider the first-order compensator with transfer function
K (s + z) Gc ( s ) = ( s + p)
The phase-lead compensation when z<p. The phase-lag compensation when z>p. (a) Phase –lead compensation
C
i
1 1 + aTs Gc ( s ) = a 1 + Ts
Байду номын сангаас
T = R2C
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Bode diagram of the phase-lag network is
dB
0
(aT )
−1
(T )−1
− 20 lg a
ω
− 20
ϕ0
−
π
2
Phase- lag design
Gc ( s ) = 1 + Ts 1 + aTs
19
step 1. Obtain the Bode diagram of the uncompensated system with the gain adjusted for the desired error constant. 2. Determine the phase margin of the uncompensated system and , if it is insufficient, proceed with the following steps. 3. Determine the frequency where the phase margin requirement would be satisfied if the magnitude curve crossed the 0-db line at this ω c' (Allow for 5 0 phase lag from the phase-lag frequency network when determining the new crossover frequency). 4. Place the zero of compensator one decade below the new crossover frequency ω c and thus ensure only 5 0 of additional phase lag at ω c