自动控制原理课件chap8

2 ˜ c (jω ) = 1 + α(ωT ) +j (1 − α)ωT G 1 + (αωT )2 1 + (αωT )2
△
=x ( ω )
△
=y ( ω )
Since x(ω ) − 1+α 2α
2
+ y 2 (ω ) =
1−α 2α
2
˜ c (jω ), ω ≥ 0, is the semi-circle in the upper half plane centered at 1+α with we see that the Nyquist locus G 2α 1−α the radius 2α , as illustrated in Fig. 8.1. ˜ c (jω ) has a maximum. To determine this maximum, we take the first The Nyquist locus indicates that ∠G ˜ c (jω ) and set it equal to 0, we get derivative of ∠G d (arctan ωT − arctan αωT ) = dω = ⇐⇒ T αT − 2 1 + (ωT ) 1 + (αωT )2 T (1 − α)(1 − αω 2 T 2 ) =0 (1 + (ωT )2 )(1 + (αωT )2 ) 1 ωm = √ . αT
15
10
5
0
−5 −2 10
10
−1
10
0
ωm
10
1
10
2
10
3
40
30
20
θm ωm
10
−1
10
0 −2 10
10
0
10
1
10
2
10
3
Figure 8.2: Bode diagram of a lead compensator
8.2.2
Lead Compensation Techniques Based on Bode Diagrams
The basic idea behind the lead compensation is using the positive phase angle to enhance the magnitude crossover frequency (hence the bandwidth) and the phase margin, so that the transient behavior can be improved. We take a simple example to illustrate this technique. Consider the system shown in Fig. 8.3. The specifications of the compensated system are • the static velocity error constant Kv = 20; • the phase margin γ ≥ 50o , the gain margin Kg ≥ 10dB.
177
178
CHAPTER 8. FREQUENCY-RESPONSE COMPENSATION DESIGN (CHAPTERS 9 AND 10)
2
1.5
Gc(jω)
1
0.5
∠ G(jω)
0
ω=0
+
(1+α)/(2*α)
ω=0−
ω=+∞ ω=−∞
−0.5
−1
−1.5
−2 −0.5
0
0.5
1
1.5
The transfer function of a lead compensator is Gc (s) = Kc α ˜ c (s) = Denote G
T s+1 αT s+1 ,
then from ˜ c (jω ) = arctan ωT − arctan αωT ∠G
˜ c (jω ) > 0 for all ω > 0. To determine the Nyquist locus we decompose G ˜ c (jω ) as it is clear that ∠G
1 jT √αT +1 1 jαT √αT +1
1 = √ >1 α
if we locate ωm at the original magnitude crossover frequency, the magnitude will be enhanced in the vicinity of this frequency so that the crossover frequency of the compensated system is shifted to the right and the phase margin will be again less than the required value. Hence, we make a reservation of 5o for the phase margin. The phase angle to be provided by the compensator is thus θm = 38o . 1−sin θm 1−α Step 3: Determine α. From θm = arcsin 1+ α we get α = 1+sin θm = 0.24.
180
CHAPTER 8. FREQUENCY-RESPONSE COMPENSATION DESIGN (CHAPTERS 9 AND 10)
−110
−120
−130
∠ G1(Байду номын сангаасω)
−140
−150
−160
−170
−180 0 10
10
1
10
2
Figure 8.5: Phase angle curve of G1 (s)
Chapter 8
Frequency-Response Compensation Design (Chapters 9 and 10)
8.1 8.2
8.2.1
Introduction Lead Compensation
Characteristics of a Lead Compensator
Ts + 1 , 0<α<1 αT s + 1
8.2. LEAD COMPENSATION
179
R(s) E (s) −6
Gc (s)
-
4 s(s + 2)
Y (s) -
Figure 8.3: A control system
Step 1: Determine the gain factor of the compensator K = Kc α. With Gc (s) = Kc α the open-loop transfer function is Q(s) = Gc (s)G(s) = Ts + 1 Kc α G(s) αT s + 1
˜ c (jω ) achieves it maximum θm . The Bode diagram of a lead which means that at the frequency ωm ∠G compensator is plotted in Fig. 8.2. Note that ωm is the geometric mean value of the two corner frequencies 1 1 T and αT .
=K K Let us denote G1 (s) = KG(s) = s(0.2 5s+1) . The specification on the static error constant demands 2K = 20, ⇐⇒ K = 10. The log-magnitude versus frequency plot of G1 (s) can be obtained by shifting the 0dB axis of the 10 = 20dB. The magnitude crossover frequency is approximately √ plot of G(s) downward by 20 log ω = 40 = 6.3. Since ∠G(j 6.3) ∼ = −163o , the phase margin is γ ∼ = 17o < 50o . Hence, a phase compensation is needed.
30
20
10
ωm
0
−10
−20
−30
−40
−50 0 10
10
1
10
2
Figure 8.6: Log-magnitude curve of G1 (s) and a lead compensator Step 4: Determine the magnitude crossover frequency ωc . From 1 20 log √ = −10 log α = 6.24dB α and 20 log |G1 (j 9)| = −6.24dB, we choose ωc = ωm = 9/sec. 1 Step 5: Calculate T . From ωm = √αT , we get T =√ Hence 1 = 0.227 αωm
2
2.5
3
3.5
4
4.5