Control Systems. Control Systems Design Lead-Lag Compensator.
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1 Design Lead-Lag Compensator
2 Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design.
3 Frequeny Domain Design Frequeny response approah an impose the transient response performane indiretly Bode diagrams: useful for the frequeny domain speifiation Nyquist plot: useful for system (relative) stability analysis.
4 Lead Compensator Appropriate improvement in transient response and a small hange in steady-state auray G s Ts ( s) K T K (0 ) Ts s T Polar plot (K C = ) jt G ( j) K jt α determines the maximum phase angle φ m whih ours at ω m Zero:- T Pole: - T sin m 2 2
5 Lead Compensator Bode diagram (K C =, α=0.) G ( j) m jt K jt T G ( j) 0. jt j0. T Lead ompensator: High-pass filter Goal of the lead ompensator: to provide suffiient phase-lead angle to offset the exessive phase lag in the unompensated system
6 Lead Compensator The maximum phase lead angle φ m ours at ω m To inrease phase margin, magnitude of the ompensated system to ross 0 db at this frequeny ω m but need to are for magnitude distortion added by lead ompensator 50 Bode Diagram Gm = 9.93 db (at 4.0 rad/s), Pm = 04 deg (at rad/s) Magnitude (db) 0-50 Phase (deg) Phase Margin PM 80 G( j ) H ( j ) ω : Gain rossover frequeny Frequeny (rad/s)
7 Lead Compensator The magnitude of the lead ontroller at: m T G ( j) jt jt m j j T This is the amount that the lead ompensator will shift the magnitude plot. To have the gain rossover point (0 db) at the right point, we have to make sure that the geometri frequeny mean falls at the point where the open-loop unompensated system is 20 log 0 ( α) below 0 db. If is absorbed into the ontrol gain, then the previous maximum magnitude would be
8 Lead Compensator Design Step 0: Assume following lead ompensator Open-loop Plant: Gs () Compensator: G Ts Ts ( s) K K ( K K) Ts Ts Loop Gain of Compensated System: Ts Ts GC ( s) G( s) K G( s) G ( s) Ts Ts with G( s) KG( s)
9 Lead Compensator Design Steps Step : Determine K to satisfy stati error onstants (K p or K v ) Step 2: Using this K, draw a Bode diagram of G (s), and evaluate the phase margin Step 3: Determine the neessary phase angle needed to meet design spes.
10 Lead Compensator Design Steps Step 4: Determine by using ( φ m =required phase lead angle + 5~2deg) Find the frequeny where: sin Selet this frequeny as the new gain rossover frequeny, where the maximum phase shift ours Step 5: Determine the lead ompensator m T m G ( s) 20log0 - T - T
11 Lead Compensator Design Steps Step 6: Solve for K K Step 7: Chek the gain margin. If it is not satisfatory, one may have to iterate.
12 Example Ex. Gs () 4 s s 2 Design a lead ompensator to give K v = 20 s -, PM 50 deg, and GM 0 db Step 0: Find low-frequeny gain: Using G () s K C C Ts Ts 4K G ( s) KG( s) K K s s 2 C
13 Example Step : Ts 4K Kv lim sgc ( s) Gs lim s 2K s0 s0 Ts s s 2 Gain requirements: 2K 20 K 0 Step 2: draw a Bode diagram of G (s), and evaluate the phase margin G ( j) j( 40 j 2)
14 Example Step 3: Determine the neessary phase angle needed to meet design spes. φ m = 33 (neessary)+5 (additional) deg (extra 5 deg for ompensation of the shift in gain rossover frequeny.) Step 4: Determine sin sin 38 m sinm 0.24 sin sin 38 m
15 Example Step 5: Determine the orner frequenies First we find ω by noting that G ( s) 20log0 The graph gives at db G ( j) 6. 2 db m 9 rad/se The zero at /T T T rad / se The pole at /T T rad / se
16 Example Step 6: Solve for K C = K/ KC G ( s) K 0/ Ts K K Ts s T s T 4.7 s s Step 7: Verify the design PM 50 deg and GM infinite
17 Example Nyquist
18 Leture Outline Lag ompensator design in frequeny domain Lag ompensator design steps. Example on lag ompensator design.
19 Lag Compensator An appreiable improvement in steady-state auray at the expense of inreasing the transient-response time G s Ts ( s) K T K ( ) Ts s T Polar plot (K C = ) Zero:- T Pole: - T G ( j) K jt jt
20 Lag Compensator Bode diagram (K C =, β=0) G ( j) 0 jt j0t G ( j) K jt jt Lead ompensator: Low-pass filter Goal of the lead ompensator: to provide attenuation in high frequeny range to give a system suffiient phase margin
21 Lag Compensator Design Steps Step 0: Assume the following lag ompensator Compensator: G Ts Ts ( s) K K ( K K) Ts Ts Loop Gain of Compensated System: G Ts Ts ( s) G( s) K G( s) ( ) Ts Ts G s with G( s) KG( s) Step : Determine K to satisfy stati error onstants.
22 Lag Compensator Design Steps Step 2: Find C for G (s) Chek PM and GM to see if they meet spes If not, find the frequeny where G( jw) = -80 deg + required PM Required PM = speified PM + 5~2 deg. for safety margin This frequeny is the new gain rossover frequeny C Step 3: Choose the orner frequeny =/T of the zero We want to hange the magnitude plot without hanging the phase plot at the new rossover frequeny Therefore, hoose the zero at /T to be around deade below the new orner frequeny C
23 Lag Compensator Design Steps Step 4: Determine and the pole loation... We now examine than 0 db. G ( j ) to find out how muh it is greater 0dB G ( j log ) 20 0 Choose and then the pole is at /T Step 5: Form the lag ompensator... The atual ompensator gain K K
24 Example Ex. G( s) s( s )(0.5s ) Design a lag ompensator K v = 5 se -, PM > 40 degrees and GM > 0 db Step 0: G s ( s) K T,( ) s T G ( s) KG( s) s( s K )(0.5s )
25 Example Step : Gain K v lim s0 lim s0 K 5 sg sg ( s) ( s) G( s) lim s0 lim s0 s sk s( s )(0.5s Ts G Ts K ) ( s) Step 2: draw a Bode diagram of G (s) G ( s) s( s 5 )(0.5s )
26 Example Step 3: The phase margin is 20 degrees whih means the unity feedbak of the gain adjusted unompensated system G (s) is unstable. If we want to have a phase margin of 40 deg, we should add another 5~2 deg to be safe. Choose orner frequeny =/T=0. rad/se (zero of the lag ompensator) The phase of G (s) is 28 deg around 0.5 rad/s Therefore, the new rossover frequeny should be about 0.5 rad/s.
27 Example Step 4: Based on the new ross over frequeny, the zero should be plaed at = /T = 0.05 rad/s about a deade below. The gain of G (s) at = 0.5 rad/s is about 20 db so that s how muh the lag ompensator must attenuate. 20log The pole of the lag ompensator 0.0 T rad/se
28 Example Step 5: G ( s) K s 0 s 00 K K G ( s) G( s) 5(0s ) s(00s )( s )(0.5s )
29 Example Lag ontroller auses slower transients
30 Outline Lead-Lag ompensator Qualitative system responses.
31 Lead-Lag Compensators Whether to use the Lead or Lag ontroller depends on the nature of your plant Lead for improved transient performane Lag for improved steady-state performane when you need improved transient performane and steady-state traking lead-lag ompensator G ( s) K s s s T T2 (, ) s T T 2
32 Lead Lag Compensators Phase lead adds phase at the unompensated gain rossover frequeny thereby inreasing the phase margin Phase lag provides attenuation allowing an inrease in gain at low frequeny to improve steady-state performane K =, γ = β = 0, T 2 = 0T. K =, γ = β.
33 Lead Lag Compensators Qualitative system responses y (t) y (t) y (t) y(t) y (t) y (t) y (t) y(t) unompensated lead lag lead-lag
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