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Controller design by R.L.

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Typical setup: C(s) Gp(s) Root Locus Based Controller Design Goal: Select poles and zero of C(s) so that R.L. pass through desired region Select K corresponding to a ... – PowerPoint PPT presentation

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Title: Controller design by R.L.


1
Controller design by R.L.
  • Typical setup

C(s)
Gp(s)
  • Root Locus Based Controller Design Goal
  • Select poles and zero of C(s) so that R.L. pass
    through desired region
  • Select K corresponding to a good choice of
    dominant pole pair

2
Types of classical controllers
  • Proportional control
  • Needed to make a specific point on RL to be
    closed-loop system dominant pole
  • Proportional plus derivative control (PD control)
  • Needed to bend R.L. into the desired region
  • Lead control
  • Similar to PD, but without the high frequency
    noise problem max angle contribution limited to
    lt 75 deg
  • Proportional plus Integral Control (PI control)
  • Needed to eliminate a non-zero steady state
    tracking error
  • Lag control
  • Needed to reduce a non-zero steady state error,
    no type increase
  • PID control
  • When both PD and PI are needed, PID PD PI
  • Lead-Lag control
  • When both lead and lag are needed, lead-lag
    lead lag

3
Proportional control design
  • Draw R.L. for given plant
  • Draw desired region for poles from specs
  • Pick a point on R.L. and in desired region
  • Use ginput to get point and convert to complex
  • Compute K using abs
  • and polyval
  • Obtain closed-loop TF
  • Obtain step response and compute specs
  • Decide if modification is needed

4
PD controller design
  • Design steps
  • From specs, draw desired region for pole.Pick
    from region, not on RL
  • Compute
  • Select
  • Select

Gpdevalfr(sys_p,pd) phipi - angle(Gpd)
zabs(real(pd))abs(imag(pd)/tan(pi-phi))
Kd1/abs(pdz)/abs(Gpd)
5
Drawbacks of PD
  • Not proper deg of num gt deg of den
  • High frequency gain ? 8
  • High gain for noise
  • Saturates circuits
  • Cannot be implemented physically

6
  • Approximation to PD
  • Same usefulness as PD
  • Lead Control
  • Draw R.L. for G
  • From specs draw region for desired c.l. poles
  • Select pd from region
  • LetPick z somewhere below pd on Re
    axisLetSelect

7
  • Alternative Lead Control
  • Draw R.L. for G
  • From specs draw region for desired c.l. poles
  • Select pd from region
  • Let
  • Select

8
Lag control design
  • It has destabilizing effect (lag)
  • Not used for improving Mp, tr, tp, ts,
  • Use it to reduce a non-zero ess
  • Use it when R.L. of G(s) go through the desired
    region but ess is too large.

9
Design steps
  1. Draw R.L. for G(s).
  2. From specs, draw desired pole region
  3. Select pd on R.L. in region
  4. Get
  5. With that K, compute error constant(Kpa, Kva,
    Kaa) from KG(s)
  6. From specs, compute Kpd, Kvd, Kad

10
  1. If Ka gt Kd , doneelse pick
  2. Re-compute
  3. Closed-loop simulation tuning as necessary

11
  • Example
  • Want
  • Solution

C(s)
Gp(s)
12
Draw region
13
  • Draw R.L.
  • Pick pd on R.L. in Region pick pd 0.35
    j0.5
  • Since there is one in G(s)

14
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15
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16
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17
A better tuning may be to go back and re-pick pd
18
  • Lag control can improve ess, but cannot eliminate
    ess
  • Use PI control to eliminate ess
  • PI

19
  • Only advantage of PI remove ess
  • It has destabilizing effect May ? MP , ? ts ,
    etc.
  • Sluggish settling, just like Lag
  • Needs trial and error tuning of Kp and KI

20
This way, just use your PDdesign program.
21
  • Second design
  • Draw R.L. for G(s)
  • From specs, draw desired region
  • Pick pd on R.L. in region
  • i. Chooseii. Choose
  • Simulate tune

22
  • Example
  • Want
  • Solution Draw R.L.

C(s)
Gp(s)
23
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24
  • Clearly, R.L. pass through desired region.
  • Pick (right on boundary)
  • Choose

25
  • Step response ess 0
  • No MP (no overshoot)
  • fast rise to 0.85, then very sluggish to 1
  • Tune 1 KP ? to 2.5 KP

26
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27
  • None unique solution
  • Design is a creative process guided by science

28
C(s)
G (s)
  • Example
  • Want

29
  • Sol G(s) is type 1Since we want finite ess to
    unit acc, we need the compensated system to be
    type 2 C(s) needs to have in it

30
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31
  • Draw R.L., it passes through the desired region.
  • Pick pd on R.L. in Region
  • pick pd 180 j160
  • Now choose z to meet Ka

32
  • Also

33
  • Pick z 0.03
  • Do step resp. of closed-loop
  • Is it good enough?

34
Design goal
35
  • If tr 0.0105 not satisfactorywe need to reduce
    tr by 5

36
Label each Root Locus with one TF choice.
J
A
E
D
F
G
H
I
B
C
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