system type steady state tracking
Download
Skip this Video
Download Presentation
System type, steady state tracking

Loading in 2 Seconds...

play fullscreen
1 / 32

System type, steady state tracking - PowerPoint PPT Presentation


  • 57 Views
  • Uploaded on

System type, steady state tracking. C(s). G p (s). Type 0: magnitude plot becomes flat as w  0 phase plot becomes 0 deg as w  0 Kv = 0, Ka = 0 Kp = flat magnitude height near w  0. Asymptotic straight line, -20dB/ dec. Kv in dB. Kv in Value.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' System type, steady state tracking' - colin-benson


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide2

Type 0: magnitude plot becomes flat as w 0

phase plot becomes 0 deg as w 0

Kv = 0, Ka = 0

Kp = flat magnitude height near w 0

slide3

Asymptotic straight line, -20dB/dec

Kv in dB

Kv in Value

Type 1: magnitude plot becomes -20 dB/dec as w 0

phase plot becomes -90 deg as w 0

Kp = ∞, Ka = 0

Kv = height of asymptotic line at w = 1

= w at which asymptotic line crosses 0 dB horizontal line

slide4

Asymptotic straight line, -40dB/dec

Ka in dB

Sqrt(Ka) in Value

Type 2: magnitude plot becomes -40 dB/dec as w 0

phase plot becomes -180 deg as w 0

Kp = ∞, Kv = ∞

Ka = height of asymptotic line at w = 1

= w2at which asymptotic line crosses 0 dB horizontal line

slide5
Margins on Bode plots

In most cases, stability of this closed-loop

can be determined from the Bode plot of G:

  • Phase margin > 0
  • Gain margin > 0

G(s)

margins on nyquist plot
Margins on Nyquist plot

Suppose:

  • Draw Nyquist plot G(jω) & unit circle
  • They intersect at point A
  • Nyquist plot cross neg. real axis at –k
stability from nyquist plot
Stability from Nyquist plot

G(s)

  • Get completeNyquist plot
  • Obtain the # of encirclement of “-1”
  • # (unstable poles of closed-loop) Z= # (unstable poles of open-loop) P + # encirclement N

To have closed-loop stable: need Z = 0, i.e. N = –P

slide10

Here we are counting only poles with positive real part as “unstable poles”

    • jw-axis poles are excluded
  • Completing the NP when there are jw-axis poles in the open-loop TF G(s):
    • If jwo is a non-repeated pole, NP sweeps 180 degrees in clock-wise direction as w goes from wo- to wo+.
    • If jwo is a double pole, NP sweeps 360 degrees in clock-wise direction as w goes from wo- to wo+.
slide11

Example:

  • Given:
    • G(s) is stable
    • With K = 1, performed open-loop sinusoidal tests, and G(jω) is on next page
  • Q: 1. Find stability margins
    • 2. Find Nyquist criterion to determine closed-loop stability
slide13

Solution:

    • Where does G(jω) cross the unit circle? ________ Phase margin ≈ ________Where does G(jω) cross the negative real axis? ________ Gain margin ≈ ________Is closed-loop system stable withK = 1? ________
slide14

Note that the total loop T.F. is KG(s).

If K is not = 1, Nyquist plot of KG(s) is a scaling of G(jω).

e.g. If K = 2, scale G(jω) by a factor of 2 in all directions.

Q: How much can K increase before GM becomes lost? ________

How much can K decrease? ______

slide15

Some people say the gain margin is 0 to 5 in this example

Q: As K is increased from 1 to 5, GM is lost, what happens to PM?

What’s the max PM as K is reduced to 0 and GM becomes ∞?

slide16

To use Nyquist criterion, need complete Nyquist plot.

    • Get complex conjugate
    • Connect ω = 0– to ω = 0+ through an infinite circle
    • Count # encirclement N
    • Apply: Z = P + N
  • o.l. stable, P = _______
  • Z = _______
  • c.l. stability: _______
slide17

Correct

Incorrect

slide18

Example:

  • G(s) stable, P = 0
  • G(jω) for ω > 0 as given.
  • Get G(jω) forω < 0 by conjugate
  • Connect ω = 0– to ω = 0+.But how?
slide19

Choice a) :

Where’s “–1” ?

# encirclement N = _______

Z = P + N = _______

Make sense? _______

Incorrect

slide20

Choice b) :

Where is“–1” ?

# encir.N = _____

Z = P + N= _______

closed-loopstability _______

Correct

slide21

Note: If G(jω) is along –Re axis to ∞ as ω→0+, it means G(s) has in it.

when s makes a half circle near ω = 0, G(s) makes a full circle near ∞.

choice a) is impossible,but choice b) is possible.

slide23

Example: G(s) stable, P = 0

  • Get conjugatefor ω < 0
  • Connect ω = 0–to ω = 0+.Needs to goone full circlewith radius ∞.Two choices.
slide24

Choice a) :

N = 0

Z = P + N = 0

closed-loopstable

Incorrect!

slide25

Choice b) :

N = 2

Z = P + N= 2

Closedloop has two unstable poles

Correct!

slide26

Which way is correct?

For stable & non-minimum phase systems,

slide27

Example: G(s) has one unstable pole

  • P = 1, no unstable zeros
  • Get conjugate
  • Connectω = 0–to ω = 0+.How?One unstablepole/zeroIf connect in c.c.w.
slide28

# encirclement N = ?

If “–1” is to the left of A

i.e. A > –1

then N = 0

Z = P + N = 1 + 0 = 1

but if a gain is increased, “–1” could be inside, N = –2

Z = P + N = –1

c.c.w. is impossible

slide29

If connect c.w.:

For A > –1N = ______

Z = P + N

= ______

For A < –1N = ______

Z = ______

No contradiction. This is the correct way.

slide30

Example: G(s) stable, minimum phase

P = 0

G(jω) as given:

get conjugate.

Connect ω = 0–

to ω = 0+,

slide31

If A < –1 < 0 :N = ______Z = P + N = ______ stability of c.l. : ______

If B < –1 < A : A=-0.2, B=-4, C=-20N = ______Z = P + N = ______ closed-loop stability: ______

Gain margin: gain can be varied between (-1)/(-0.2) and (-1)/(-4),

or can be less than (-1)/(-20)

slide32

If C < –1 < B :N = ______Z = P + N = ______ closed-loop stability: ______

If –1 < C :N = ______Z = P + N = ______ closed-loop stability: ______

ad