System type steady state tracking
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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.

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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

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


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


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


  • 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+.


  • Example: “unstable poles”

  • 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


  • Solution: “unstable poles”

    • 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? ________


Note that the total loop T.F. is “unstable poles”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? ______


Some people say the gain margin is 0 to 5 in this example “unstable poles”

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 ∞?



Correct “unstable poles”

Incorrect


  • Example: “unstable poles”

  • 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?


Choice a) : “unstable poles”

Where’s “–1” ?

# encirclement N = _______

Z = P + N = _______

Make sense? _______

Incorrect


Choice b) : “unstable poles”

Where is“–1” ?

# encir.N = _____

Z = P + N= _______

closed-loopstability _______

Correct


Note: If “unstable poles”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.


Incorrect “unstable poles”


  • Example: “unstable poles”G(s) stable, P = 0

  • Get conjugatefor ω < 0

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


Choice a) : “unstable poles”

N = 0

Z = P + N = 0

closed-loopstable

Incorrect!


Choice b) : “unstable poles”

N = 2

Z = P + N= 2

Closedloop has two unstable poles

Correct!


Which way is correct? “unstable poles”

For stable & non-minimum phase systems,


  • Example: “unstable poles”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.


# encirclement “unstable poles”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


If connect c.w.: “unstable poles”

For A > –1N = ______

Z = P + N

= ______

For A < –1N = ______

Z = ______

No contradiction. This is the correct way.


Example: “unstable poles”G(s) stable, minimum phase

P = 0

G(jω) as given:

get conjugate.

Connect ω = 0–

to ω = 0+,


If “unstable poles”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)


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

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


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