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System type, steady state trackingPowerPoint Presentation

System type, steady state tracking

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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

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

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

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 ∞?

- To use Nyquist criterion, need complete Nyquist plot. “unstable poles”
- 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: _______

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.

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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