System type steady state tracking Bode plot At

System type, steady state tracking, & Bode plot At very low frequency: gain plot slope = – 20 N d. B/dec. phase plot value = – 90 N deg If LF gain is flat, N=0, Kp = DC gain, Kv=Ka=0 If LF gain is -20 d. B/dec, N=1, Kp=inf, Kv=w. LFg_tan_c , Ka=0 If LF gain is -40 d. B/dec, N=2, Kp=Kv=inf, Ka=(w. LFg_tan_c)2

System type on Nyquist plot Kp

System relative order

Margins on Bode plots G(s) In most cases, stability of this closed-loop can be determined from the Bode plot of G: – Phase margin > 0 – Gain margin > 0

Margins on Bode plot

If never cross 0 d. B line (always below 0 d. B line), then PM = ∞. If never cross – 180° line (always above – 180°), then GM = ∞. If cross – 180° several times, then there are several GM’s. If cross 0 d. B several times, then there are several PM’s.

Example: Bode plot on next page.

Example: Bode plot on next page.

1. Where does cross the – 180° line Answer: _____ at ωpc, how much is 2. Closed-loop stability: _____

1. crosses 0 d. B at _____ at this freq, 2. Does cross – 180° line? ____ 3. Closed-loop stability: _____

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 The complete Nyquist plot: – Plot G(jω) for ω = 0+ to +∞ – Get complex conjugate of plot, that’s G(jω) for ω = 0– to –∞ – If G(s) has pole on jω-axis, treat separately – Mark direction of ω increasing – Locate point: – 1

Encirclement of the -1 point • As you follow along the G(jω) curve for one complete cycle, you may “encircle” the – 1 point • Going around in clock wise direction once is +1 encirclement • Counter clock wise direction once is – 1 encirclement

Nyquist Criterion Theorem # (unstable poles of closed-loop) Z = # (unstable poles of open-loop) + # encirclement or: Z = P + N To have closed-loop stable: need Z = 0, i. e. N = –P P N

That is: G(jω) needs to encircle the “– 1” point counter clock wise P times. If open loop is stable to begin with, G(jω) cannot encircle the “– 1” point for closed -loop stability In previous example: 1. No encirclement, N = 0. 2. Open-loop stable, P = 0 3. Z = P + N = 0, no unstable poles in closed-loop, stable

Example:

As you move around from ω = –∞ to 0–, to 0+, to +∞, you go around “– 1” c. c. w. once. # encirclement N = – 1. # unstable pole P = 1

i. e. # unstable poles of closed-loop = 0 closed-loop system is stable. Check: c. l. pole at s = – 3, stable.

Example: 1. Get G(jω) for ω = 0+ to +∞ 2. Use conjugate to get G(jω) for ω = –∞ to 0– 3. How to go from ω = 0– to ω = 0+? At ω ≈ 0 :

# encirclement N = _____ # open-loop unstable poles P = _____ Z = P + N = ____ = # closed-loop unstable poles. closed-loop stability: _______

Example: Given: 1. G(s) is stable 2. 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: 1. 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 with K = 1? ____

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? ______

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 ∞?

2. To use Nyquist criterion, need complete Nyquist plot. a) Get complex conjugate b) Connect ω = 0– to ω = 0+ through an infinite circle c) Count # encirclement N d) Apply: Z = P + N o. l. stable, P = _______ Z = _______ c. l. stability: _______

Incorrect Correct

Example: G(s) stable, P = 0 G(jω) for ω > 0 as given. 1. Get G(jω) for ω < 0 by conjugate 2. Connect ω = 0– to ω = 0+. But how?

Choice a) : Incorrect Where’s “– 1” ? # encirclement N = _______ Z = P + N = _______ Make sense? _______

Choice b) : Where is “– 1” ? # encir. N = _____ Z=P+N = _______ closed-loop stability _______ Correct

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.

Incorrect

Example: G(s) stable, 1. Get conjugate for ω < 0 2. Connect ω = 0– to ω = 0+. Needs to go one full circle with radius ∞. Two choices. P=0

Choice a) : N=0 Z=P+N=0 closed-loop stable Incorrect!

Choice b) : N=2 Z=P+N =2 Closed loop has two unstable poles Correct!

Which way is correct? For stable & non-minimum phase systems,

Example: G(s) has one unstable pole P = 1, no unstable zeros 1. Get conjugate 2. Connect ω = 0– to ω = 0+. How? One unstable pole/zero If connect in c. c. w.

# 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

If connect c. w. : For A > – 1 N = ______ Z=P+N = ______ For A < – 1 N = ______ Z = ______ No contradiction. This is the correct way.

Example: G(s) stable, minimum phase P=0 G(jω) as given: get conjugate. Connect ω = 0– to ω = 0+,

If A < – 1 < 0 : N = ______ Z = P + N = ______ stability of c. l. : ______ If B < – 1 < A : A=-0. 2, B=-4, C=-20 N = ______ 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 C < – 1 < B : N = ______ Z = P + N = ______ closed-loop stability: ______ If – 1 < C : N = ______ Z = P + N = ______ closed-loop stability: ______