System type, steady state tracking, & Bode plot R(s) C(s) Gp(s) Y(s) Type = N At very low frequency: gain plot slope = –20N dB/dec. phase plot value = –90N deg

Kv = 0 Ka = 0 Low freq phase = 0o

Kp = ∞ Ka = 0 Low freq phase = -90o =Kv

Kp = ∞ Kv = ∞ Low freq phase = -180o

System type, steady state tracking, & Nyquist plot C(s) Gp(s) As ω → 0

Type 0 system, N=0 Kp=lims0 G(s) =G(0)=K Kp w0+ G(jw)

Type 1 system, N=1 Kv=lims0 sG(s) cannot be determined easily from Nyquist plot winfinity w0+ G(jw)  -j∞

Type 2 system, N=2 Ka=lims0 s2G(s) cannot be determined easily from Nyquist plot winfinity w0+ G(jw)  -∞

System type on Nyquist plot

System relative order

Examples System type = Relative order = System type = Relative order =

In most cases, stability of this closed-loop 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)

If never cross 0 dB line (always below 0 dB 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 dB several times, then there are several PM’s.

Example: Bode plot on next page.

Example: Bode plot on next page.

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

crosses 0 dB at __________ at this freq, Does cross –180° line? ________ 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 c.w. once is +1 encirclement c.c.w. once is –1 encirclement

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

That is: G(jω) needs to encircle the “–1” point c.c.w. P times. If open loop is stable to begin with, G(jω) cannot encircle the “–1” point for closed-loop stability In previous example: No encirclement, N = 0. Open-loop stable, P = 0 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

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

Example: Get G(jω) for ω = 0+ to +∞ Use conjugate to get G(jω) for ω = –∞ to 0– 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: _______

Q: 1. Find stability margins 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

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 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 ∞?

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: _______

Correct Incorrect

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

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

Correct 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, P = 0 Get conjugate for ω < 0 Connect ω = 0– to ω = 0+. Needs to go one full circle with radius ∞. Two choices.

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

Correct! 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 Get conjugate 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 G(jω) as given: get conjugate. Connect ω = 0– to ω = 0+,

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