ME190L Nyquist Stability Criterion UC Berkeley Fall

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ME190L Nyquist Stability Criterion UC Berkeley Fall 2010 http://jagger ME190L Nyquist Stability Criterion UC Berkeley Fall 2010 http://jagger.me.berkeley.edu/~pack/me190L Copyright 2007-10, Andy Packard. This work is licensed under the Creative Commons Attribution-ShareAlike License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/2.0/ or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAA

Nyquist Harry Nyquist Hendrik Bode Relate closed-loop stability to the shape of the frequency-response curves of P and C, called the Nyquist Stability Criterion Today’s Proof: Assumes linear differential equations govern P and C Result is true for more complex linear systems (linear partial differential equations, convolution systems, Desoer-Callier algebra) Natural notion of robustness margins emerges Since frequency-responses can be obtained experimentally, closed-loop stability of interconnections can be ascertained from the individual blocks’ frequency response functions without ever explicitly finding the governing ODE (simply must assume that system is governed by an ODE) Harry Nyquist Hendrik Bode

Phase change and encirclements Bounded closed contour, Σ , not passing through the origin 0, traversed by s. The origin, 0, is either Inside Σ Σ encircles 0 clockwise (CW) Σ encircles 0 counterclockwise (CCW) Outside Σ What is the net change in as s traverses Σ? Σ Inside Σ, CW Inside Σ, CCW Outside Σ

Phase change and encirclements Outside Σ (does not encircle 0)

Phase change and encirclements Inside Σ (Σ encircles 0 one time, CW)

Phase change and encirclements Inside Σ (Σ encircles 0 two times, CW)

Phase change and encirclements Inside Σ (Σ encircles 0 three times, CW)

Phase change and encirclements Inside Σ (Σ encircles 0 two times, CW)

Phase change and encirclements Inside Σ (Σ encircles 0 one time, CW)

Phase change and encirclements Inside Σ (Σ encircles 0 one time, CW)

Net Phase change versus encirclements of 0 Summary: The phase change as a complex number traverses a bounded, closed path Σ is equal to 2πN , where N is the number of counterclockwise encirclements of 0 by Σ Remark: negative clockwise encirclements are the same as positive counterclockwise encirclements

pre-Nyquist: Argument Principal Basic principle, preliminary to Nyquist stability criterion Rational function G(s) Simple, closed contour Γ, in the complex plane, not passing through any poles or zeros of G Map points s on Γ (complex numbers) into their value G(s) The closed contour G(Γ ) is defined by mapping all of Γ by G Convention: Γ is traversed clockwise by mapping, this then defines the direction that G(Γ ) is traversed Γ s Re Im G G(s) Re Im G Γ G(Γ)

pre-Nyquist: Argument Principal Basic principle, preliminary to Nyquist stability criterion Rational function G(s) Simple, closed contour Γ, not passing through any poles or zeros of G The closed contour G(Γ ) is defined by mapping Γ by G, Γ is traversed clockwise, and this then defines the direction that G(Γ ) is traversed Define The number of poles of G inside Γ is denoted nP The number of zeros of G inside Γ is denoted nZ Result The closed contour G(Γ ) encircles the origin nZ-nP times, clockwise called “Cauchy’s Argument Principle” Equivalently closed contour G(Γ ) encircles the origin nP-nZ times, counterclockwise G Γ G(Γ)

Argument Principal: simplest case Simple closed contour, Γ , traversed clockwise by s. Fixed complex number r, two possibilities Inside Γ Outside Γ What is the net phase change in (s-r) as s traverses Γ? Γ r inside Γ r outside Γ

Argument Principal: general case Map Γ by G, giving G(Γ) Here G is the product/quotient of many phasors. At any value of s, the angle of G(s) is Γ G(Γ)

Argument Principal Map Γ by G, giving G(Γ) Since Γ does not pass through any poles or zeros of G, the curve G(Γ) is nonzero, and bounded. Define Theorem (argument principal): Observation for later: If Γ passes through a zero of G, then the curve G(Γ) passes through 0 Γ G(Γ)

Application to Closed-loop Stability analysis The open-loop gain is Define Clearly Apply argument principal with: G(s):= 1 + L(s) Γ is a contour that “encloses the entire right-half-plane” -

Nyquist contour, “enclosing the right-half plane” Standard ΓR contour, 3 segments, Segment 1 corresponds to s=jω, with ω ranging from 0 to R. on this segment, G(Γ) is just G(jω), with ω ranging from 0 to R, so all frequency-response Segment 2 corresponds to s=Rejθ with θ ranging from π/2 to –π/2 on this segment, G(Γ) collapses to a single point, since R is large Segment 3 corresponds to s=jω, with ω ranging from -R to 0. on this segment, G(Γ) is just G(-jω), with ω ranging from 0 to R. But G has real coefficients, so G(-jω) is the complex-conjugate of G(jω), so that G(Γ3) is the reflection (across real axis) of G(Γ1) 1 2 R 3

Nyquist Stability Theorem - 1 2 R Nyquist Stability Theorem: Assume L has no poles on imaginary axis (we’ll deal with that next time). Then 3 Observation: If L(ΓR) passes through -1, then the closed-loop system has a pole on ΓR

Nyquist Stability Theorem: Proof - 1 2 R Nyquist Stability Theorem: Assume L has no poles on imaginary axis. Then 3

Nyquist contour, “enclosing the right-half plane” Nyquist analysis w/ Standard ΓR contour determines the presence (or lack thereof) of closed-loop poles inside ΓR But the goal is to determine existence of poles in right-half-plane, which is larger than ΓR Solution: Do the analysis for arbitrarily large R If there are any closed-loop poles in RHP, then a large enough R will enclose them, and the analysis (which detects closed-loop poles in ΓR) will detect them. If there are no closed-loop poles in RHP, then for every R>0, the analysis will conclude that there are no closed-loop poles inside ΓR Analysis for arbitrarily large R is easy Segment 2 corresponds to s=Rejθ with θ ranging from π/2 to –π/2 on this segment, as R takes on larger and larger values, G(Γ) collapses to a single point R3 R2 R1 2 2 2