《信号与系统 Signals and Systems》课程教学资料（英文版）lecture21 The Concept of a root locus

H AA Signals and systems Fall 2003 Lecture #21 25 November 2003 Feedback a) Root locus Tracking c) Disturbance rejection d) The Inverted Pendulum 2. Introduction to the z-transform

Signals and Systems Fall 2003 Lecture #21 25 November 2003 1. Feedback a) Root Locus b) Tracking c) Disturbance Rejection d) The Inverted Pendulum 2. Introduction to the Z-Transform

The concept of a root locus x()(+) C(s) H(s) G(s Q(s) C(SH(s 1+C(SG(sH(S) C(s), G(s)-Designed with one or more free parameters Question: How do the closed-loop poles move as we vary these parameters?-Root locus of 1+C(SG(sH(S)

The Concept of a Root Locus • C(s), G ( s) — Designed with one or more free parameters • Question: How do the closed-loop poles move as we vary these paramet ers? — Root locus of 1+ C(s) G ( s ) H( s )

The“ Classical” Root locus problem C(s)=K-a simple linear amplifier K H(S G(s) Q(s) KH(s 1+KH(s)G(s) Closed-loop H(s) y(t) poles are same Q(s)= H(s) 1+KH(s)G(s)

The “Classical” Root Locus Problem C ( s) = K — a simple linear amplifier Closed-loop poles are the same

A Simple example +2 G(s)=1 K K (a)Q( +k s+K+ (b)Q(s) 1+ +K+2 In either case. pole is at s=-2-K Sketch where gm pole moves 9x as k increases. K>0 2 Becomes more stable Becomes less stable

A Simple Example Becomes more stable Becomes less stable Sketch where pole moves as |K| increases... In either case, pole is at so = -2 - K

What Happens More generally For simplicity, suppose there is no pole-zero cancellation in G(s)H(s h( Q()=1+KG(s)H() Closed-loop poles are the solutions of 1+KGSH(S That is G(SH(s Difficult to solve explicitly for solutions given any specific value of K, unless G(s)H(s)is second-order or lower Much easier to plot the root locus the values of s that are solutions for some value of k because 1)It is easier to find the roots in the limiting cases for K=0.± 2) There are rules on how to connect between these limiting points

What Happens More Generally ? • For simplicity, suppose there is no pole-zero cancellation in G ( s ) H( s ) — Difficult to solve explicitly for solutions given any specific value of K, unless G ( s ) H( s) is second-order or lower. That is Closed-loop poles are the solutions of — Much easier to plot the root l ocus, the values of s that are solutions for some value of K, because: 1) It is easier to find the roots in the limiting cases for K = 0, ± ∞. 2) There are rules on how to connect between these limiting points

Rules for Plotting root Locus G(s)H(s) K End points AtK=0,G(s0)H(S)=∞ s, are poles of the open-loop system function G(s)H(s) At|K=∞,G(0)H(S)=0 so are zeros of the open-loop system function G(s)H(s). Thus Rule #1: A root locus starts(atK=0) from a pole of G(s)H(s)and ends(at K=o)at a zero ofG(s)H(s Question: What if the number of poles t the e number oi zeros Answe Start or end at±o

Rules for Plotting Root Locus • End points — A t K = 0, G ( s o ) H( s o) = ∞ ⇒ s o are poles of the open-loop system function G ( s ) H( s). — A t |K| = ∞, G ( s o ) H( s o) = 0 ⇒ s o are zeros of the open-loop system function G ( s ) H( s). Thus: Rule #1: A root locus starts (at K = 0) from a pole of G ( s ) H( s) and ends (at |K| = ∞) at a zero of G ( s ) H( s). Question: What if the number of poles ≠ the number of zeros ? Answer: Start or end at ± ∞

Rule #2: angle criterion of the root locus K real number Thus, So is a pole for some positive value of K if: K≥0→∠G(s0)H(S0)=(27+1)r: In this case, So is a pole ifK=1/G(so H(so)l Similarly so is a pole for some negative value of K if: K≤0→∠G(0)H(s0)=2n丌 In this case, So is a pole ifK=-1/G(so) H(soI

Rule #2: Angle criterion of the root locus • Thus, s0 is a pole for some positive value of K if: In this case, s0 is a pole if K = 1/|G(s0) H(s0)|. • Similarly s0 is a pole for some negative value of K if: In this case, s0 is a pole if K = -1/|G(s0) H(s0)|

Example of root locus +2 One zero at -2 s(s+1)two poles at 0 locus g(sh(s)) K≥0 K≤0 oX 0K=0 2 Real Axis

Example of Root Locus. One zero at -2, two poles at 0, -1

Tracking x(→(+ e(t) P(s) Compensator Plant In addition to stability we may want good tracking behavior,1.e e(t)=m(t)-y(t)≈0ast→ for at least some set of input signals E(s)=1+ C(SP(s) X(S E(w) 1+C(u)Pgw X We want C(oP(o) to be large in frequency bands in which we want good tracking

In addition to stability, we may want good tracking behavior, i.e. for at least some set of input signals. Tracking ⇓ + = ( ) 1 ( ) ( ) 1 ( ) X s C s H s E s ( ) 1 ( ) ( ) 1 ( ) ω ω ω ω X j C j H j E j + = We want to be large in frequency bands in which we want good tracking C ( j ω) P ( j ω)

Tracking(continued Using the final-value theorem lim e(t)=lim sE(s)=lim t→∞ s→0 8=01+C(5NX(s Basic example: Tracking error for a step input Suppose(t)=u(t)←→X(s) Then lim e(t)=lim 8-01+C(s)P(s

Tracking (continued) Using the final -value theorem Basic example: Tracking error for a step input