Root Locus sketching rulesPoles and zeros at infinityRoot Locus sketching rulesRoot Locus sketching rulesRoot Locus sketching rulesRoot Locus sketching rulesRoot Locus sketching rulesLecture 18 – Friday, Oct. 192.004 Fall ’07 Root Locus sketching rulesWednesday• Rule 1: # branches = # poles• Rule 2: symmetrical about the real axis• Rule 3: real-axis segments are to the left of an odd number of real-axis finite poles/zeros• Rule 4: RL begins at poles, ends at zerosToday• Rule 5: Asymptotes: angles, real-axis intercept• Rule 6: Real-axis break-in and breakaway points• Rule 7: Imaginary axis crossings (transition to instability)Next week• Using the root locus: analysis and design examplesLecture 18 – Friday, Oct. 192.004 Fall ’07 Poles and zeros at infinityT (s)hasazero at infinity if T (s →∞) → 0.T (s)hasapole at infinity if T (s →∞) →∞.ExampleKG(s)H(s)=Ks(s +1)(s +2).Clearly, this open—loop transfer function has three poles, 0, −1, −2. It has nofinite zeros.For large s,wecanseethatKG(s)H(s ) ≈Ks3.So this open—loop transfer function has three zeros at infinity.Lecture 18 – Friday, Oct. 192.004 Fall ’07 Root Locus sketching rules• Rule 5: Asymptotes: angles and real-axis interceptσa=Pfinite poles −Pfinite zeros#finite poles − #finite zerosθa=(2m +1)π#finite poles − #finite zerosm =0, ±1, ±2,...In this example, poles = {0, −1, −2, −4},zeros = {−3} soσa=[0 + (−1) + (−2) + (−4)] − [(−3)]4 − 1= −43θa=(2m +1)π4 − 1=½π3, π,5π3¾Nise Figure 8.12-j3-j2-j1j10 1-1-2-3-4AsymptoteAsymptoteAsymptotes-planejωσ2j2j3X X X XσaθaFigure by MIT OpenCourseWare.Lecture 18 – Friday, Oct. 192.004 Fall ’07 Root Locus sketching rules• Rule 6: Real axis break-in and breakaway pointsNise Figure 8.13For each s = σ on a real—axissegment of the root locus,KG(σ)H(σ)=−1 ⇒ K = −1G(σ)H(σ)(1)Real—axis break—in & breakaway pointsare the real values of σ for whichdK(σ)dσ=0,where K(σ)isgivenby(1)above.Alternati vely, we can solveX1σ + zi=X1σ + pi.for real σ.Figure by MIT OpenCourseWare.-j3-j2-j1j10 1-1-2jωσ2 3 4 5j2j3j4X X-σ1σ2maxK forthis real—axissegmentminK forthis real—axissegmentLecture 18 – Friday, Oct. 192.004 Fall ’07 Root Locus sketching rules• Rule 6: Real axis break-in and breakaway pointsNise Figure 8.13In this example,KG(s)H( s)=K(s − 3)(s − 5)(s +1)(s +2)so on the real—axis segments we haveK(σ)=−(σ +1)(σ +2)(σ − 3)(σ − 5)= −σ2+3σ +2σ2− 8σ +15Taking the derivative,dKdσ= −11σ2− 26σ − 61(σ2− 8σ +15)2and setting dK/dσ =0wefindσ1= −1.45 σ2=3.82Alternatively, poles = {−1, −2},zeros = {+3, +5} so we must solve1σ − 3+1σ − 5=1σ +1+1σ +2⇒11σ2− 26σ − 61 = 0.This is the same equation as before.Figure by MIT OpenCourseWare.-j3-j2-j1j10 1-1-2jωσ2 3 4 5j2j3j4X X-σ1σ2maxK forthis real—axissegmentminK forthis real—axissegmentLecture 18 – Friday, Oct. 192.004 Fall ’07 Root Locus sketching rules• Rule 7: Imaginary axis crossingsIf s = jω is a closed—loop poleon the imaginary axis, thenKG(jω)H(jω)=−1(2)The real and imaginary parts of (2)provide us with a 2 × 2systemof equations, which we can solvefor the two unknowns K and ω(i.e., the critical gain beyond whichthe system go es unstable, and theoscillation frequency at the critical gain.)Note:Nise suggests using the Ruth—Hurwitz criterion for the same purpose.Since we did not cover Ruth—Hurwitz,we present here an alternativebut just as effective method.-j3-j2-j1j10 1-1-2-3-4AsymptoteAsymptoteAsymptotes-planeσ2j2j3X X X Xsystem responsecontains undampeterms at this pointjωdFigure by MIT OpenCourseWare.Lecture 18 – Friday, Oct. 192.004 Fall ’07 Root Locus sketching rules• Rule 7: Imaginary axis crossingsIn this example,KG(s) H(s)=K(s +3)s(s +1)(s +2)(s +4)=Ks +3Ks4+7s3+14s2+8s⇒KG(jω)H(jω)=jKω +3Kω4− j7ω3− 14ω2+ j8ω.Setting KG(jω)H(jω)=−1,−ω4+ j7ω3+14ω2− j(8 + K)ω − 3K =0.Separating real and imaginary parts,½−ω4+14ω2− 3K =0,7ω3− (8 + K) ω =0.In the second equation, we can discard thetrivial solution ω = 0. It then yieldsω2=8+K7.Substituting into the first equation,−µ8+K7¶2+14µ8+K7¶− 3K =0⇒K2+65K − 720 = 0.Of the two solutions K = −74.65, K =9.65 w e candiscard the negative one (negative feedback ⇒ K>0).Thus, K =9.65 and ω =p(8 + 9.65)/7=1.59.-j3-j2-j1j10 1-1-2-3-4AsymptoteAsymptoteAsymptotes-planejωσ2j2j3X X X Xsystem responsecontains undampedterms at this pointFigure by MIT
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