Slide 1

Definition: The root locus is the path of the roots of the characteristic equation traced out in the s-plane as a system parameter is varied.The roots of the closed-loop characteristic equation define the system characteristic responses.Their location in the complex s-plane lead to prediction of the characteristics of the time domain responses in terms of:damping ratio, znatural frequency, wndamping constant, s first-order modesConsider how these roots change as the loop gain is varied from 0 to .

Root Locus AnalysisThe General Root Locus MethodConsider the general system

whereThe characteristic equation is

or

orR(s)C(s)+ G(s)H(s))(1)()()(sGHsGsRsC+=L2,1,0=k1)(-=sGH0)(1=+sGH)12()(1)(+==ksGHsGHpAll values of s which satisfy ; ;are roots of the closed-loop characteristic equation.Consider the following general formThe General Root Locus MethodL2,1,0=k1)(=sGH)12()(+=ksGHp zero. bemay :Noteip)())(()())(()(2121nmpspspszszszsKsGH++++++=LLThenThe General Root Locus Method1=i1)(1=++==pszsKsGHnimiiL2,1,0=k)12()()()(11+=+-+===kpszssGHniimiipRoot Locus Method:Geometric InterpretationConsider the example

Then the values of s = s1 which satisfy

))(()()(321pspsszsKsGH+++=p)12())()(()(321+=++++-+kpspsszs1321=+++pspsszsKare on the loci and are roots of the characteristic equation.jwsXX-z1XO-p1-p3-p2s1qp1qp2qp3qz1ABCDRoot Locus Method:Geometric InterpretationIn terms of the vectors, the condition for s = s1 to be on the root loci are

and

jwsXX-z1XO-p1-p3-p2s1qp1qp2qp3qz1ABCDKBCDABCDAK1or 1==L,2,1,0)12()(3211=+=++-kkpppzpqqqqRoot Locus MethodWhen plotting the loci of the roots as K = 0 , the magnitude condition is always satisfied.Therefore, a value of s = s1 that satisfies the angle condition, is a point of the root loci.The magnitude condition may then be used to determine the gain K corresponding to that value s1 .How can we easily determine if the angle condition is satisfied?

Root Locus Construction Rules1. The loci start (K = 0) at the poles of the open-loop system. There are n loci .2. The loci terminate(K ) at the zeroes of the open-loop system (include zeroes at infinity).For our example system

Therefore, as K 0 ,GH(s) , the poles of the loop transfer function.As K , GH(s) 0 , the zeroes of the loop transfer function. KpspsszssGH1)(321=+++=3. The root loci are symmetrical about the real axis.4. As K the loci approach asymptotes. There are q = n m asymptotes and they intersect the real axis at angles defined byThe roots with imaginary parts always occur in conjugate complex pairs.When the loci approach infinity, the angles from all the poles and zeroes are equal. The angle condition then ismq nq = (2k + 1)pRoot Locus Construction RulesL,2,1,0 , )12(=+kqkpThe angles from poles and zeroes to the left of s1 are zero. The angles from poles and zeroes to the right are p. An odd number are required to satisfy the angle condition.Root Locus Construction Rules5. The asymptotes intersection point on the real is defined by

6. Real axis sections of the root loci exist only where there is an odd number of poles and zeroes to the right.qsGHsGH-=)( of zeroes)( of polessaConsider our example with z1 = 4 , p12 = 1 2j

Asymptotes:Root Locus Construction RulesExample)21)(21()4()(jsjsssKsGH-++++=[][]113)4()21()21(0+=-----+--=jjas213)12(=-+kppangles =jwsXX4XO22j2j1+1real axislocusasymptote7. The angles of departure, qd from poles and arrival, qa to zeroes may be found by applying the angle condition to a point very near the pole or zero.The angle of arrival at the zero, -z1 is obtained fromRoot Locus Construction Rulesp)12()(11+=+=kp-zniiq)(211-++=z-zmiiazDeparture angle from p2 .qz1 = tan-1(2/3) = 33.7 qp1 = tan-1(2/1) = 116.6 qp3 = 90Then 33.7 (90 + 116.6 + qp2 ) = 180 qp2 = 352.9 = + 7.1 Root Locus Construction RulesExamplejwsXX4XO22j2j+1-p2116.633.790-p3-p1-z1qp28. The imaginary axis crossing is obtained by applying the Routh-Hurwitz criterion and checking for the gain that results in marginal stability. The imaginary components are found from the solution of the resulting auxiliary equation.Marginal stability refers to the point where the roots of the closed-loop system are on the stability boundary, i.e. the imaginary axis.Root Locus Construction RulesImaginary axis crossing:Characteristic equationFor marginal stability,K = 5 and the auxiliary equation is

Therefore, the imaginary axis intersection is

Root Locus Construction RulesExample04)5(20)4()21)(21(23=++++=++-+++KsKsssKjsjsss3 1 5+K 0s2 2 4K 0s 5K 0 0s0 4K 0Routh tablejjss16.31002022===+j16.3Summary:There are three root loci. One on the real axis from -p1 to -z1 , and a pair of loci from -p2 and -p3 to zeroes at infinity along the asymptotes. The departure angle from these poles is 7.1 and an imaginary axis crossing at s = 3.16j .Root Locus Construction RulesExamplejwsXX4XO22j2j+1-p2-p3-p1-z13.16 j7.1Breakaway Points:When two or more loci meet, they will breakaway from this point at particular angles. The point is known as a breakaway point. It corresponds to multiple roots.Root Locus Construction Rulesxxooxxxxxxxx45Some examples9. The angle of breakaway is 180/k where k is the number of converging loci.The location of the breakaway point is found fromNote:

Also, Root Locus Construction Rules[]0)(or 0==dssGHddsdK[][]0)()(2==-dssGHdsGHdsdK[])(1-=-sGHK0)()()()(=-sDsNsNsD[][]0)()()()()()()()(2=-==sDsDsNsDsNdssDsNddssGHdRoot Locus Plot:Breakaway Point ExampleConsider the following loop transfer function.

Real axis loci exist for thefull negative axis.Asymptotes:angles = (2k+1)p = p/3 , p , 5p/3

2)3()(+=ssKsGH23)0()033(-=----=asjwsXX4X22j2j+160asymptotes3Determine the breakaway points from

thenRoot Locus Plot:Breakaway Point ExamplejwsXX4X22j2j+13,10)3)(1(342--==++=++sssss0)96()9123(96)3(2232232=++++-=++=+sssssKsssKdsdssKdsd10. For a point on the root locus, s =s1 calculate the gain, K fromAlternately, K may be determined graphically from the root locus plotRoot Locus Construction RulesLL21112111zszspspsK++++=jwsXXXOs1ABCDABCDK=Summary of Root Locus Plot ConstructionPlot the poles and zeros of the open-loop system.Find the section of the loci on the real axis (odd number of poles an zeroes to the right).Determine the asymptote angles and intercepts.+pqkmnqqka-==-==zeroespoles,2,1,0, , )12(anglessLDetermine departure angles. For a pole -p1

Check for imaginary axis crossings using the Routh-Hurwitz criterion.Determine breakaway points.

Complete the plot.Summary of Root Locus Plot Constructionpq)12()()()(2112111+=-+--++++kp-pz-pz-ppLL==[]0)( fromlocation loci converging of # k , /angle=dssGHdkpRoot Locus Plot Example 3Loop Transfer function:

Roots:s = 0, s = 4, s = 2 4jReal axis segments:between 0 and 4 .Asymptotes:angles = )204)(4()(2+++=ssssKsGH24)0224(-=----=as47,45,43,404)12(=-+kpppppasymptotesjwsX4X22j2j+1454j4jXXBreakaway points:

Three points that breakaway at 90 . solving,jsb45.22,2--=ssss)ss(ssssKssssKdsd020186or 080368)8072244(803682323423234=+++=++++++-=+++jwsX4X22j2j+1454j4jXXRoot Locus Plot Example 32The imaginary axis crossings:Characteristic eqn.

Routh tables4 1 36 Ks3 8 80 0s2 26 K 0s 80-8K/26 0 0s0 K 0 0Condition for critical stability80-8K/26 > 0 or K