7.1 Root Locus (RL) Principle We introduce the RL through an example. Consider servo motor system shown bellow The closed loop transfer function is motor compensator K  R (s)  A (s) – + The CE is s 2 + 2s + K = 0 The system is stable for K > 0. It is not evident how K affect the transient response. To have more understanding how K affect the system characteristic we will plot the locus of the roots in the s plane as K is varied from 0 to infinity. The roots are K=0. s K=1 –1 –2 K= j –j–j RL is a plot of the closed loop poles as some parameters of the system is varied.

7.1 Root Locus Principle We generally consider a system as in the following figure The CE of this system is 1+KG(s)H(s)=0 (1) A value s 1 is a point in the RL if and only if satisfies (1) for real 0<K< . For this system we call KG(s)H(s) the open loop function (OLF) Equation (1) can be written as K= –1 / G(s)H(s) (2) Since G(s) and H(s) are generally complex and K is real then there are 2 condition must be met |G(s)H(s)|=1/|K| and(3)  G(s)H(s) = r , r = ±1, ±3, ±5 …(4) We call (3) the magnitude criterion of the RL, and (4) the angle criterion. Any value s on the RL must meet both criterions. K – + G(s)G(s) H(s)H(s) s G(s) includes plant and compensator TF. The closed loop TF is For 2 nd order system the RL appears as a family of 2 paths or branches traced out by 2 roots of CE.

7.1 Root Locus Principle To illustrate this criterion let us consider s1s1 11 s 1 - z 1 22 s 1 - p 1 33 s 1 - p 2 s z1z1 p2p2 p1p1 Suppose that s 1 on the RL. Thus the angle condition becomes  1 –  2 –  3 = ±  The locus of all points meet this condition forms the complete RL of the system From the preceding discussion, it is seen that the condition of a point to be on the RL is that  (all zero angles) –  (all pole angles) = r  where zero angle is the angle of s – z k and pole angle is the angle of s – p k. z k and p k are zero and pole of the OLF Calculating and plotting RL digitally is available and it is a convenient way to do it. However a good knowledge of the rule for plotting the RL will offer insight effect of changing parameter and adding poles and/or zeros in the design process. From mason gain formula, the CE is  (s) = 1 + F(s) = 0 where F(s) is the OLF. Many design procedures are based on the OLF.

7.2 Root Locus Rules Since the TF is real function then its complex roots must exist in conjugate pairs, then RL is symmetrical with respect to the real axis Rule 1. RL is symmetrical with respect to the real axis With the z k and p k are zeros and poles of the OLF. The CE may be expressed as Therefore for K = 0 the roots of CE are simply the poles of OLF  RL start at the poles of G(s)H(s). It is always that the number of zeros, N Z,, equals the number of poles, N P. If the number of finite zeros < N P then the rest of zeros lies at infinity. If K  but s remain finite then the branches of RL must approach the zeros of the OLF, else there must be zeros at infinity. This mean that RL finish at zeros of G(s)H(s). Rule 2. RL originates on the poles of G(s)H(s) for K=0 and terminates on the zeros of G(s)H(s) as K , including zeros at infinity. When there are zeros at infinity then the RL will extend to infinity as well. The asymptotes can be determined. If there is  zeros at infinity then the OLF can be written as

7.2 Root Locus Rules As s , the polynomial becomes Remember that the values of s must satisfy the CE 1-KG(s)H(s)= 0 regardless the value of s hence we have hence we can write for large s s  + Kb m =0 The angle of the roots are principal values of angle (asymptote angles)  = r  /  r = ±1, ±3,… The asymptote intersect the real axis at  asymptote angle 0No asymptote ±90 0 3±60 0, ±45 0, ± Where P is the finite poles and Z is the finite zeros

7.2 Root Locus Rules Rule 3 If OLF has  zeros at infinity, the RL approach  asymptote as K approach infinity. The asymptote intersect real axis at  a with angle . Example. Consider the OLF There are 2 zeros at infinity, hence there are 2 asymptotes with angle ±90 0. The asymptote intersect the real axis at Consider point a to the right of p 1. The angle condition states that if a is on the RL then  (all zero angles) –  (all pole angles) = r . Since this not the case then any points on the real axis to the right of p 1 are not on the RL. By the same argumentations we conclude that the points between p 1 and p 2 are on the RL, the points between p 2 and z 1 are not on the RL, the points to the left of z 1 are on the RL. Rule 4 The RL includes all points to the left of an odd number of real critical frequencies (poles and zeros) 7.3 Additional Techniques a z1z1 p2p2 p1p1 s -- 

Multiple roots on the real axis. If CE has multiple roots a on the real axis then we can write the characteristic polynomial as 1+KG(s)H(s) =(s-a) k Q(s) (1) Differentiating (1) with respect to s gives For s =a it must be (2) Hence the multiple roots a can be found by solving (2). Since G(s)H(s) is rational we can represent G(s)H(s)= N(s) / D(s) And (2) can be written as N(s)D’(s)- N’(s)D(s)=0 (3) The multiple points are breakaway points, they are points at which branches of RL leave or enter the real axis. Rule 5. The breakaway points on RL will appear among the roots of polynomial obtained either from (2) or (3). Since on the RL K = –1/ G(s)H(s) = – D(s)/N(s) we conclude that K’= N(s)D’(s)- N’(s)D(s)=0 Hence at breakaway points K is either maximum or minimum.

7.3 Additional Techniques Angle of Departure from Complex Poles Suppose the open loop poles are as shown in the figure below, and s 1 is on the RL.  11 22 s1s1 33 p1p1 p2p2 p3p3 We are interested in finding the angle  1 when  is very small. This is the angle of departure. From angle criterion we know that –  1 –  2 –  3 =180 0 For very small ,  2 =90 0, hence  1 = – 90 0 –  3 Rule 6 Loci will depart from a pole p j (arrive at z j ) at angle of  d (  a ) where  d =   zi    pi + r   a =   pi    zi + r 

7.3 Additional Techniques Example 1 We consider a system with OLF -4/ j j with 3 finite poles and 3 infinite zeros. Rule 2. The RL start from pole s =1, s = – 2, and s = – 3. Rule 3. The asymptote intersect the real axis at (1-2-3)/(3-0) = – 4/3, with angle ±60 0, and Rule 4. The RL occurs on the real axis for – 2< s <1 and s<–3. As K is increased p 1 and p 3 move left, but p 2 move right. Rule 5. Determine the breakaway point by differentiating (1) and set the result to zero. 3s 2 +8s+1=0  s 1 = and s 2 = (ignore s 2 ) p1p1 p2p2 p3p3 (1) Routh-Hurwitz method can be used to find values of K for stability, this is found to be 6 < K< 10. K=10 K=6

7.3 Additional Techniques Example 2 -2 The RL start from double pole s =0 and stop at zero s =  1 and s =  One asymptote at The RL occurs on real axis for s <  1 Breakaway point at s = 0 and s =  2 Example 3 A simplified ship steering system modeled as a system of order 2 The RL start from s =0 and s =  0.1 and stop at s = . Two asymptotes at 90 0 intersect the real axis at s =  0.05 The RL occurs on real axis for  0.1< s <  1 Breakaway point at s =   0.05  0.1

7.3 Additional Techniques Example 4 The order-3 model of ship steering system is The RL start from poles s =0, s =  0.1, & s =  2 and stop at 3 zeros at s =  3 asymptotes intersect real axis at s =  0.7 The real axis RL, for  0.1<s<0 and s<  2 Breakaway point at s =   2  0.1s =  0.7 j K a =42 The CE is The Routh array is s s K a s 1 (0.42  0. 01K a )/2.1  K a < 42 s K a  K a > 42 For stability 0<K a <42. The auxiliary polynomial for K a =42 is Q a = 2.1 s (42) = 2.1 (s ) Setting Q a to zero we find s = ±j At this point the system is oscillating with frequency = rad/s

7.3 Additional Techniques Example 5: The order-4 model of ship steering system is Breakaway point at s =  , s =  For stability 0<K a < j K a =38.03  2 0  20  s =  The RL start from poles s =0, s =  0.1, s =  2, & s =  2 and stop at 4 zeros at s =  4 asymptotes intersect real axis at s =  The real axis RL, for  0.1<s<0 and  20<s<  2 For K a =38.03 the system is oscillating with frequency

7.3 Additional Techniques Table 1. Stabilities of the last 3 example example K a for marginal stability Oscillation frequency 3 (order 2) Always stable … 4 (order 3) (order 4) The last 3 example illustrate RL construction and the reduction of order of system. From table 1 and 2 we see that for small K a (=1) model of order 2 is adequate. For moderate K a (= 20) model of order 3 is adequate. For large K a (= 40) model of order 4 is required. Six rules for RL construction have been developed many additional rule are available for accurate RL graph, however we should use digital computer to draw accurate RL. exK a =1K a =20K a =40 3–0.05±j0.05–0.05±j0.312–0.05±j –0.04±j0.05 – –0.025±j0.31 – –0.002±j0.4 – –0.04±j0.05 – – –0.022±j0.31 – – –0.023±j0.4 – – Table 2. Poles of example 3, 4, and 5.

7.4 Additional RL properties Sketching RL relies on experiences, followings are some low order RL p1p1 p 1 p 2 z1z1 p 1 p 2 z1z1 p 1 p 2 p 1 p 2 p 3 p 1 p 2 p 3 p 1 p 2 p 3 p 1 z1z1 p 2

7.4 Additional RL properties The CE can be written as D(s) + KN(s) =0 For a given K 1 we have D(s) + K 1 N(s) =0 Suppose that K is increased by K 2 from K 1 then the CE becomes D(s) + (K 1 + K 2 )N(s) =0 which can be expressed as With respect to K 2 the locus appears to originate on roots as placed by letting K = K 1 but still terminates on the original zeros. -2 Example suppose we have OLF The CE can be written as s 2 + K(s+1)=0 For K =K 1 =2, we have s 2 +2s+2=(s+1) 2 +1=0 With roots s = 1±j. Thus the RL of OLF is the same with RL of (1) but start from 1±j. (1) (2)

7.4 Additional RL properties The CE can be written as 1+ KG(s)H(s) =0 (1) Consider replacing s with s 1 1+ KG(s-s 1 )H(s-s 1 ) =0 (2) then for K 0, (s-s 1 ) = s 0 satisfied (2). This is the efect of shifting all poles and zeros of the OLF by a constant amount s 1. Example. The RL of OLF -2 (3) is as shown below The RL of OLF (3) is as shown below The last property is that a RL has symmetry at breakaway points. Suppose 4 branches come together at a common breakaway point, the the angle between 4 branches will be /4=90 0.

7.5 Other Configuration So far we plot RL of CE 1+ KG(s)H(s) =0 (1) by varying K from 0 to . Other system configuration should be converted to this form first. As an example let us vary  from 0 to  for the following CE (2) or We rewrite the equation to yield and (3) Now (3) is the same form as (1) with K=  and G(s)H(s) = s/(s 2 +5) In general the procedure is as follows 1.Write the CE in a polynomial of s 2.Grouped the terms that are multiplied by  and that are not, that is we express D c (s) +  N c (s)=0 Example Let us design PI controller using RL. motor compensator – +

7.5 Other Configuration The system CE is We have to rewrite this equation as The sketch of th RL is as follow Supposed that K I =1, and we want to vary K P and plot the RL. Eq. (1) becomes (2) j0.5 -j0.5 Now the RL is available, it is easier to specify the roots. Suppose we want critical damping condition then s =-0.5 and from (2) K P is (1) The required PI compensator is then G c (s) = /s and the design is complete