Root Locus Root Locus Unit 7: Part 1: Sketching the Root Locus Engineering 5821: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland 1 Root Locus Vector Representation of Complex Numbers Defining the Root Locus Properties of the Root Locus Sketching the Root Locus March 14, 2010 ENGI 5821 Unit 7: Root Locus Techniques ENGI 5821 Unit 7: Root Locus Techniques Root Locus The Root Locus is a graphical method for depicting location of the closed-loop poles as a system parameter is varied. It is applicable for first and second-order systems, but also to higher order systems. Changes in a parameter, such as the gain K, affects the location of the equivalent closed-loop system poles. Root locus allows us to determine the movement of these poles as K is varied. Vector Representation of Complex Numbers We will need to represent complex numbers and complex functions F (s) as vectors. Typically the complex functions we are concerned with have the following form: (s + z1)(s + z2) F (s) (s + p1)(s + p2) Consider the complex number s + q, where q will stand in for either a zero or a pole. We can represent s + q as a vector in the complex plane. The magnitude and angle of this vector are given by the complex exponential representation, s + q re jθ Where r is the vector length and θ is the angle from the real-axis.
Normally we would draw s + q as a vector out of the origin. Normal: s s + q q Complex number drawn from its own zero: However, we can recognize s q as a zero of s + q and draw the same vector with its tail at q (see above right). s + q -q s (s + z1)(s + z2) F (s) (s + p1)(s + p2) We can replace each term s + zi or s + pi with their corresponding complex exponential forms: F (s) rz1eθz 1 rz2 eθz 2 rp1 eθp 1 rp2 eθp 2 rz1rz2 rp1 rp2 eθz 1 +θz 2 + θp 1 θp 2 The magnitude and phase of F (s) are as follows: F (s) rz1rz2 rp1 rp2 s + z1 s + z2 s + p1 s + p2 F (s) θz1 + θz2 + θp1 θp2 (s + z1) + (s + z2) + (s + p1) (s + p2) e.g. Evaluate the following complex function when s 3 + j4, (s + 1) F (s) s(s + 2) Defining the Root Locus Consider the following system, designed to track a visual target. The magnitude and phase of the zero is: 20 116.6 o. The pole at the origin evalutes to 5 126.9 o. The pole at -2 evaluates to 17 104.0 o. 20 F (s) F (s) 5 17 [116.9o 129.9 o 104.0 o ] For a unity feedback system such as this we will refer to K1K2 s(s+10) as the open-loop transfer function (if the unity feedback signal were cut the system would be open-loop). Thus, the open-loop poles are at s 0 and s 10. However, depending upon K we will obtain different closed-loop poles, which are the roots of s 2 + 10s + K. We can utilize the quadratic formula to obtain these roots for values of K 0. We can then plot the positions of these poles...
The path of the closed-loop poles as K varies is the root locus. Observations: for K < 25 the system is overdamped for K 25 the system is critically damped for K > 25 the system is underdamped Properties of the Root Locus The transfer function for a general closed-loop system is, KG(s) T (s) 1 + KG(s)H(s) We are concerned with the poles of T (s). We will have a pole whenever KG(s)H(s) 1. The root locus is the locus of points in the s-plane for which this is true. We can express this equality as follows, More Observations: In the underdamped portion the real-part of the pole, σd is constant. Therefore so is Ts 4/σd. The root locus never crosses the jω axis. Therefore the system is stable. KG(s)H(s) KG(s)H(s) 1 (2k+1)180 o k 0, ±1, ±2, ±3,... A particular s is on the root locus if its magnitude is unity and its angle is an odd multiple of 180. Satisfying both of these requirements means that s is a closed-loop pole of the system.
e.g. Consider the following system, We will test a couple of points to see if they are closed-loop poles. We evaluate KG(s)H(s) graphically for s 2 + j3, If we test a point and find it is on the RL (e.g. s 2 + j 2/2) we may then want to determine the corresponding value of K. Since KG(s)H(s) 1 on the RL, 1 pole lengths K G(s)H(s) zero lengths θ1 + θ2 θ3 θ4 56.31 + 71.57 90 108.43 70.55 Therefore s s + j3 is not on the root locus. Sketching the Root Locus One possibility is to sweep through a sampling of points in the s-plane and test each one for inclusion in the RL. It is much more preferable to utilize some insights about the RL to identify its major characteristics and therefore obtain a rough sketch. The following rules will help achieve this... 1. Number of branches: Consider a branch to be the path that one pole traverses. There will be one branch for every closed-loop pole. e.g. Two branches for the previous quadratic example. 2. Symmetry: The poles for real physical systems either lie on the real-axis or come in conjugate pairs. Hence... The root locus is symmetrical about the real axis. 3. Real-axis segments: Consider evaluating the anglular contribution of the open-loop poles and zeros at points P1, P2, P3, and P4 below: The angular contribution of a pair of complex open-loop poles (or zeros) is zero The contribution of poles or zeros to the left of the point is zero Using only the real-axis poles or zeros to the right of the point, we find that the angle sum alternates between 0 o and 180 o. On the real axis, for K > 0 the root locus exists to the left of an odd number of real-axis, finite open-loop poles and/or finite open-loop zeros.
4. Starting and ending points: Where does the RL begin and end? It begins at K 0 and ends at K. Consider the closed-loop transfer function: e.g. G(s) K(s+3)(s+4) (s+1)(s+2), H(s) 1 T (s) KG(s)H(s) 1 + KG(s)H(s) NG (s) NH(s) K DG (s) DH(s) NG (s) NH(s) 1 + K DG (s) DH(s) KNG (s)nh(s) DG (s)dh(s) + KNG (s)nh(s) If we let K 0 the poles of T (s) approach the combined poles of G(s) and H(s). If K the poles of T (s) approach the combined zeros of G(s) and H(s). Thus, the RL begins at the open-loop poles of G(s)H(s) and ends at the zeros of G(s)H(s). e.g. G(s) K(s+3)(s+4) (s+1)(s+2), H(s) 1 Note that we don t know yet what the exact trajectory of the root locus will be. What if the number of open-loop poles and zeros is mismatched? e.g. K F (s) s(s + 1)(s + 2) A function can have both poles and zeros at infinity. For example, as s. F (s) K s s s We therefore consider F (s) to have three zeros at infinity. If we include both finite and infinite poles and zeros every function has an equal number of poles and zeros. The root locus for F (s) (i.e. KG(s)H(s) F (s)) would start at the three finite poles and go towards the zeros at infinity. Yet how do we get to these zeros at infinity?
5. Behaviour at infinity: The root locus approaches straight lines as asymptotes for zeros at infinity. These asymptotes are defined as lines with real-axis intercept σa and angle θa. finite poles finite zeros σa (2k + 1)π θa where k is an integer and θa is the angle (in radians) to the positive real-axis. We get as many asymptotes as there are branches corresponding to zeros at infinity. The derivation for these formulae can be found at www.wiley.com/college/nise under Appendix L.1. e.g. Sketch the RL for the following system: We first apply rules 1-4: The following is our complete RL sketch. We now compute the real-axis intercept and the angles of all asymptotes: finite poles finite zeros σa (0 1 2 4) ( 3) 4 1 4/3 (2k + 1)π θa π/3 for k 0 π for k 1 5π/3 for k 2 Notice that we have one real-axis intercept but multiple angles. There are three asymptotes one for each infinite zero. We obtain three unique values for θa before the angles start to repeat.