Unit 7: Part 1: Sketching the Root Locus. Root Locus. Vector Representation of Complex Numbers

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1 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.

2 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: o. The pole at the origin evalutes to o. The pole at -2 evaluates to o. 20 F (s) F (s) 5 17 [116.9o o 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 s + 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...

3 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.

4 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 θ 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 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.

5 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?

6 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 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 ( ) ( 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.