Unit 7: Part 1: Sketching the 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 March 14, 2010 ENGI 5821 Unit 7: Root Locus Techniques

Root Locus 1 Root Locus Vector Representation of Complex Numbers Defining the Root Locus Properties of the Root Locus Sketching the Root Locus 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: F (s) = (s + z 1)(s + z 2 ) (s + p 1 )(s + p 2 ) 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 + q Complex number drawn from its own zero: s q s + q s -q 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).

F (s) = (s + z 1)(s + z 2 ) (s + p 1 )(s + p 2 ) We can replace each term s + z i or s + p i with their corresponding complex exponential forms: F (s) = r z 1 e θz 1 r z2 e θz 2 r p1 e θp 1 r p2 e θp 2 = r z 1 r z2 r p1 r p2 eθz 1 +θz 2 + θp 1 θp 2 The magnitude and phase of F (s) are as follows: F (s) = r z 1 r z2 r p1 r p2 = s + z 1 s + z 2 s + p 1 s + p 2 F (s) = θ z1 + θ z2 + θ p1 θ p2 = (s + z 1 ) + (s + z 2 ) + (s + p 1 ) (s + p 2 )

e.g. Evaluate the following complex function when s = 3 + j4, F (s) = (s + 1) s(s + 2) 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 ]

Defining the Root Locus Consider the following system, designed to track a visual target. For a unity feedback system such as this we will refer to K 1K 2 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

More Observations: In the underdamped portion the real-part of the pole, σ d is constant. Therefore so is T s = 4/σ d. The root locus never crosses the jω axis. Therefore the system is stable.

Properties of the Root Locus The transfer function for a general closed-loop system is, T (s) = KG(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, 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, θ 1 + θ 2 θ 3 θ 4 = 56.31 + 71.57 90 108.43 = 70.55 Therefore s = s + j3 is not on the root locus.

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, K = 1 G(s)H(s) = pole lengths zero lengths

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 P 1, P 2, P 3, and P 4 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.

e.g. G(s) = K(s+3)(s+4) (s+1)(s+2), H(s) = 1

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: T (s) = = = KG(s)H(s) 1 + KG(s)H(s) K N G (s) N H (s) D G (s) D H (s) N H (s) D G (s) D H (s) 1 + K N G (s) KN G (s)n H (s) D G (s)d H (s) + KN G (s)n H (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 σ a and angle θ a. σ a = θ a = finite poles finite zeros #finite poles #finite zeros (2k + 1)π #finite poles #finite zeros where k is an integer and θ a 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:

We now compute the real-axis intercept and the angles of all asymptotes: σ a = finite poles finite zeros #finite poles #finite zeros = (0 1 2 4) ( 3) = 4/3 4 1 θ a = (2k + 1)π #finite poles #finite zeros = 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 θ a before the angles start to repeat.

The following is our complete RL sketch.