16.06 Principles of Automatic Control, Lecture 11

16.06 Principles of Automatic Control Lecture 11 The Root Locus Method Often, it is useful to find how the closed-loop poles of a system change as a single parameter is varied. To do this, we use the root locus method. Root - root of s polynomial equation Locus - Set of points (plural - loci) Consider a typical feedback loop r + G(s) K(s) y If both Kpsq and Gpsq are rational, then the loop gain may be expressed as KpsqGpsq “ kLpsq where npsq dpsq npsq “sm ` b1 sm´1 ` ... ` bm “ps ´ z1 qps ´ z2 q...ps ´ zm q m ź “ ps ´ zi q Lpsq “ i“1 n dpsq “s ` a1 sn´1 ` ... ` an n ź “ ps ´ pi q i“1 1 Then the roots of the closed-loop system occur at: 1 ` KpsqGpsq “ 0 p‹q or 1 ` kLpsq “ 0 p‹q or Lpsq “ ´ 1 k p‹q or dpsq ` knpsq “ 0 p‹q The root locus is the set of values s for which p‹q holds, and k is any positive real value. (For reasons that will become clear later, this is the definition of the positive or 180 degree locus. Will later define the negative, or 0 degree locus.) Example: + 1 s(s+1) k - In this case, 1 , npsq “ 1 sps ` 1q dpsq “sps ` 1q “ s2 ` s zeros: none poles: pi “ 0, ´1. Lpsq “ The characteristic equation is: s2 ` s ` k “ 0 2 Because characteristic equation is quadratic, we can find the roots using the quadratic for­ mula: 1 s“´ ˘ 2 ? 1 ´ 4k 2 When 0 ď k ď 14 , the roots are real, and between -1 and 0. For k ą 14 , the roots ? are complex, 1 with real part ´ 2 , and imaginary part that increases (asymptotically) as k. Im(s) θ Re(s) Suppose our goal is to choose k so that ζ “ sin θ “ 0.5 ñ θ “ 30˝ . Looking at the geometry in the figure, the imaginary part is ´Repsq tan θ 1 Rpsq “ ´ 2 1{2 1 sin 30˝ tan θ “ “? “? ˝ cos 30 3{2 3 ? ñ Impsq “ 3{2 ? 4k ´ 1 But Impsq “ 2 6 k “1 Impsq “ Example: What is the root locus of 3 + - 1 (s+1)(s+2) k s+3 s+8 In this problem, Lpsq “ s`3 npsq “ ps ` 8qps ` 1qps ` 2q dpsq The characteristic equation is ps ` 8qps ` 1qps ` 2q ` kps ` 3q “ 0 Because the polynomial is cubic, we can’t find the roots (easily) in closed form. Nevertheless, can sketch the root loci using root loci sketching rules: Im(s) Re(s) With a little practice, you should be able to sketch root loci very rapidly. Guidelines for Sketching Root Locus Will give rules for k ą 0. For k ą 0 and 1 ` kLpsq “ 0 must have that Lpsq “ ´ 1 “ negative real number k That is, the phase of the L(s) must be: 4 =Lpsq “ 180˝ ` l ¨ 360˝ , where l is an integer. This is the root locus phase condition, and the reason we call the locus for k ą 0 the 180˝ locus. Consider the example above: Im(s) φ1 φ3 Re(s) ψ2 φ2 The phase of L(s) is given by =Lpsq “ Ψ1 ´ φ1 ´ φ2 ´ φ3 To see if a given point is on the locus, could measure all the angles, add/subtract, and test result. This used to be done mechanically with a “spirule”. However, it’s only important to be able to sketch general shapes; Matlab can do the rest. Root Locus Rules Rule 1 The n branches of the locus start at the n points of L(s). m branches end at the zeros of L(s). n ´ m branches end at s “ 8. Rule 2 The loci cover the real axis to the left of an odd number of poles and zeros. To the left of the pole, φ “ 180˝ To the left of a zero, Ψ “ 180˝ . 5