Lecture 25: System Stability, Routh-Hurwitz Criterion

Massachusetts Institute of Technology Department of Mechanical Engineering 2.004 Dynamics and Control II Spring Term 2008 Lecture 251 Reading: • Nise: Chap. 6 1 System Stability g r a v ita tio n a l fo r c e a c ts to r e s to r e a d is p la c e d b a ll to its e q u ilib r iu m p o s itio n ( a ) a s ta b le s y s te m g r a v ita tio n a l fo r c e a c ts to m o v e a d is p la c e d b a ll a w a y fr o m its e q u ilib r iu m p o s itio n ( b ) a n u n s ta b le s y s te m n o fo rc e s a c t to m o v e a d is p la c e d b a ll ( c ) a n e u tr a lly s ta b le s y s te m The figure above illustrates three stability conditions using a rolling ball on an undulating surface as a graphic example. Assume that the horizontal position and velocity of the ball are a pair of state variables describing this system. In a concave region of the surface, as shown in in (a) the base of the hollow is an equilibrium point. If the ball is displaced a small distance from this position and released it oscillates but ultimately returns to its rest position at the base as it loses energy due to frictional losses; this is therefore a stable equilibrium point. (Without energy dissipation the ball would roll back and forth forever and exhibit neutral, or marginal, stability.) On a convex portion of the surface, as shown in (b), the ball is in equilibrium if placed exactly at the top of the surface, but if it is displaced an infinitesimal distance to either side the net gravitational force acting on the ball causes it to roll down the surface and never to return to the equilibrium point. This equilibrium point is therefore unstable. If the ball is displaced along the flat portion of the surface, as shown in (c) it neither moves away nor returns; the flat portion represents a neutrally stable equilibrium region. There are many formal definitions of stability of a system. • From the solution of the linear differential equation y(t) = yh (t) + yp (t) a system is defined to be asymptotically stable, if for all initial conditions, the steadystate homogeneous response returns to zero. If the initial condition response increases 1 c D.Rowell 2008 copyright � 25–1 without bound the system is said to be unstable, and if the steady-state response is constant or purely oscillatory the system is defined to be marginally stable. • From the forced response, a system is defined to be bounded-input bounded-output (BIBO) stable only if every bounded input produces a bounded output. These two definitions are not the same. A system G(s) = s2 1 + ωns has a pair of imaginary poles at s = ±jωn and will exhibit a purely oscillatory response from any finite initial conditions. On this basis the system is therefore marginally stable. On the other hand, if excited by an input sin(ωn t) the response amplitude will tend to infinity, therefore the system is BIBO unstable. From our previous work on linear system response, we have seen that if any system pole lies in the r.h. s-plane the system response will grow without bound as a result of any finite initial condition on the output of its derivatives. A fundamental definition of system stability is therefore A linear system is defined to be stable only if all of its poles have negative real parts. A system with one or more poles that are purely imaginary are defined to be marginally stable. The following figure shows a system with r.h. plane poles and the response components that result. jw s - p la n e X X s X Example 1 Determine the stability of the two systems G1 (s) = s3 + 2s2 15 + 5s + 15 and G2 (s) = 25–2 s3 + 2s2 5 . + 5s + 5 MATLAB’s roots() function was used to find the roots of the characteristic equation for each system, to determine the system poles: 1. For G1 (s): roots([1 2 5 15]) gave the system poles as s = −2.4357, s = 0.2268 ± j2.462. 2. For G2 (s): roots([1 2 5 5]) gave the system poles as s = −1.233, s = −0.384 ± j1.977. System G1 (s) has a pair of complex conjugate poles in the r.h. plane and is therefore unstable. System G2 (s) has all of its poles in the l.h. plane and is therefore stable. The task of determining system stability is therefore that of determining the roots of the characteristic equation D(s) = 0, or determining whether any root has a positive real part. Before the advent of numerical software packages, such as MATLAB, this was a very difficult problem. 2 The Routh-Hurwitz Criterion The Routh-Hurwitz criterion is a method for determining the presence the number of roots of a polynomial with positive real parts. • It is based on the coefficients of the polynomial. • It involves placing the coefficients into a 2-D Routh array, and then systematically extending the array by creating new rows. When the array is complete, the number of r.h. plane roots can be found by inspection. The method has two basic steps: Step 1: A preliminary step is the following Given a polynomial, a sufficient condition for the existence of at least one r.h. plane root is that there is a sign change in the coefficients. For example the system G(s) = s5 + 47s4 27 − 23s3 + 19s2 + 6s + 9 is unstable because there is a sign change in the coefficients of D(s). Put another way, for a system to be stable there must be no sign changes in the coefficients of the denominator polynomial D(s). Step 2: The Routh array is created and examined, as described below. 25–3 2.1 The Routh Array Let the denominator polynomial be D(s) = an sn + an−1 sn−1 + . . . + a1 s + a0 the initial step is to place the coefficients in the first two rows of the array: sn sn−1 an an−1 an−2 an−3 an−4 an−5 ... ... Additional rows are then created, based on the two rows above. For example, let the next row be labelled sn−2 and sn−3 with entries bi and ci : sn sn−1 sn−2 sn−3 an an−1 b1 c1 an−2 an−3 b2 c2 an−4 an−5 b3 c3 ... ... ... ... • Each row is constructed from left to right, until all further entries are zero. • Each entry is based on entries in the two rows above, using values from (i) the first (leftmost) column, and the column to its right. • The entries us a 2 × 2 determinant. In the above example the bi are computed from the rows labelled sn and sn−1 : � � � � � 1 �� an 1 �� an 1 �� an an−2 �� an−4 �� an−6 b1 = − , b2 = − , b3 = − � � � � � an−1 an−1 an−3 an−1 an−1 an−5 an−1 an−1 an−7 and the ci are computed � 1 �� an−1 an−3 c1 = � b2 b1 b1 from the rows labelled sn−1 and sn−2 : � � � � � an−1 an−5 � 1 � , c2 = − � � , c3 = − 1 � � � b3 b1 b1 b1 � � an−1 an−7 � � b1 b4 � � �, � � � �, � ... ... The process is repeated until the row that will be labelled s0 . Note that in each determinant the left column is taken from the left column of the Routh array, the second column is taken from the column to the right in the Routh array. When the table is complete, the number of r.h. plane roots is the number of sign changes in the first column. Example 2 Find the number of r.h. plane poles in the system G(s) = s3 + s2 25–4 37 + 2s + 24 The first two rows of the Routh array are taken directly from D(s): s3 s2 1 2 0 1 24 0 where an additional zero has been added at the end of each row. Then construct the s1 row: � � � � 1 �� an 1 �� 1 2 �� an−2 �� b1 = − =− � = −22 an−1 � an−1 an−3 � 1 1 24 � � � � � 1 �� an 1 �� 1 0 �� an−4 �� b2 = − =− � =0 an−1 � an−1 an−5 � 1 1 0 � and the Routh array, with the new row, becomes s3 s2 s1 1 2 0 1 24 0 -22 0 0 The s0 row is now computed: � � � � 1 �� 1 24 �� 1 �� an−1 an−3 �� c1 = − � � = − 22 � −22 0 � = 24 b2 b1 b1 � � � � 1 �� an − 1 an−3 �� 1 �� 1 0 �� c2 = − � � = − 22 � −22 0 � = 0 b3 b1 b1 and the completed Routh array is s3 s2 s1 s0 1 1 -22 24 2 0 24 0 0 0 The next step is to examine the first column; in this case we note there are two sign changes 1 → −22 → 24, and we conclude that there are therefore two unstable r.h. plane poles in this system. √ √ In fact, for this system D(s) = (s + 3)(s − 1 + j 7)(s − 1 − j 7). Example 3 Use the Routh-Hurwitz method to develop a simple test for the stability of a third-order system. Let the denominator polynomial be D(s) = a3 s3 + a2 s2 + a1 s + a0 25–5 and assume that all coefficients are positive. The complete Routh array is s3 s2 s1 s0 a3 a2 −(a3 a0 − a2 a1 )/a1 a0 a1 a0 0 0 0 0 and to ensure stability we require all terms in the first column to be positive, that is a1 a2 > a0 a3 In other words a third-order system will be stable if the product of the inner coefficients is greater than the product of the outer coefficients. Note: The Routh-Hurwitz method has some special cases to deal with situations when zeros appear in the first column. See Nise, Chapter 6, for details. 25–6