MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.g r a v i t a t i o n a l f o r c e a c t s t om o v e a d i s p l a c e d b a l l a w a yf r o m i t s e q u i l i b r i u m p o s i t i o n( b ) a n u n s t a b l e s y s t e m( a ) a s t a b l e s y s t e m( c ) a n e u t r a l l y s t a b l e s y s t e mn o f o r c e s a c t t o m o v e ad i s p l a c e d b a l lg r a v i t a t i o n a l f o r c e a c t s t or e s t o r e a d i s p l a c e d b a l l t o i t s e q u i l i b r i u m p o s i t i o n� 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 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 p oint. 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 steady-state homogeneous response returns to zero. If the initial condition response increases 1copyright cD.Rowell 2008 25–1XXXsjws - p l a n ewithout 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 1 G(s) = s2 + ωs n 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(ωnt) 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. Example 1 Determine the stability of the two systems 15 5 G1(s) = and G2(s) = . s3 + 2s2 + 5s + 15 s3 + 2s2 + 5s + 5 25–2MATLAB’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 27 G(s) = s5 + 47s4 − 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. 25–32.1 The Routh Array Let the denominator polynomial be nD(s) = ans + an−1s n−1 + . . . + a1s + a0 the initial step is to place the coefficients in the first two rows of the array: sn an an−2 an−4 . . . sn−1 an−1 an−3 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 an an−2 an−4 . . . …
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