Stabilit of Linear Control Sstem Concept of Stabilit Closed-loop feedback sstem is either stable or nstable. This tpe of characterization is referred to as absolte stabilit. Given that the sstem is stable, the degree of stabilit of the sstem is referred to as relative stabilit. A stable sstem is defined as a sstem with bonded response to a bonded inpt. Bonded inpt signal sstem Otpt signal implse step Bonded otpt signal - sstem stable Otpt signal nbonded nstable sstem Consider the concept of stabilit for cones shown below.
Example 3.1 Obtain the time response of the sstem shown below for a nit step inpt U(s) = 1/s (an example of a bonded inpt). Determine whether the sstem is stable or nstable. U(s) 1 s + 1 Y(s) Example 3.2 Obtain the time response of the sstem shown below for a nit step inpt U(s) = 1/s (an example of a bonded inpt). Determine whether the sstem is stable or nstable. U(s) 1 s 1 Y(s)
Stabilit of a Closed-loop sstem When a closed-loop is designed, the problem of stabilit ma arise if the controller is not properl designed. A stable open-loop sstem ma become an nstable closed-loop sstem. In some cases, an nstable open-loop sstem can be stabilized sing a feed-back control sstem. Consider a closed-loop sstem shown below. + e G(s) - H(s) The transfer fnction is G = 1 + GH The eqation 1 + GH = 0 is known as the characteristic eqation. In general, the transfer fnction of a closed loop sstem can be written as Ns () = = Ds () m m 1 b s + b s + L+ b m m 1 n n 1 s + a s + L+ a n 1 0 0, n > m where Ds ()= 0 is the characteristic eqation. This transfer fnction can be written in pole-zero configration as ( s + z1)( s + z2 ) L( s + zm ) = ( s + p )( s + p ) L( s + p ) 1 2 n where the poles are the roots of the denominator of the transfer fnction and zeros are the roots of the nmerator of the transfer fnction.
The response of this sstem to a nit implse inpt U(s) = 1 can be obtained as The response is bonded if the poles are negative. Stabilit of the sstem is determined b the poles onl. Ths, the sfficient condition for stabilit of a feedback control sstem is all poles of the closed loop transfer fnction mst have a negative real vales. Stabilit region on the s-plane is shown below. The sstem is stable if all the poles are located on the left hand side of the imaginar axis.
Example 3.3 Determine the stabilit of a sstem with a characteristic eqation q(s) = s 3 + 4s 2 + 6s + 4 Example 3.4 For a marginall stable open-loop sstem shown below, determine the stabilit of a closed-loop sstem if a proportional controller gain K p = 24 is sed. m 1 s( s + 1)( s + 2)
Roth Stabilit Criterion The Roth Stabilit Criterion is a method which can determine the existence of positive poles. This criterion is sfficient if the designer onl wish to determine the range of control parameter that will ensre closed loop stabilit. Consider a closed loop sstem with the characteristic eqation n n 1 q( s) = a s + a s + + a0 n n 1 L = 0 The stabilit of this sstem can be tested b constrcting the Roth table as shown below. This table is filled horizontall and verticall ntil the remaining elements are zeros. The characteristic eqation has all negative roots if the signs of all elements in the first colmn are the same. The nmber of positive roots is eqal to the nmber of the signs change.
Example 3.5 Determine the stabilit of a sstem with a characteristic eqation 4 3 2 q ( s) = s + 5s + 20s + 40s + 50 Example 3.6 Determine the stabilit of a closed-loop sstem shown below. + 24 3 2 s + s + 2s -
Example 3.7 Determine the stabilit of a sstem with a characteristic eqation 5 4 3 2 q ( s) = s + 2s + 2s + 4s + 11s + 10 Example 3.8 + K s( s + 3)( s + 10) - Determine the range of K which ensre the closed loop sstem to be stable.
The Root Locs Method The Root Locs Concept The relative stabilit and the transient performance of a closed-loop control sstem are directl related to the location of the closed-loop roots of the characteristic eqation in the s-plane. It is freqentl necessar to adjst one or more parameters in order to obtain sitable root locations. The root locs is the path of the roots of the characteristic eqation traced ot in the s- plane as a sstem parameter is changed. Consider a feedback control sstem shown below. + e KG(s) - H(s) The closed-loop transfer fnction, the open-loop transfer fnction, and the characteristic eqation can be written as:
It can be seen that the vales of the roots of the characteristic eqation will change if the vale of the parameter K is changed. When K = 0 the locs starts at the poles of the open-loop transfer fnction and the locs ends at the zeros of the open-loop transfer fnction when K =. Magnitde and Angle Criteria Ever point on the locs mst satisf the magnitde and angle criteria and can be formall written as:
Example 3.9 Draw the locs of the roots of the characteristic eqation of the control sstem shown below when K varies from 0 to. + K s( s + 2) -
Example3.10 For a control sstem shown below, show that the root locs starts at the poles of the open-loop transfer fnction when K = 0 and it ends at the zeros of the open-loop transfer fnction when K =. + K s( s + 2) -
Example 3.11 Using the magnitde and angle criteria, verif that s 1 = -1 + j is one of the roots of the characteristic eqation of the control sstem shown below when K = 4 + K s( s 2 + 4s + 6) -
The Root Locs Procedre The Root Locs Procedre 1. Locate the poles and zeros of the open-loop transfer fnction and plot them sing x for poles and o for zeros. 2. The root locs on the real axis alwas lies in a section of the real axis to the left of an odd nmber of poles and zeros. 3. The loci begin at the poles and end at zeros or zeros at infinit,, along asmptotes. 4. The nmber of asmptotes is eqal to the nmber of poles mins the nmber of zeros, p z. The direction of the asmptotes is define b the asmptote angle φ, where 0 (2r + 1)180 φ = with r = 0, 1, 2, p z 5. All asmptotes intersect the real axis at a single point σ A, often called the asmptote centroid, defined b σ A where pi p z = z i p i is the vales of the poles z i is the vales of the zeros 6. Points of breakawa from or arrival at the real axis ma also exist and can be obtained b rearranging the characteristic eqation to isolate the mltipling factor K in the form of K = f(s) On the real axis, the breakawa point happens when K is maximm, i.e. dk ds = 0
7. The loci are smmetrical abot the real axis. The loci approach or leaving the real axis at an angle of ± 90 o. 8. The angle of departre, θ d, form a complex pole is obtained from the angle criterion θ d = 180 0 - Σθ p + Σθ z where θ p and θ z are the angles from other poles and zeros to the pole in qestion. 9. The intersection of the locs with the imaginar axis is obtained sing the Roth stabilit criterion.
Example 3.12 Draw the root locs for a control sstem with an open-loop transfer fnction given as KGH ( s) = K s( s + 2)( s + 6) Determine the nmber of asmptotes, the asmptote angles, and the asmptote centroid.
Example 3.13 Draw the root locs for a control sstem with an open-loop transfer fnction given as K KGH ( s) = ( s + 2)( s + 4) Determine the breakawa point.
Example 3.14 Draw the root locs for a control sstem with an open-loop transfer fnction given as K( s + 2) KGH ( s) = 2 ( s + 2s + 4) Determine the breakawa point.
Example 3.15 Draw the root locs for a control sstem with an open-loop transfer fnction given as K( s + 2) KGH ( s) = 2 s( s + 2s + 2) Determine the departre angle from the complex poles.
Example 3.16 Draw the root locs for a control sstem with an open-loop transfer fnction given as KGH s) = s( s ( 2 K + 6s + 25) Determine the intersection of the locs with the imaginar axis.
Interpretation of Root Locs Interpretation of Root Locs and controller design When the design of a feedback control sstem is ndertaken, the controller mst be chosen sch that the closed-loop sstem is stable and its transient response is satisfactor and all specifications are satisfied. For a stable sstem, all roots mst lie on the left-hand side of the imaginar axis. Meanwhile, the transient response of the closed-loop sstem is determined b the damping ratio, ξ, natral freqenc, ω n, damped natral freqenc, ω d, and time constant, T. The vales of these parameters can be estimated from the location of the dominant roots from the root locs in the s-plane.
Example 3.17 A control sstem is represented b a block diagram shown below. + K 1 2 ( s + 1)( s + 4s + 8) - a) Draw the root locs when K increases from zero. b) It is given that when K = 10, the roots of the characteristic eqation are 3 dan 1 ± j2.24. Determined the damping ratio, natral freqenc, damped natral freqenc, and time constant, T for K = 10.
Example 3.18 For a control sstem shown below, chose the vale of the controller K sch that the maximm vale of the damping ratio is 0.5 and the minimm vale of the time constant is 1s. + K 5 s(0.25s + 1)(0.1s + 1) -