Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19IllustrationsThe issue of ensuring the stability of a closed-loop feedback system is central to control system design. Knowing that an unstable closed-loop system is generally of no practical value, we seek methods to help us analyze and design stable systems. A stable system should exhibit a bounded output if the corresponding input is bounded. This is known as bounded-input, bounded-output stability and is one of the main topics of this chapter. The stability of a feedback system is directly related to the location of the roots of the characteristic equation of the system transfer function. The Routh–Hurwitz method is introduced as a useful tool for assessing system stability. The technique allows us to compute the number of roots of the characteristic equation in the right half-plane without actually computing the values of the roots. Thus we can determine stability without the added computational burden of determining characteristic root locations. This gives us a design method for determining values of certain system parameters that will lead to closed-loop stability. For stable systems we will introduce the notion of relative stability, which allows us to characterize the degree of stability. Chapter 6 – The Stability of Linear Feedback SystemsIllustrationsThe Concept of StabilityA stable system is a dynamic system with a bounded response to a bounded input.Absolute stability is a stable/not stable characterization for a closed-loop feedback system. Given that a system is stable we can further characterize the degree of stability, or the relative stability.IllustrationsThe Concept of StabilityThe concept of stability can be illustrated by a cone placed on a plane horizontal surface.A necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts.A system is considered marginally stable if only certain bounded inputs will result in a bounded output.IllustrationsThe Routh-Hurwitz Stability CriterionIt was discovered that all coefficients of the characteristic polynomial must have the same sign and non-zero if all the roots are in the left-hand plane.These requirements are necessary but not sufficient. If the above requirements are not met, it is known that the system is unstable. But, if the requirements are met, we still must investigate the system further to determine the stability of the system.The Routh-Hurwitz criterion is a necessary and sufficient criterion for the stability of linear systems.IllustrationsThe Routh-Hurwitz Stability Criterion 31311131421331211321110531353125311420122111110------------nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnbbaabcaaaaabaaaaaaaaaabhscccsbbbsaaasaaasasasasasa Characteristic equation, q(s)Routh arrayThe Routh-Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array.IllustrationsThe Routh-Hurwitz Stability CriterionCase One: No element in the first column is zero.Example 6.1 Second-order systemThe Characteristic polynomial of a second-order system is:q s( ) a2s2 a1s a0The Routh array is w ritten as:w here:b1a1a0 0( ) a2a1a0Therefore the requirement f or a stable second-order system is simply that all coef ficients be positive or all the coefficients be negative.001011022bsasaasIllustrationsThe Routh-Hurwitz Stability CriterionCase Two: Zeros in the first column while some elements of the row containing a zero in the first column are nonzero.If only one element in the array is zero, it may be replaced w ith a small positiv e number that is allow ed to approach zero after completing the array.q s( ) s52s4 2s3 4s2 11s 10The Routh array is then:w here:b12 2 1 420 c14 2 612d16 c1 10c16There are tw o sign changes in the first column due to the large negative number calculated for c1. Thus, the system is unstable because tw o roots lie in the right half of the plane. 0010000100610421121011121345sdscsbsssIllustrationsThe Routh-Hurwitz Stability CriterionCase Three: Zeros in the first column, and the other elements of the row containing the zero are also zero.This case occurs w hen the polynomial q(s) has zeros located sy metrically about the origin of the s-plane, such as (s+)(s -) or (s+j)(s-j). This c ase is solved using the auxiliary polynomial, U(s), w hich is located in the row above the row containing the zero entry in the Routh array.q s( ) s32 s2 4s KRouth array:For a stable system w e require that0 s 8For the marginally stable case, K=8, the s^1 row of the Routh array contains all zeros. The auxiliary plynomial comes from the s^2 row . U s( ) 2s2Ks0 2 s2 8 2 s24 2 s j 2( ) s j 2( )It c an be proven that U(s) is a f actor of the characteris tic polynomial:q s( )U s( )s 22Thus, w hen K=8, the factors of the characteristic polynomial are:q s( ) s 2( ) s j 2( ) s j 2( )00241028123KssKssKIllustrationsThe Routh-Hurwitz Stability CriterionCase Four: Repeated roots of the characteristic equation on the jw-axis.With simple roots on the jw-axis, the system will have a marginally stable behavior. This is not the case if the roots are repeated. Repeated roots on the jw-axis will cause the system to be unstable. Unfortunately, the routh-array will fail to reveal this instability.IllustrationsExample 6.4IllustrationsExample 6.5 Welding controlUsing block diagram reduction we find that:The Routh array is then:KascsKabsKsKas0313234)6(6111For the system to be stable bothb3and c3must be positive.Using these equations a relationship can be determined for K and a .where: b360 K6and c3b3K 6( ) 6 Kab3IllustrationsThe Relative Stability of Feedback Control SystemsIt is often necessary to know the relative damping of each root to the characteristic equation. Relative system stability can be measured by observing the relative real part of each root. In this diagram r2 is relatively more stable than the pair of roots labeled r1.One method of
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