# EE Control Systems LECTURE 9

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1 Updated: Sunday, February, 999 EE - Control Systems LECTURE 9 Copyright FL Lewis 998 All rights reserved STABILITY OF LINEAR SYSTEMS We discuss the stability of input/output systems and of state-space systems The Routh Test is given for finding the number of right-half plane roots of a prescribed polynomial Stability of Input/Output Systems Input/output systems may be described by a transfer function H(s) or equivalently by an impulse response h(t) The system is said to be: Figure Asymptotically Stable (AS) if u(t)= 0 for all time t implies that y(t) goes to zero with time Marginally Stable (MS) if u(t)= 0 for all time t implies that y(t) is bounded for all time In terms of time signals, one could say that a decaying exponential is AS, a sine wave is MS, the unit step is MS, and an increasing exponential is unstable A system is AS if and only if (iff) the impulse response goes to zero with time, and MS iff the impulse response is bounded The natural modes that occur in h(t) depend on the locations of the poles, defined as the roots of the denominator of H(s) Thus, the system is AS iff all poles are in the open left-half plane (OLHP) (eg strictly in the LHP with none on the jω-axis) The system is MS iff all poles are in the left-half plane (eg

2 they may be in the OLHP or on the jω-axis), with any poles occurring on the jω-axis nonrepeated Examples: The unit step has transform /s, which has one pole at s=0 and is MS The unit ramp has transform /s, which has two poles at s=0 and is unstable β The transfer function H ( s) = has poles at s = α ± jβ and is AS Its ( s + α) + β impulse response is e t α sin βt, an exponentially decaying function The transfer function H ( s) = s has poles on the jω-axis at s = ± jβ and is MS s + β Its impulse response is h( t) = cos βt, a bounded function s The transfer function H ( s) = has repeated poles at s = ± jβ and so is ( s + β ) unstable Its impulse response is h( t) = t cos βt, an unbounded function Routh Test Given a polynomial p(s), the number of poles in the right-half plane may be determined without finding the roots by using the Routh test Given L s + a, the Routh Array is given by n n = a0s + as + + an n s n a 0 a a a 6 s n- a a a 5 a 7 s n- b b b : : : : s 0 The third row and below are each computed from the two rows immediately preceding it by using the matrix determinantal equations, eg for the third row, a0 a a0 a a0 a6 a a a a5 a a7 b =, b =, b a a = a

3 Element a is called the pivot element for row two To simplify computations by avoiding fractions, one can at any point multiply any row by a positive constant before proceeding to the next row Routh Theorem The number of roots of p(s) in the right-half plane equals the number of sign changes in column one To examine the input/output stability of a system, one applies the Routh test to the characteristic polynomial, the denominator of the transfer function The Routh test was extremely useful for analyzing system stability in the days before calculators and computers Finding the roots of high-order polynomials by hand is very difficult Example Let a system have characteristic polynomial = s + s + 6s + 0s + 7 How many roots are in the right-half plane? The Routh array is manufactured as s 6 7 s 0 s 7 s - s 0 7 There are two sign changes in column one, so the polynomial has two unstable roots Example Let 7 = s s How many roots are in the right-half plane? The Routh array is manufactured as s s 7 s 0

4 There is one sign change in column one, so the polynomial has one unstable root Note that, for stability, a second-order polynomial must have all signs the same Example - Stability as a Function of a Parameter A very important application of the Routh test is determining the stability of a closed-loop system as a function of a variable parameter, which could be the feedback gain In the figure, determine for what values of feedback gain K the system is stable u(t) /s /s /s y(t) K To solve this problem one simply determines the characteristic polynomial and then constructs the Routh Array Using Mason's Formula, one finds that the determinant is K s + s + s + K ( s) = = s s s s The characteristic polynomial (denominator of the transfer function) is the top portion of this, so that = s + s + s + K The Routh Array is constructed to be s s K s 6 K s 0 K This has no sign changes in column one if 6 K > 0 K > 0 Thus, for stability one requires

5 0 < K < 6 Note that too much gain in this problem will destabilize the system, evidently making the poles unstable Routh Test Problem : Two Successive Rows Proportional The Routh test can have two problems: Successive rows proportional Pivot element equal to zero (eg a zero in column one) If two successive rows are proportional then, using the determinants, one will get the next row of the Routh Array equal to zero In this case, one must take the second, linearly dependent, row out of the Routh Table, differentiate it, and put it back to obtain the row for the next lower power of s Example - Two Successive Rows Proportional Let there be prescribed = s + s + 5s + 5s + 6s + 9s + 8s + One computes the first part of the Routh Array as: s s s (This row was multiplied by for simplicity) s s Now all progress stops as one has obtained a zero row To remedy the problem and finish the Routh Array, take out the polynomial corresponding to s, which is = s + s + Note the way one writes every other power of s to fabricate the polynomial Now, differentiate this to obtain = s s + This is used as the row for s in the array, since the original s row came out as zero Proceeding to compute the rest of the Routh Array, one obtains: 5

6 s s s (This row was multiplied by for simplicity) s Take row out and differentiate s Put s row back s s - s There are two signs in column one, corresponding to two unstable roots This problem of two successive rows proportional occurs when the second row corresponds to an exact divisor of p(s) In this example, the polynomial = s + s exactly divides p(s) In fact, all the sign changes in the first column + below the s row refer to the roots of ( s p ) Thus, the polynomial s + s + has two roots in the right-half plane Routh Test Problem : Zero in Column One If the pivot element of a row is equal to zero, one cannot divide by it to find the next row In this event, one places a small positive nonzero number as the pivot element, represented by a Greek symbol such as ε, and then proceeds One can have several of the Greek letters in the same Routh Array if the problem occurs more than once in the same array Example 5- Zero Pivot Element 6 5 Let there be prescribed = s + s + s + 6s + s + 6s + One computes the first part of the Routh Array as: s 6 s s 0 One now has a zero in column one (but not an entire zero row, which occurs when two rows are proportional) Place ε in column one, representing a small positive constant, and proceed to compute the Routh Array In doing this, one may neglect terms in ε and higher, as they are very small One also neglects terms in ε when added to finite 6

7 numbers Finally, one makes use of the ability to multiply an entire row by a positive constant to simplify the computations s 6 s s ε s 6ε 6ε 9 9 This row was multiplied by ε s s ξ Zero pivot: Place a small positive const ξ in col one s 0 This row was multiplied by ξ Note that in the s row one again encounters a zero pivot element, so it is necessary to put another small positive variable into the table; here ξ was used Since the variables are assumed positive, there are two sign changes in the first column and this polynomial has two unstable roots Stability of State Variable Systems The linear time invariant (LTI) state-space form is x& = Ax + Bu, x(0) y = Cx + Du The transfer function is given by H ( s) = C( si A) B + D The denominator of this is the characteristic polynomial ( s) = si A The system poles are the roots of the characteristic equation ( s ) = si A = 0 A state-space formulation allows one to get more information about the system than the input/output formulation, which is described only by a transfer function Specifically, if A, B, C, D are known, then the internal states x(t) can be computed in addition to the input u(t) and output y(t) Thus, one is effectively able to look inside the black box in Fig The (internal) system poles (eg roots of (s)) should be distinguished from the (input/output) poles of the transfer function There may be some pole/zero cancellation in computing the transfer function Then the denominator of H(s) is not the same as (s), 7

8 since some system poles do not occur in H(s) Therefore, the definitions of stability need to be modified to differentiate internal stability from the external (eg input/output stability) just defined The system is said to be: (Internally) Asymptotically Stable (AS) if u(t)= 0 for all time t implies that x(t) goes to zero with time for all initial conditions x(0) Bounded-Input/Bounded-Output Stable (BIBOS) if u(t) bounded for all time t implies that y(t) is bounded for all time when x(0)=0 Though the first of these is usually simply called AS, it is not the same definition as the AS defined for input/output systems, where only the output y(t) is required to go to zero Here, we require all the internal states to go to zero AS as defined here is concerned with the effects of the initial state x(0) on y(t), while BIBOS is concerned with the throughput effects of u(t) on y(t) The system is (internally) AS iff all system natural modes decay with time This occurs iff all the system poles are in the open-left half plane For AS all the roots of (s) must be in the OLHP To find a condition for BIBOS, recall that when x(0)= 0 the output may be found by convolving the impulse response h(t) with the input, t y( t) = h( t τ ) u( τ ) dτ 0 If the impulse response is a decaying exponential, then the output is bounded for all bounded inputs On the other hand, suppose h(t) is the unit step for instance Convolution of any u(t) with the unit step simply gives the area under u(t) Thus, if u(t) is, eg, also the unit step, then the output would be the unit ramp, which is not bounded It turns out that the output y(t) is bounded for all bounded u(t) (when x(0)=0) if and only if the impulse response goes to zero with time Therefore, the system is BIBOS iff the poles of the transfer function are in the OLHP To study the AS of a system, one may apply the Routh test to the characteristic polynomial (s) To study the BIBOS of a system, one may apply the Routh test to the denominator of the transfer function after pole/zero cancellation, if any 8

9 Note: AS requires ALL the system poles to be in the OLHP BIBOS only requires the poles remaining in the transfer function, after pole/zero cancellation, to be in the OLHP Therefore, it is clear that AS implies BIBOS A single-input/single-output (SISO) system is said to be minimal if there is no pole/zero cancellation In this event, the poles of H(s) are the same as the roots of (s), so that the system is AS if and only if it is BIBOS That is, for minimal systems, BIBOS also implies AS Example 6 0 s Let x& = Ax = x Then ( s ) = = s + s + = ( s + ) so that s + t t both poles are at s= - Therefore the system is AS The natural modes are e, te Example Let x& = Ax + Bu = x + u, y = Cx = [ ]x 0 a The characteristic polynomial is s ( s ) = = s = ( s + )( s ) The poles are at s=-, s=, so the system is s not AS It is unstable The natural modes are b The transfer function is s H ( s) = C( si A) B = =, ( s )( s + ) s + which has poles at s=- Therefore, the system is BIBOS Note that the unstable pole at s= has cancelled with a zero at s= t e, e t 9

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