Outline. Time Delays in Control and Estimation (038806) lecture no. 4. Crossing frequencies and delays. Idea. arctan.! i i / C k, k 2 Z C

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1 Outline Time Delays in Control and Estimation (3886) e sh lecture no. 4 Leonid Mirkin Faculty of Mechanical Engineering Technion T dea Consider transfer function R.s/ s C s Crossing frequencies and delays Let h.s/ D P.s/ C Q.s/e sh : Substituting e sh! R.s/ and multiplying by C s we obtain polynomial As runs from to the frequency response R.j!/ at every! > covers the complete unit circle T in C i.e., jr.j!/j and arg R.j!/ runs continuously from to. This covers exactly the same area as e j!h as h changes from to. Moreover, for every h i and! i > there exists unique i 2 R such that R i.j! i / D e j! i h i and it is i D!i h i tan :! i 2 This suggests that we may replace e sh with R.s/ in any pure frequency-domain analysis where h sweeps large enough range..s/. C s/p.s/ C. s/q.s/: Obvious (yet important) fact is that for any h h.j! i / D iff 9 2 R such that.j! i / D Thus, j!-axis roots of h.s/ and corresponding delays obtained as follows:. find all 2 R for which.s/ has j!-axis roots 2. find corresponding frequencies! i 3. find positive crossing delays by h i;k D 2! i arctan.! i i / C k, k 2 Z C

2 j!-axis roots of.s/ Can be found from Routh s array (provided it is nonsingular): Then s n ~ n;./ ~ n;2./ ~ n;3./ s n ~ n ;./ ~ n ;2./ ~ n ;3./ : : : : s 2 ~ 2./ ~ 22./ s ~./ s ~./ required i are real zeros of ~./ such that ~ 2. i /~ 22. i / > crossing frequencies are! i D q ~ 22. i / ~ 2. i / Crossing frequencies and delays: example Consider P.s/ D s 2 C :s C and Q.s/ D :4. Then Routh s array.s/ D s 3 C. C :/s 2 C.: C :6/s C :4 s 3 : C :6 s 2 C : :4 s :6 2 7:9C.C:/ s :4 Then ~./ has two real zeros, at ;2 D 7:9p 6:, which produce crossing q :2 frequencies! ;2 D :5.9:9 p6:/ or, equivalently, i :28 3:39! i :76 :78 2! i arctan.! i i / D h i; :254 3:778 exactly as by delay sweeping. Crossing directions Denote by s h and s roots of h.s/ and.s/, respectively. t can be shown that ds dsh d.p.j!/p. j!/ Q.j!/Q. j!// sgn Re D sgn Re D sgn d dh d! (does not depend on h, ). This implies that root crossing directions can be determined from the root locus of.s/, which based on D s.p.s/ Q.s// µ P aux.s/: P.s/ C Q.s/ Crossing directions: example For P.s/ D s 2 C :s C and Q.s/ D :4, Then.5 Positive P aux.s/ D s.s2 C :s C :6/ s 2 C :s C :4.5 Negative t confirms that higher frequency is a switch and lower one is a reversal.

3 Pros and cons Pros: readily extends to commensurate delay case Outline Cons: increases order of polynomials in analysis some tricks not straightforward in singular cases multiple j!-axis roots! i h D C 2k, in which case i D (think of h.s/ D s 2 C s C C se sh ) dea Consider 2D polynomial dea 2 Consider 2D polynomial 2D.s; / P.s/ C Q.s/ Clearly, h.j! i / D ) 2D.j! i ; / D at D e j! i h mportant fact is that if for some! i 2D.j! i ; / has root at 2 T ) 9h such that h.j! i / D. 2D.s; / P.s/ C Q.s/ Clearly, h.j! i / D ) 2D.j! i ; / D at D e j! i h mportant fact is that if for some i 2 T 2D.s; i/ has root at s 2 jr ) 9h such that h.j! i / D. This suggests that we may look for! i 2 R C at which 2D.j! i ; / has unit circle roots. Toward this end, discrete stability criteria for complex polynomials (like Schur-Cohn, etc) may be used. This suggests that we may look for i 2 T at which 2D.s; i/ has imaginary axis roots s. Toward this end, continuous stability criteria for complex polynomials (like???) may be used.

4 Outline dea Consider h.s/ D det.s A A h e sh / We already saw that for any!, h.j!/ D pair.j! A ; A h / has eigenvalue in T Thus, to find j! roots of h.s/ we may check eigenvalues of.j! A ; A h / on some dense enough grid of! (in fact, absolute values of these eigenvalues). Application to delay-independent test Theorem (delay-independent stability) h.s/ D det.s A A h e sh / has no NC roots for all h 2 R C iff. A C A h is Hurwitz 2. A is Hurwitz 3..j! A / A h < for all! 2 R Proof (outline).,2 A C A h and A have to be Hurwitz for stability of.s/ and.s/ 3 if A is Hurwitz, j! A is nonsingular 8! and then h.j!/ D det j! A A h e j!h D det.j! A / det.j! A / A h e j!h : and condition is necessary (it holds at!! ) and sufficient for the absence of crossing frequencies ) delay independent. Outline

5 Problem Consider quasi-polynomial mx h;m.s/ P.s/ C Q i.s/e sih ; where, for simplicity, we assume that deg P.s/ > deg Q i.s/, 8i D ; : : : ; m. id Delay sweeping: example Let h;2.s/ D s C e sh C e s2h : Like in single-delay case, we look for j!-crossings of roots. As h;2.s/ real, its j!-axis roots coincide with those of h;2. s/, i.e., they satisfy both s C e sh C e s2h D and s C e sh C e s2h D From the first equation, e s2h D s e sh, then from the second equation: We want to find h for which all roots of h;m.s/ are in C n NC. D C e sh se s2h D C e sh C s.s C e sh / D C s 2 C. C s/e sh This is a single-delay quasi-polynomial with.!/ D.! 2 / 2. C! 2 / D! 2.! 2 3/ from which crossing! D p 3 (switch) and crossing h ; D As ;2.s/ stable, h;2.s/ stable iff h 2 Œ; 3 p 3 /. p Cj arg 3 p 2 3 D 3 p. 3 Delay sweeping: more general observations Let h;2.s/ D P.s/ C Q.s/e sh C Q 2.s/e s2h Clearly, s c D j! c is root of h;2.s/ iff it is root of e s2h h;2. s/ D Q 2. s/ C Q. s/e sh C P. s/e s2h Delay sweeping: more general observations (contd) f s c D j! c is crossing point of h;.s/ such that jp.j! c /j jq 2.j! c /j; then it also crossing point of h;2.s/. Hence, this s c D j! c is also root of h;.s/ P. s/ h;2.s/ Q 2.s/e s2h h;2. s/ D P.s/P. s/ Q 2.s/Q 2. s/ C Q.s/P. s/ Q 2.s/Q. s/ e sh which is single-delay quasi-polynomial. Yet converse not necessarily true: h;.s/ might have more j!-roots than h;2.s/, which complicates the analysis. Moreover, crosing directions at h;.j! c / and h;2.j! c / coincide iff jp.j! c /j > jq 2.j! c /j opposite iff jp.j! c /j < jq 2.j! c /j n fact, j!-roots of h;.s/ are roots of either h;2.s/ or P.s/P. s/ Q 2.s/Q 2. s/.

6 Bilinear transformation Extends to commensurate-delay case literally: h;m.s/!.s/. C s/ m P.s/ C mx. C s/ m i. s/ i Q i.s/; which has order deg P.s/ C m. With a little help of Routh s array we could routinely count j!-roots of.s/, etcetera (doable, yet not such a fun). id Outline Lyapunov s method for finite-dimensional systems Consider LT system Px.t/ D Ax.t/; x./ D x and assume there is (differentiable) V.x/ W R n 7! R C such that 2. V./ D 2. V.x/ > for all dx 3. derivative along trajectory dt called Lyapunov function. Then system is stable (in the sense of Lyapunov). f 3 replaced with 3. PV.x/ < system is asymptotically stable. Yet another way to end up with asymptotic stability is via LaSalle nvariance Principle 3. PV.x/ and PV.x/ implies x.t/. 2 n principle, statements like V.x/ > should be more specific, yet we proceed with this sloppiness to simplify exposition. Lyapunov s method for finite-dimensional systems (contd) Let s choose for some P >. Then V.x/ D x.t/p x.t/ PV.x/ D Px.t/P x.t/ C x.t/p Px.t/ D x.t/a P x.t/ C x.t/pax.t/ D x.t/.a P C PA/x.t/ f we can choose P > satisfying A P C PA D C C for some C, PV.x/ D x.t/c Cx.t/ ; which implies stability. For asymptotic stability we then need observability of.c; A/ as in that case Cx.t/ implies x.t/. n fact and P D 9P > such that A P C PA < A Hurwitz Z e A C C e A d > is observability Gramian of.c; A/.

7 Adding delays: developing intuition via discrete systems Consider NxŒk C D A NxŒk C A NxŒk As state vector here Nx a D Nx Œk Nx Œk Nx Œk h, quadratic Lyapunov function should look like 2 3 ƒ 2 3 PN PN PN h NxŒk NV. Nx a / D Nx Œk Nx Œk Nx Œk h N P PN PN h NxŒk 6 4 : : : :: : 5 4 : 5 PN h PN h PN hh NxŒk h hx hx D Nx Œk i PN ij NxŒk j id j D > h Adding delays: Lyapunov-Krasovskiĭ functional Consider Px.t/ D A x.t/ C A x.t h/ Quadratic Lyapunov function (actually, functional fœ h; 7! R n g 7! R C ) for this system could be of the form V.x / D Z h Z h x.t /P.; /x.t /dd for some function P.; /, where ; h, such that P.; / D P.; / and R h R h./p.; /./dd > for all./ 6. This functional called Lyapunov-Krasovskiĭ functional. Alternative expression: V.x / D Z t t h Z t t x./p.t ; t /x./dd h Adding delays: Lyapunov-Krasovskiĭ functional (contd) Choice regarded sufficiently general is P.; / D P ı./ı./ C P./ı./ C P./ı./ C P 2./ı. / C P 3. / for some matrix P, matrix functions P./ and P 2./ defined in 2 Œ; h, and matrix function P 3.t/ defined in t 2 Œ h; h. n this case V.x / D x.t/p x.t/ C C 2x.t/ Z h Z h Z h P./x.t the derivative of which is a mess 3... x.t /P 3. /x.t /dd /d C Z h x.t /P 2./x.t /d Preliminary: Leibniz integral rule d dt Z b.t/ a.t/ f.; t/d D Z db.t/ f.; t/d C f.b.t/; t/ dt da.t/ f.a.t/; t/ dt 3 t is possible to choose P.; / of this form by reverse engineering: choose observable measurement operator for state vector and construct P.; / as observability Gramian.

8 Delay-independent conditions via LK approach Consider Px.t/ D A x.t/ C A h x.t h/ and choose Z t V.x / D x.t/p x.t/ C x./p 2 x./d t h for some P > and P h >. Then, using Leibniz integral rule, PV.x / D Px.t/P x.t/ C x.t/p Px.t/ C x.t/p 2 x.t/ x.t h/p 2 x.t h/ D x.t/ x.t h/ A P C P A C P 2 P A h x.t/ A h P P 2 x.t h/ Thus, if there are P > and P 2 > such that A P C P A C P 2 P A h A h P < ; P 2 PV.x / and system asymptotically stable (mind LaSalle). This is Linear Matrix nequality (LM), which can be efficiently solved. Adding delays: Razumikhin approach Consider again Px.t/ D A x.t/ C A x.t h/ Lyapunov function for it, V.x /, doesn t have to be quadratic. We may take V.x / D max 2Œ h; QV.x.t C // for some Lyapunov function QV.x/. n this case P QV.x/ may be even positive at points where QV.x/ < V.x /. This, relaxed, condition reads as follows: Theorem (Razumikhin) System is asymptotically stable if there is V.x/ W R n 7! R C such that. V./ D 2. V.x/ > for all x 3. PV.x/ < whenever V.x.t// V.x.t C //, 2 Œ h;, for some > Delay-independent conditions via Razumikhin approach Consider Px.t/ D A x.t/ C A h x.t h/ and choose V.x/ D x.t/p x.t/ for some P >. For every > and > we can define function.t/ PV.x/ C V.x.t// V.x.t h// D Px.t/P x.t/ C x.t/p Px.t/ C x.t/p x.t/ x.t h/p x.t h/ D x.t/ x.t h/ A P C PA C P PA h x.t/ A h P P x.t h/ At x.t/ W V.x.t C // V.x.t// we have that PV. Thus, if <, the system is stable by Razumikhin arguments. Hence, if there is matrix P > and scalar > such that A P C PA C P PA h A h P P < ; Lyapunov-Krasovskiĭ vs. Razumikhin (not generic) f there are P > and > such that A P C PA C P PA h A h P P < ; then A P C P A C P 2 P A h A h P < for P P D P and P 2 D P : 2 Thus, the LK condition holds whenever so does the Razumikhin condition, but not necessarily vise versa. Hence, in this case Razumikhin approach is potentially more conservative (solvability of LK LM does not necessarily mean that P h D P ). then the system asymptotically stable.

9 Delay-independent stability in scalar case Consider Px.t/ D a x.t/ C a h x.t h/ for x 2 R. Delay-independent stability would require stability at h D and no positive crossing frequencies. These read a C a h < and! 2 C a 2 D a2 h or, equivalently, a 2 a2 h and a < (the latter can be interpreted as stability under h! ). has no positive real solutions Scalar case: Lyapunov-Krasovskiĭ and Razumikhin LK solvability LM becomes 2a p 9p > ; p 2 > such that C p 2 a h p < : a h p p 2 Taking Schur complement of the.2; 2/ term, the last condition equivalent to 2a p C p 2 C a 2 h p2 =p 2 < p 2 2 C 2a p p 2 C a 2 h p2 < : The latter reads a < and a 2 p2 a 2 h p2 a2 a2 h ; i.e., we recover exact conditions (LK conservative in general). n scalar case 2a p C p 2 a h p 2a p C p a < h p < a h p p 2 a h p p and Razumikhin result coincides with Lyapunov-Krasovskiĭ one. Outline Problem y P r.s/e sh u C.s/ e - r Given: P.s/ D P r.s/e sh for some rational P r.s/ and loop delay h > Design: a fixed-structure controller C.s/ (typically, P, P, or PD) Main tool: parametric search (brute force)

10 Example : P controller for unstable dead-time plant We need to know, y e u sh k s p C ki s for which k p and k i closed-loop system is stable for given h. n principle, we may consider P.s/ D yet it is a matter of normalization with a s a e sh ; a > ; h! ah and k i! k i a e - r Example : solution Characteristic quasi-polynomial From magnitude equality: with the only positive root h.s/ D s.s / C k p.s C k i /e sh :! 2 c D 2.!/ D! 4.k 2 p /! 2 k 2 p k2 i k 2 p q C.kp 2 / 2 C 4kp 2k2 i which is a switch. Hence, it is necessary to have stability at h D H) we assume that k p > and k i >. n this case the phase equality reads kp.k i C j! c /! c h D arg D arg k i C j! c j! c. C j! c /! c C j D arctan!2 c k i. C k i /! c : Example : solution (contd) Thus, the closed-loop system is stable iff k p >, k i >, and where! 2 c D 2 k 2 p h < arctan!2 c k i < ;! c. C k i /! c q C.kp 2 / 2 C 4kp 2k2 i. This expression is unsolvable in controller parameters k p and k i analytically. Even as k i! (P controller), expression for stabilizing h, h < unsolvable in k p analytically. q arctan kp 2 q < kp 2 Example : stability contours in k p k i plane k i :5 :5 h < : h < :2 h < :5 h < :3 :5 2 2:5 3 k p This is only tangible for 2 parameters...

11 Example 2: PD controller for unstable dead-time plant y e sh u e r k s p C k d s - Example 2: solution Characteristic quasi-polynomial: h.s/ D s C.k d s C k p /e sh We need to know, for which k p and k d closed-loop system is stable for given h. n principle, we may again consider P.s/ D yet it is a matter of normalization with a s a e sh ; a > ; h! ah and k d! k d a As a D k d, we must keep jk d j <. Now, with the only positive root.!/ D. k 2 d /!2.k 2 p /! 2 c D k2 p k 2 d which is a switch. Hence, delay-free system must be stable, which requires k p >. n this case the phase equality reads! c h D arg k p C j! c k d D arctan k d! c C arctan! c j! c k p Example 2: solution (contd) Thus, the closed-loop system is stable iff k p >, jk d j <, and h < arctan k d! c C arctan! c < 2;! c k p where! c 2 D.k2 p /=. k 2 /. This expression is again d unsolvable in controller parameters k p and k d analytically. Example 2: stability contours in k p k d plane :5 k d h < h < :75 h < :5 h < :25 :5 Yet as k p!, stability conditions read h < C k d : :5 2 2:5 3 3:5 k p This is again only tangible for 2 parameters...