Exercise 1 (MC Question)

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1 D J Tani -9- Contol Systems II (Sping 8 Solution Execise Set 9 Linea Quadatic Regulato Gioele Zadini, gzadini@ethzch, 7th May 8 Execise (MC Question (I-D, (II-A, (III-C, (IV-B (I-C, (II-A, (III-D, (IV-B (I-A, (II-B, (III-C, (IV-D (I-B, (II-A, (III-C, (IV-D Explanation: Case (III is the only case whee thee exists a cost on x (t Since x (t epesents ẋ (t, this means that the contolle will ty to avoid apid vaiations (deivative in x (t The only plot without oveshoot is plot D, ie the paiing is (III-D The weight on the contol signal (Q defines the speed of the system A low weight on the contol signals esults in a fast esponse: we cae less if we ae using moe contol signal (eg you don t cae how much oil you bun Based on that, one can pai the othe plots as (I-C (II-A (IV-B

2 Execise (LQR a Using quadatic foms, one can identify Q and R to be (using the null matix fo N 7 Q ( and R 4 ( The state-space desciption of the system can be e-witten in standad fom as (ẋ (t x (t + u(t ẋ (t x (t }{{}}{{} B y(t A 7 4 }{{} C x (t x (t + }{{} D u(t In ode to find the contolle K, one has to compute the symmetic, positive definite solution of the Riccati equation elated to this poblem Fist, one has to look at the fom that this solution should have Hee B R This means that since Φ Φ we ae dealing with Φ R of the fom ϕ ϕ Φ (4 ϕ ϕ With the Riccati equation, it holds Φ (B R B Φ Φ (A BR N (A BR N Φ Q Φ B R B Φ Φ A A Φ Q 7 Φ 4 Φ Φ Φ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ 7 ϕ ϕ ϕ ϕ ϕ ϕ ϕ ( ϕ ϕ ϕ 6ϕ ϕ ϕ ϕ + 7 ϕ ϕ + ϕ ϕ ϕ + ϕ 6ϕ 4ϕ + ( Hence, one gets equations (two elements ae equal because of symmety: ( ϕ 6 ϕ 7 ϕ ϕ ϕ + ϕ ϕ ϕ 6 ϕ + 4 ϕ (I (II (III (6 Sylveste s Citeion: An Hemitian (hee symmetic matix M C m m is positive definite if and only if all the uppe-left i i submatices (leading minos, i,, m has positive deteminant Applying this to Φ one gets the conditions: ϕ >

3 ϕ ϕ ϕ > Fom the Equation (I, one gets ϕ 6 ± 64 {, 7} (7 Since we cannot discad a specific value, we pusue with ϕ, and ϕ, 7: Case ϕ, : Plugging this into the Equation (III, one gets and ϕ ϕ ϕ + 4ϕ +, ϕ 4 ± 4 {, } In ode fo these two values to fulfill the second Sylveste condition, it should hold ϕ <, which violates the fist condition Fo this eason this is not a possible choice (8 (9 Case ϕ, 7: Plugging this into the Equation (III, one gets and ϕ 4 + 4ϕ ϕ + 4ϕ 4, ϕ 4 ± 96 { 9, } ϕ is the only value which does not violate the two Sylveste s conditions Plugging the values into Equation (II one gets ϕ + 4 The solution of the Riccati equation hence is 4 Φ 7 7 The contolle K can be computed as ( ( ϕ 4 ( ( K R B Φ ( 4 ( (4

4 b Befoe applying the Hamiltonian method, one need to check (A, B stabilizable (all unstable modes ae eachable The eachability matix fo this pai ( R B AB ( ( has full ank, hence the system is eachable (A, Q detectable The obsevability matix fo the pai Q O QA (6 has full column ank, hence the system is obsevable In ode to use the Hamiltonian method, one needs to build the Hamiltonian matix H whee ( Ã R Q Ã, (7 Ã A BR N R BR B Q Q NR N (8 Fo this specific example (N, one has Ã A ( R 4 Q Q (9 Theefoe, the Hamiltonian is H 7 ( 4

5 In ode to have the eigenvalues, one computes λ λ det (H λi det 7 λ λ λ λ det λ det 7 λ λ λ [ ] [ ] λ λ λ 7 λ λ ( + λ det + det 9 det det λ λ (λ λ + λ 9 det λ + 6 λ (λ 9 λ(λ 9 + λ λ + 6 λ 4 λ + 44 Theefoe, the eigenvalues ae λ, ±, λ,4 ±4 ( Since we only cae about stable eigenvalues (in LHP, we compute the eigenvectos fo λ and λ 4 4 It holds: E λ E : fom (H λ I x one gets the linea system of equations 7 Using the fist ow as efeence and subtacting the coect multiples of it fom the othe ows, one gets the fom 7 Using the second ow as efeence and subtacting the coect multiples of it fom the othe ows, one gets the fom Since one has one zeo ow, one can intoduce a fee paamete Let x 4 s, then x 6 s, x s, x s, s R This defines the fist eigenspace, which is (multiplying eveything by 6 { } E ( 6 (

6 E λ4 E 4 : fom (H λ 4 I x one gets the linea system of equations Using the fist ow as efeence and subtacting the coect multiples of it fom the othe ows, one gets the fom Using the second ow as efeence and subtacting the coect multiples of it fom the othe ows, one gets the fom Since one has one zeo ow, one can intoduce a fee paamete Let x 4 s, then x 6s, x 4s, x s, s R This defines the second eigenspace, which is (multiplying eveything by 6 { 4 } E 4 (4 6 Stacking the eigenvectos one gets X X 4 6 ( 6 It holds Φ X X , 7 (6 which confims the esult of a 6

7 c The closed-loop matix to analyse is A B K (7 The eigenvalues of the closed loop system ae given by det((a B K λ I λ + 7λ + fom which it follows: λ and λ 4 d No Since it holds J new J, K emains the same 7

8 Execise (Fondue Contol a The signal flow diagam is depicted in Figue u (t u (t s x(t k Figue : Signal flow diagam b The diffeential equation becomes with dx(t dt u(t k x(t + u (t + u (t x(t + }{{} u(t, }{{} A B u (t u (t In ode to find the state feedback contolle we need to use the fomula K R B Φ, whee Φ is the solution of the Riccati equation Φ B R B Φ Φ A A Φ Q ( Since R is a diagonal matix, the invese is R ( 4 Fo the dimensions of the Riccati equation, Φ Φ must be a scala becomes (N is the zeo matix: Φ (B R B Φ Φ (A BR N (A BR N Φ Q Φ B R B Φ Φ A A Φ Q Φ Φ Φ Φ + 4 Φ 4Φ Φ Φ {, } (8 (9 ( The equation ( 8

9 Since > and Φ must be positive definite, the solution is Φ The contolle is then ( K R B Φ 4 ( 4 (4 c The signal flow diagam of the closed-loop system is depicted in Figue u (t u (t s x(t k 4 Figue : Signal flow diagam of the closed loop system d The dynamics of the closed loop system ae descibed by A BK ( 4 ( The poles of the system ae the eigenvalues of A BK In this case we ae dealing with a scala poblem: the only pole of the system is π (6 e In ode to minimize the costs, we conside only u (t Since the two inflows have the same influence on the system we only need to make sue that the sum of the elements of K is the same as the one of the elements K This means K (7 9

10 It holds A BK (, (8 ie the pole emais at π Execise 4 (SISO LQR a One stats by solving the Riccati equation to find Φ (N is the zeo matix: Φ (B R B Φ Φ (A BR N (A BR N Φ Q fom which it follows Φ a ± 4a + 4 b q b Φ B R B Φ Φ A A Φ Q b Φ aφ q, (9 a ± a + b q b (4 Since Φ must be positive definite, we only take into account Φ a + a + b q b (4 The feedback contol law hence is u(t bφx(t ba + a + b q b x(t a + a + b q b } {{ } K x(t (4 b The autonomous system s dynamic is ẋ(t ax(t (4 The closed loop system dynamic is ẋ(t (a bkx(t a b a + a + b q b a + b q x(t (44 c The two cases ae:

11 q : in this case it holds ẋ(t a x(t, (4 ie the bandwidth emains the same (we cae much moe about the cost of the input, we use eally small input, the system is pactically open loop q : in this case it holds q ẋ(t b x(t, (46 ie the bandwidth inceases (we cae much moe about the cost on the state

where and are polynomials with real coefficients and of degrees m and n, respectively. Assume that and have no zero on axis.

where and are polynomials with real coefficients and of degrees m and n, respectively. Assume that and have no zero on axis. function whee is an unknown constant epesents fo the un-modeled dynamics The pape investigates the position contol of electical moto dives that can be configued as stuctue of Fig 1 This poblem is fomulated