30 11 2013 11 DOI: 10.7641/CTA.2013.21325 Control Theory & Applications Vol. 30 No. 11 Nov. 2013, ( ;, 310027) :. (LQG),.,,,.,,,.. : LQG ; ; ; : TP273 : A Linear-quadratic-Gaussian control of linear system with Rayleigh flat fading channel TIAN Yin, ZHANG Hui (National Key Laboratory of Industrial Control Technology; Department of Control Science and Engineering, Zhejiang University, Hangzhou Zhejiang 310027, China) Abstract: In networked control systems, communication constrains have been the focus of study for a long time. This paper studies the linear quadratic Gaussian (LQG) control of linear systems under fading constraints, where the outputs of systems are assumed to be amplified and transmitted to the controller over a Rayleigh flat fading channel. Aiming at the tradeoff between controller design and signal amplification, we add an item concerning communication power into the LQG cost function and then derive the controller and amplification factor in a finite horizon, based on the off-line design of amplification factor. With the proposed method, the designed amplification factor turns to a constant for linear timeinvariant systems as time approaches infinity. This result inspires us to extend our method to infinite horizon. Simulation results illustrate the effectiveness and performance of our method. Key words: LQG control; signal amplification; fading; tradeoff 1 (Introduction),,.,,, [1].,,,,..,., [2] [3] [4].,,,.,,, [5]., [6], : 2012 12 30; : 2013 04 28.. Email: zhanghui@iipc.zju.edu.cn. : (61174063).
11 : 1385,, [7 9] ;, [10] ; [11].,. [12]., [13], ; [14] ; [15] ; [16 17] ; [18] (linear quadratic Gaussian, LQG). : [19]; [20 21]. [18].,,,,., LQG,,,., [19],,. : 2, 3, 4 5. 2 (Problem statement) 1,. : x(k + 1) = Ax(k) + Bu(k) + w(k), (1) y(k) = Cx(k) + v(k), (2) : x(k) R n, u(k) R, y(k) R, w(k) N(0, Σ w ), v(k) N(0, σ 2 v), x(0) N(m 0, Σ 0 ). N(m, Σ) m, Σ. y(k).,,, [2] z(k) = + τ= h(k; τ)α(k τ)y(k τ) + n(k), (3) : α(k), n(k), z(k), h(k; τ),.,, H(k; f) [2], H(k; f) = H(k; 0) = h(k)e jφ(k). h(k; τ) = H(k; 0)δ(τ), (3) z(k) = H(k; 0)α(k)y(k) + n(k). (4), H(k; 0) [13], (4) e jφ(k) z(k) = h(k)α(k)y(k) + n(k), (5) : z(k)=re[ z(k)e jφ(k) ], h(k), n(k)=re[ n(k)e jφ(k) ], n(k) N(0, σ 2 n). [2], H(k; 0), h(k), E{h 2 (k)} = σ 2 h. e jφ(k). Z k {z(0),, z(k)} u(k). [19],. 1 Fig. 1 System structure (5) (2), x(k + 1) = Ax(k) + Bu(k) + w(k), (6) z(k) = C (k)x(k) + v (k), (7) : C (k) h(k)α(k)c, v (k) N(0, σ 2 v (k)), σ 2 v (k) h 2 (k)α 2 (k)σ 2 v + σ 2 n.
1386 30 LQG JM c = E{x T (M)Q(M)x(M)+ [x T (k)q(k)x(k) + R(k)u 2 (k)]}. (8) 3 [22] :. 3α(k), α(k),,, α(k), α(k)., [16]. α(k). E{S(k)α 2 (k)y 2 (k)},, u(k) α(k). : u(k) α(k), J M E{x T (M)Q(M)x(M) + [x T (k)q(k)x(k) + R(k)u 2 (k) + S(k)α 2 (k)y 2 (k)]} (9), Q(k) 0, R(k) > 0, S(k) > 0. 1, 2, 3, 4. 1 [13] : 1 1) (A, B), (A, Σw 1/2 ), (A, C), (A, σ v ) ; 2) A ; 3) {h(k)} (i.i.d.) ; 4) x(0), w(k), v(k), n(k), h(k). 1 1) 4) ; 2),,, 2) ; 3)., Clarke H(t; 0) U [23],,, [2]., { H(k; 0)} i.i.d., {h(k)} { H(k; 0)}, i.i.d.. 3 (Design of control signal and amplification factor) u(k) α(k). u(k) α(k),. 3.1 (Finite horizon) (2) (9), J M = E{x T (M)Q(M)x(M)+ [x T (k)q (k)x(k) + R(k)u 2 (k) + S(k)α 2 (k)v 2 (k)]}, (10) Q (k) Q(k) + S(k)α 2 (k)c T C. (1) (10),, J M = E{x T (0)P (0)x(0)+ {S(k)α 2 (k)v 2 (k) + : w T (k)q (k)w(k)+[u(k)+f (k)x(k)] T [B T P (k+1)b+r(k)][u(k)+f (k)x(k)]}}, (11) F (k)= [B T P (k + 1)B+R(k)] 1 B T P (k + 1)A, P (k) = A T {P (k + 1) P (k + 1)B[B T P (k + (12) 1)B + R(k)] 1 B T P (k + 1)}A + Q (k), (13) P (M) = Q(M). (14) LQG, (11) x(k) x(k) = ˆx(k k) + x(k k), ˆx(k k) k, x(k k). x(k k) Z k,, u(k) ˆx(k k) Z k, x(k k) u(k) + F (k)ˆx(k k). (11) J M = E{x T (0)P (0)x(0)+ {S(k)α 2 (k)v 2 (k) + w T (k)q (k)w(k) + x T (k k)f T (k) [B T P (k + 1)B + R(k)]F (k) x(k k) + [u(k) + F (k)ˆx(k k)] T [B T P (k + 1)B + R(k)][u(k) + F (k)ˆx(k k)]}}. (15) (15), u(k),, u (k) = F (k)ˆx(k k). (16) J M (u ) = E{x T (0)P (0)x(0)+ {S(k)α 2 (k)v 2 (k)+
11 : 1387 w T (k)q (k)w(k)+ x T (k k)f T (k) [B T P (k+1)b+r(k)]f (k) x(k k)}}. (17) α (k) = {α(k) : min α(k) J M(u )}. (18) (16),. ˆx(k k)=e{x(k) Z k } α(k),. (12) (14) (16), u (k) α (k),, α (M 1),.,,.,.,,,. ˆx(k k) Kalman : ˆx(k k) = ˆx(k k 1) + K(k)[z(k) C (k)ˆx(k k 1)], (19) Σ(k k) = Σ(k k 1) K(k)C (k)σ(k k 1), (20) ˆx(k k 1) = Aˆx(k 1 k 1)+Bu(k 1), (21) Σ(k k 1) = AΣ(k 1 k 1)A T + Σ w, (22) K(k) = Σ(k k 1)C T (k) [C (k)σ(k k 1)C T (k)+σ 2 v (k)] 1, (23) : Σ(k k) k, ˆx(k k 1) k 1 k, Σ(k k 1). ˆx(0 1)=m 0, Σ(0 1) = Σ 0.. ζ T ζ = tr{ζζ T }, ζ R n, J M (u ) J M (u ) = m T 0 P (0)m 0 + tr{p (0)Σ 0 } + S(k)α 2 (k)σv+ 2 tr{q (k)σ w }+ M 1 tr{ψ(k)e Hk {Σ(k k)}}, (24) Ψ(k) F T (k)[b T P (k+1)b +R(k)]F (k). E Hk {Σ(k k)} H k, H k {h(0),, h(k)},,. [13], J M (u )α s (k). Dey [13] E Hk {Σ(k+1 k)}, E Hk {Σ(k+ 1 k)} E Hk 1 {Σ(k k 1)} : Σ k+1 k E h(k) {Σ w + A Σ k k 1 A T h 2 (k)α 2 (k) h 2 (k)α 2 (k)(c Σ k k 1 C T + σv) 2 + σn 2 A Σ k k 1 C T C Σ k k 1 A T }, (25) Σ k+1 k E Hk {Σ(k + 1 k)}. (25), Dey E Hk {Σ(k + 1 k)} {Ω(k + 1)}: Ω(k + 1) = E h(k) {Σ w + AΩ(k)A T h 2 (k)α 2 (k) h 2 (k)α 2 (k)(cω(k)c T + σv) 2 + σn 2 AΩ(k)C T CΩ(k)A T }. (26) Ω(0) = Σ 0.. h(k), h 2 (k), E{h 2 (k)} = σ 2 h, Ω(k+1) = Σ w + AΩ(k)A T AΩ(k)CT CΩ(k)A T CΩ(k)C T + σv 2 [1 γ k e γ k E 1 (γ k )], (27) : γ k σ 2 n σ 2 h α2 (k)[cω(k)c T + σ 2 v], E 1(x) e t : E 1 (x) = x t dt. {Ω(k + 1)} E Hk {Σ(k k)}. (22), Σ(k k)=a 1 (Σ(k+1 k) Σ w ) (A 1 ) T, E Hk {Σ(k k)} = A 1 (E Hk {Σ(k + 1 k)} Σ w )(A 1 ) T A 1 (Ω(k + 1) Σ w )(A 1 ) T Γ (k). (28) (28), J M (u ) J M(u )= m T 0 P (0)m 0 + tr{p (0)Σ 0 } + S(k)α 2 (k)σv+ 2 tr{q (k)σ w }+ M 1 tr{ψ(k)γ (k)}, (29) α s (k) = {α(k) : min α(k) J M(u )}. (30) α s (k)., Hooke Jeeves [24], :,.,
1388 30,,. MATLAB patternsearch. 2 α s (k), [13] : (1) (2)(5) : A=1.25, B =C =1, Σ w =1, σ 2 v =4, σ 2 n =4, σ 2 h =0.01; : m 0 = 1, Σ 0 =1; J M Q(k)=R(k)=1, S(k)=0.1, M 40, 60 80. α(k), α(k). 2 α s (k) : M,, α s (k), 4.6172. J (ū ) = tr{ Q Σ w } + Sᾱ 2 σ 2 v + tr{ Ψ Σ}. (33) : ᾱ = {ᾱ : min ᾱ J (ū )}, (34) Ψ = F T [B T P B + R] F, F = [B T P B + R] 1 B T P A, (35) P = A T { P P B[B T P B + R] 1 B T P }A + Q, (36) Q = Q + Sᾱ 2 C T C. (37) 1 2 M = 40, 60 80 α s (k) Fig. 2 Value of α s (k) with M = 40, 60 and 80, S(k)., S(k), α s (k). (27) e γ k, α s (k), γ k, e γ k., S(k), Ω(k). 3.2 (Infinite horizon), Q(k) = Q, R(k) = R, S(k) = S. 3.1 α(k) = ᾱ. J = lim M 1 M E{M 1 [x T (k) Qx(k) + Ru 2 (k) + Sᾱ 2 y 2 (k)]}. (31) E Hk {Σ(k k)}, 2. 2 1) ᾱ 0; 2) max(0, log h(0)c ). [13], 1 2, E Hk {Σ(k k)} Σ, Γ (k) Γ. ū (k) = F ˆx(k k). (32) 3.1, ᾱ s : ᾱs = {ᾱ : min J ᾱ (ū )}, (38) J (ū ) = tr{ Q Σ w } + Sᾱ 2 σv 2 + tr{ Ψ Γ }, (39) Γ Γ = A 1 ( Ω Σ w )(A 1 ) T, (40) Ω = Σ w + A ΩA T A ΩC T C ΩA T C ΩC T + σv 2 [1 γe γ E 1 ( γ)], (41) γ = σ 2 n σ 2 hᾱ2 [C ΩC T + σ 2 v]. (42), P, Ω, ᾱs. 3 S ᾱ s, 3.1, A = 1.25, B = C = 1, m 0 = 1, Σ 0 = 1, Σ w = 1, σv 2 = 4, σn 2 = 4, σh 2 = 0.01, J Q = R = 1, ᾱs., S = 0.1, ᾱs = 4.6177, 3.1. 3 S ᾱ s Fig. 3 Value of ᾱ s with different S LQG,
11 : 1389, Riccati, (A, D) (detectable), D Q = D T D [25].,, (A, D ), Q = D T D. (A, D), [25] λ, λ 1, [ ] λi A rank( ) = n. D rank((λi A) T (λi A) + D T D) = n, (λi A) T (λi A) + Q > 0. (λi A) T (λi A) + Q = (λi A) T (λi A) + Q + Sᾱ 2 C T C > 0. rank((λi A) T (λi A) + D T D ) = n, [ ] λi A rank( ) = n, (A, D ). D, LQG,. 3.3 (Summary of the method) (1) (2)(5)., (9),, LQG, (16), Kalman (19) (22);, (30), (29),.,, (31), (32), (39)(38).. 4 (Simulation).,,,,,., : [ ] [ ] 0.7 1 0 x(k + 1) = x(k) + u(k) + w(k), 0 0.9 1 y(k) = [1 0]x(k) + v(k), [ ] [ ] 0 1 0 : w(k) N(, ), v(k) N(0, 1), 0 0 1 [ ] [ ] 1 1 0 x(0) N(, ). 1 0 1 z(k) = ᾱh(k)y(k) + n(k), : h(k), E{h 2 (k)} = 1, n(k) N(0, 1). J Q = diag{1, 1}, R = 1, S = 0.1, diag. D = diag{1, 1}, (A, D),. (38) (42), ᾱ s = 1.1230, Q = diag{1.1123, 1}, D = diag{1.0546, 1}, (A, D ),. ᾱ=1.1230 J (ū )=12.4127. Σ, [13], 50000. 4 ᾱ = 1.1230 tr1(k)=tr{e Hk {Σ(k k)}} tr2(k) = tr{γ(k)}, tr1(k), tr2(k)., E Hk {Σ(k k)}., 10 Σ. 4 Fig. 4 Trace of the expected estimate error covariance and its bound of the stable example, ᾱ
1390 30 J (ū ). 5 ᾱ J (ū ) J (ū ), J (ū ), J (ū )., ᾱ ᾱ s. 7 ᾱ J (ū ) J (ū ) Fig. 7 Value of J (ū ) and J (ū ) with different ᾱ of the unstable example 5 ᾱ J (ū ) J (ū ) Fig. 5 Value of J (ū ) and J (ū ) with different ᾱ of the stable example, [ ], 1.3 1 A=, ᾱs =2.3913. 6 ᾱ= 0 1.7 2.3913 tr1(k) tr2(k), tr1(k), tr2(k). 10 Σ J (ū ) = 427.4744,,,. 6 Fig. 6 Trace of the expected estimate error covariance and its bound of the unstable example 7 ᾱ J (ū ) J (ū ), J (ū ), J (ū ). ᾱ 2.25, ᾱ s = 2.3913,, Σ Γ.,, ᾱ ᾱ s ᾱ J (ū ).,,. 2, 2 :,, ;,,, 2.,,., S,,, 2. 5 (Conclusions).,,.,, LQG,,.,.,,.,.,,,.
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