]> 3.3.3 Example 3: Automatic control system with PID controller II

3.3.3 Example 3: Automatic control system with PID controller II

In accordance with the theory presented in the Section 3.2 the integral (3.15) exists if c < 1 and all roots of the characteristic quasipolynomial

(s2 as) (s2c + sb + d)esr (3.37)

have negative real parts.

The observed output is

z1 = v1 + cψ1(r) ,

so we have

P = 10 00 ,Q = c 0 00 ,R = c20 0 0 .

The equation (3.24) yields

γ = c2 1 c20 0 0 .

Taking into account the nature of the initial conditions we conclude that

J(x0)(b,c,d) = x0,x0H = v0T αv 0 = z02α 11 (3.38)

Hence, to find the matrix α we need to solve the vector differential equation

β11(θ) β12(θ) β21(θ) β22(θ) ϑ11(θ) ϑ12(θ) ϑ21(θ) ϑ22(θ) = I II A0T A0T II I 1 A1T I I A 2T A2T I I A 1T β11(θ) β12(θ) β21(θ) β22(θ) ϑ11(θ) ϑ12(θ) ϑ21(θ) ϑ22(θ) = = a + bc 1 c2 0 cd 1 c2 0 b + ac 1 c2 d 1 c2 0 0 0 a 0 0 0 0 0 0 1 0 0 0 0 0 b + acd 0 1 0 0 0 0 0 0 b + ac c2 1 0 d c2 1 0 a + bc c2 1 cd c2 1 0 0 0 b ac 0 d 1 0 0 0 0 0 0 0 0 0 a 0 0 0 0 0 0 0 1 0 β11(θ) β12(θ) β21(θ) β22(θ) ϑ11(θ) ϑ12(θ) ϑ21(θ) ϑ22(θ) (3.39)

with initial conditions

β11(0) = 1 2(1 c2) aα11 β12(0) + β21(0) = α11 aα12 β22(0) = α12 ϑ11(0) = c 1 c2 + (ac + b)α11 + dα12 + cβ11(0) ϑ12(0) =(ac + b)α12 + dα22 + cβ21(0) ϑ21(0) =0 ϑ22(0) =0 (3.40)

which follow from (3.25) and (3.27). From (3.39) and (3.40) we obtain ϑ21(θ) 0, ϑ22(θ) 0, and by (3.27): β12(θ) 0, β22(θ) 0. Consequently α12 = 0 and therefore the system (3.39) reduces to the form

β11(θ) β21(θ) ϑ11(θ) ϑ12(θ) = a + bc 1 c2 cd 1 c2 b + ac 1 c2 d 1 c2 1 0 0 0 b + ac 1 c2 d 1 c2 a + bc 1 c2 cd 1 c2 0 0 1 0 β11(θ) β12(θ) ϑ11(θ) ϑ12(θ) (3.41)

with initial conditions

β11(0) β21(0) ϑ11(0) ϑ12(0) = aα11 1 2(1 c2) α11 c 2(1 c2) + bα11 dα22 cα11 (3.42)

The characteristic polynomial of the system (3.41) is biquadratic

λ4 + λ2 b + ac 1 c2 2 a + bc 1 c2 2 2cd 1 c2 d2 1 c2 = λ2 + G2 λ2 H2

where

G2 =1 2 k1 + k1 2 + 4k2 ,H2 =1 2 k1 + k1 2 + 4k2 k1 =b2 a2 2cd 1 c2 , k2 = d2 1 c2 .

Hence the fundamental matrix of (3.41) takes a form

eθT = Z 1 cos Gθ + Z2 sin Gθ + Z2 cosh Hθ + Z4 sinh Hθ

where T stands for the matrix of the system (3.41), and

Z1 Z3 = 1 G2 + H2 H2I T2 G2I + T2 , Z2 Z4 = 1 GH(G2 + H2) H(H2 T3) G(G2 + T3) .

The unknown elements α11, α22 can be determined from the equality

ϑ11(r) ϑ12(r) =  the last two rows of the matrix erT 2×4  a vector of initial conditions (3.42) 4×1 = β11(0) β21(0) (3.43)

arising from (3.27). After tedious but elementary calculations, we get from (3.43)

σ + δ + aG2 + aH2 H2σ G2δ Σ + Δ + G2 + H2 H2Σ G2Δ α11 α22 = = 1 2(1 c2) G2 H2 + Σ + Δ 1 G2σ + 1 H2δ (3.44)

where:

σ =bG2 cos Gr + (cG3 dG) sin Gr , δ =bH2 cosh Hr (cH3 + dH) sinh Hr , Σ =(d G2c) cos Gr + bG sin Gr , Δ =bH sinh Hr (d + cH2) cosh Hr .

Solving (3.44) with respect to α11 and substituting the result into (3.38) we find the desired expression for the performance index J in terms of the system parameters.

In what follows the case of the PD controller will be investigated. In this case d = 0 and by (3.14) we have z2(t) 0. The characteristic quasipolynomial (3.37) reads as s s a (cs + b)esr where the expression in the square brackets is the characteristic quasipolynomial of the remaining part of the system (3.14). It is not difficult to establish that if a < 0 then the domain of stability in the space of parameters (b,c) is the bounded part of the semistrip c < 1, b < a cut by the arc

b(ω) = a cos ωr ω sin ωr c(ω) = cos ωr asin ωr ω ,ω0 ω π r

where ω0 denotes the smallest positive solution of the equation ω = a2 + ω2 2a sin ωr. Moreover,

H = 0, G = b2 a2 1 c2 ifb2 a2 H = a2 b2 1 c2 ,G = 0 ifb2 a2

and the system (3.43) reduces to the first equation only. Finally,

α11 = 1 c cos kr + b k sin kr 2(1 c2)(b cos kr + ck sin kr + a) ifb2 > a2 1 c cosh kr + b k sinh kr 2(1 c2)(b cosh kr ck sinh kr + a)ifb2 < a2 1 + c ar 4(1 c2)(a) if b = a ,k = b2 a2 1 c2 .

The results of minimization of the performance index (3.38) for a = 5, r = 0.5 are presented at Figure 3.5. The minimal value of z02J(x 0) is 0.099329936 and it is achieved at b = 0.42234051, c = 0.078988818.


PIC

Figure 3.5: The level curves of the performance index z02J(x 0) as a function of b and c

Exercise 3.3.1. For the time–delay system of the neutral type

v̇(t) = v(t) v(t) =z(t) 1 2z(t 1) z(θ) =φ(θ), 1 θ 0 v(0) =v0

calculate the integral performance criterion J =0z2(t)dt in the two following cases:

1o
φ(θ) = eθ,v 0 = 1 e 2,
2o
φ(θ) 0,v0 = 1.

In the first case we have z(t) = et and therefore by direct calculations J = 1 2. In the second case z(t) 1 2z(t 1) = et and using the method of steps we obtain

z(t) = et m=0i1 e 2 m = et1 e 2 i 1 e 2 ,i 1 t < i,i .

Hence

0z2(t)dt = i=1i1iz2(t)dt = i=1i1ie2t 1 e 2 i 1 e 2 2dt =

= i=11 e 2 i 2 2 e 2 2 e2 1 2 e2i = 2(e2 1) (2 e)2 i=1e2i 2 i=1(2e)i + i=1 1 4i =

= 2(e2 1) (2 e)2 e2 1 e2 2 (2e)1 1 (2e)1 + 1 4 1 1 4 = 2(2e + 1) 3(2e 1) .

The above results can be confirmed by the method presented in the previous section. We identify parameters in (3.5) and (3.16):

A1 = 1,A2 = A0 = 1 2,r = 1,P = 1,Q = 1 2,R = 1 4 .

The polynomial (3.29) has two roots λ = ±1. Thus taking λ = 1 we find from (3.30): L = 0, M = 1. The unique solution of (3.24) is γ = 1 3. Hence, the system (3.33) takes the form

2 2e1 0 1 2e1 + 1 α κ = 4 3 2 3 .

Its solution is α = 4e + 2 3(2e 1), κ = 4e 3(2e 1). From (3.32) and (3.26) we find: β(θ) = 4eθ 3(2e 1), Φ(θ) = 0 for 1 θ 0. Confirm the results using Remark 3.3.1. Now

J v0 φ = v0 φ ,v0 φ H

where is given by (3.21). Substituting the initial data we get the same values of J as by direct calculations.

An advantage of the general attempt to evaluating J is that the value of the performance index is calculated for a wide variety of initial conditions, while the direct calculations are carried out separately for particular initial data.