: state space and polynomial es Lecture 3 O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.fr www.gipsa-lab.fr/ o.sename 10th January 2014
Outline
The classical structure
Towards 2
The RST structure
Notations r is the reference signal, u(t) the input and y(t) the system output d u (t) and d y (t) are the disturbance signals, on the system input and output respectively n is the measurement noise (output) Here the RST design ology is presented for discrete-time, but it could be applied to continuous-time systems as well.
System data The system is given by : A(z) = a 0 z n + a 1 z n 1 +... + a n B(z) = b 0 z n + b 1 z n 1 +... + b n Usually A is assumed to be monic, i.e. a 0 = 1. The input is given by: S(z)u k = T (z)r k R(z)y k where R, S and T are some polynomials in z.
Closed-loop specifications Since the is a pole placement one, the specifications concern the characteristic polynomial of the closed-loop system. First: The aim is that the closed-loop system will follow a reference model given by: B m (z),such that: A m (z) deg(a m ) deg(b m ) and A m is monic and stable, Second: The underlying objective is to place the closed-loop poles to handle to transient behavior.
Other specifications As usually, specifications may concern disturbance rejection, noise effect attenuation, robustness (+ actuator saturation)... We then add some constraints synthesis as for instance Integral Action. Constraints: the closed-loop steady state errors in tracking and regulation should be zero. With the given action, the closed-loop system becomes: y k = B(z)T (z) A(z)S(z) + B(z)R(z) r k
Remarks The reference model denominator A m is often chosen of a lower degree than AS + BR B m is chosen to specify the closed-loop zeros.
RST form The polynomials R, S and T are defined by: R(z) = r 0 z n r + r 1 z n r 1 +... + r nr S(z) = s 0 z n s + s 1 z n s 1 +... + s ns T (z) = t 0 z n t + t 1 z n t 1 +... + t nt Thus the implementation of that ler requires: n s n r et n s n t Then both following cases are usually considered 1. the computational time can be neglected w.r.t the sampling period. Then : n r = n t = n s 2. The computational times equals the sampling period. Then: n r = n t = n s 1
As said before, an Integralaction is often necessary. On the other hand, the order of the closed-loop system may become very large according to the choice of the order of the polynomials R, S and T. The system under consideration is : H(z) = B(z) A(z) Assumption: B and A are co-prime and A is monic (a 0 = 1). (1)
It is usually required that the steady state gain between d y and y r is zero, which aims at providing: A(1)S(1) = 0 Then, if the system does not include an integrator (A(1) 0), S(z) must satisfy : S(1) = 0 and then include z 1. If many integrators are required then S(z) will include (z 1) l, with l > 1.
The system output y should follow (in closed-loop) the reference signal r, which corresponds to B(1)T (1) A(1)S(1) + B(1)R(1) = 1 Then T (z) should satisfy the following constraint: T (1) = R(1)
Step 1 Let decompose B as : B = B + B (2) where B + represents some zeros which are also some zeros of AS + BR (in order to get simplifiactions), thus being stable zeros, and B the remaining zeros (not being zeros of AS + BR) which may be unstable (or on the unitary circle) zeros. Let assume B + monic. The closed-loop system can be written as: B + B T AS + BR = B m A m B + B TA m = (AS + BR)B m Then the closed-loop stability implies that AS + BR have B no common zeros. Hence B m = B B m (3)
Step 2 B + is a factor from AS + BR and AS + BR = AS + B + B R, Therefore the zeros of B + are zeros of AS, and: where S, B + and S are monic polynomials. Then we obtain: S = B + S (4) B + B T B + (AS + B R) = B B T = A m AS + B R = B A m It can be conclude that T equals B m and AS + B R equals A m to a factor polynomial: m m T = B ma 0 (5) AS + B R = A m A 0 (6) where A 0 is usually referred to as the observer polynomial.
Step 3 To summarize BT and AS + BR both contain polynomials A 0 and B +, which leads to some simplification (leading to the tracking model). When an (or manu) integrator are included, S(z) should be replaced by (z 1) l S(z). The polynomial equation to be solved is then: B(z)T (z) A(z)(z 1) l S(z) + B(z)R(z) = B m(z) A m (z) Applying the previous ology, it leads: T (z) = B m(z)a 0 (z) (7) A(z)(z 1) l S (z) + B (z)r(z) = A m (z)a 0 (z) (8) Usaually B m(z) can be chosen as a simple constant gain, which ensures tracking performances (without steady state error) i.e. R(1)/A 0 (1).
The following theorem states the existence of RST. Theorem There exists a proper (deg(s) deg(t ) and deg(s) deg(r)) if the following inequalities are satisfied: deg(a m ) deg(b m ) deg(a) deg(b) (9) deg(a 0 ) 2deg(A) deg(a m ) deg(b + ) 1 (10) When one (or more) integrators are included, there exists a RST ler such that (deg(s) + l deg(t ) and deg(s) + l deg(r)) if : deg(a m ) deg(b m ) deg(a) deg(b) (11) deg(a 0 ) 2deg(A) deg(a m ) deg(b + ) 1 + l (12)
The following theorem ensures the uniqueness of RST. Theorem The degree of R equals deg(a) 1 or, in the presence of an integrator deg(a) 1 + l. Furthermore the polynomials S and R are unique.
Two cases 1. the computational time can be neglected w.r.t the sampling period: deg(s) = deg(r) = deg(t ),then deg(a m ) deg(b m ) = deg(a) deg(b) (13) deg(a 0 ) = 2deg(A) deg(a m ) deg(b + ) 1 + l (14) deg(r) = deg(a) 1 + l (15) with l = 0 without integrator 2. The computational times equals the sampling period. Then: deg(s) 1 = deg(r) = deg(t ),then deg(a m ) deg(b m ) = deg(a) deg(b) + 1 (16) deg(a 0 ) = 2deg(A) deg(a m ) deg(b + ) + l (17) deg(r) = deg(a) 1 + l (18) with l = 0 without integrator
Design procedure Data Polynomials A(z) and B(z) Polynomials A m (z), B m (z) and A 0 (z) Conditions A(z) and B(z) have no common factors. B m (z) = B (z)b m(z) deg(a m ) deg(b m ) deg(a) deg(b) deg(a 0 ) 2deg(A) deg(a m ) deg(b + ) 1 + l deg(r ) = deg(a m ) + deg(a 0 ) deg(a) l deg(s) = deg(a) 1 + l Step 1 Factorization B(z) = B + (z)b (z) Step 2 Solve A(z)(z 1) l S (z) + B (z)r(z) = A m (z)a 0 (z) Step 3 Compute S(z) = B + (z)s (z) and T (z) = B m(z)a 0 (z)
Bezout equation The Bezout identity (or diophantine equation) is given by: A(z)X(z) + B(z)Y (z) = C(z) (19) Theorem The equation (19) has a solution if and only if the largest common divisor of A(z) and B(z) is a divisor of C(z). If there exists a solution (X 0 (z), Y 0 (z)) then : X(z) = X 0 (z) + S(z)B(z) and Y (z) = Y 0 (z) S(z)A(z) where S(z) is an arbitrory polynomial (solution as well). The Bezout equation can be rewritten as a linear system to be solved (using the equality of powers in z).
Bezout equation In the general case where: deg(a) = deg(b) = n and deg(r) = deg(s) 1 = n + l 1, (20) i.e. deg(a m A 0 ) = 2n + l. Denote: A (z) = A(z)(z 1) l = a 0 z n+l + a 1 z n+l 1 +... + a n+l B(z) = b 0 z n + b 1 z n 1 +... + b n S (z) = s 0 z n + s 1 z n 1 +... + s n R(z) = r 1 z n+l 1 + r 2 z n+l 2 +... + r n+l C(z) = c 0 z 2n+l + c 1 z 2n+l 1 +... + c 2n+l
Bezout equation The linear system to be solved is: MX = C,where a 0 0... 0 0...... 0 0...... 0 a 1 a 0 0 0 b 0 0............. 0............................. 0 an a n 1... a 0 b n 1... b 0 0 0...... 0 a n+1 an... a 1 bn b n 1... b 0 0 0... 0 a n+2 a n+1... a 2 0 bn b n 1... b 0 0... 0.............................. M =. 0............ a n+l a n+l 1... a l 0 bn bn 1 b0............... 0 a n+l... a l+1 0 bn b1.............................................. 0 an+l a n+l 1 0. 0 bn bn 1 0... 0 a n+l 0............... 0 bn (21) C = ( c 0 c 1... c n+l...... c 2n+l ) T and X = ( s 0 s 1... s n r 1... r n+l ) T for which the resolution is done by gaussian elimination.
Bezout equation Remark When the degrees of polynomials do not satisfy equalities (20) we need to cancel the unuseful terms. If deg(b) = n 1 then b 0 = 0. There exists an algorithm based on polynomial division, well adapted to solve such an equation.
Equivalence and The use of an ler is equivalent to the following ler : u k = F d (zi n A d + B d F d + L d C d ) 1 L d y k +[I n F d (zi n A d + B d F d + L d C d ) 1 B d ]G d r k which corresponds to a two-degrees of freedom ler u k = R(z) S(z) y k + T (z) S(z) r k and this can be implemented in an RST form.
Sensitivity functions The two structure requires to define specific sensitivity functions. with S (z) = A(z)S(z) P c (z) K S (z) = A(z)R(z) P c (z) and T (z) = B(z)R(z) P c (z) and S G (z) = B(z)S(z) P c (z) P c (z) = A(z)S(z) + B(z)R(z) (22) (23)
Sensitivity functions Figure: Input and Output sensitivity functions