UDC 62.501.12

CYBERNETICS AND CONTROL THEORY

A. G. BUTKOVSKII

FINITE CONTROL OF DISCRETE LINEAR SYSTEMS

(Presented by Academician B. N. Petrov, February 21, 1969)

In \((^{1-4})\), problems of finite control of linear systems with lumped and distributed parameters were considered. Here we consider the control of discrete linear systems of the form

\[ x(n+1)=Ax(n)+Bu(n)+F(n),\qquad n=0,1,2,\ldots, \tag{1} \]

where \(x\) is a single-column vector (matrix) of the state in the \(k\)-dimensional complex phase space \(X_k\); \(u\) is a scalar complex control (vector control is considered analogously); \(A\) is a \((k\times k)\)-matrix with complex constant elements; \(B\) is a single-column vector (matrix) with complex constant elements; \(F\) is a single-column vector (matrix) with complex elements.

Problem of finite control. Let \(N\) be given and

\[ x(0)=x_c. \tag{2} \]

Find a lattice function \(u(n)\), \(n=0,1,2,\ldots,N\), such that

\[ x(N+1)=0. \tag{3} \]

To solve this problem we shall use the \(Z\)-transform \((^5)\). The \(Z\)-transform \(\tilde y(z)\) of a function \(y(n)\), \(n=0,1,2,\ldots\), is defined by the formula

\[ \tilde y(z)=Z[y(n)]=\sum_{n=0}^{\infty} y(n)z^n. \tag{4} \]

Apply the \(Z\)-transform to (1), (2). We obtain

\[ \frac{1}{z}\tilde x(z)-\frac{1}{z}x_0=A\tilde x(z)+B\tilde u(z)+\tilde F(z). \tag{5} \]

Hence

\[ \tilde x(z)= \frac{C(1/z)\,[B\tilde u(z)+x_0/z+\tilde F(z)]} {\left|(1/z)E-A\right|} = \frac{z^k C(1/z)\,[B\tilde u(z)+x_0/z+\tilde F(z)]} {\left|E-Az\right|}, \tag{6} \]

where \(C(w)\) is the adjugate matrix of the matrix \(A\) \((^6)\), and \(E\) is the identity matrix.

In order that \(x(n)\) be a finite function, i.e., have finite support \(n=0,1,2,\ldots,N\), and \(x(n)=0\) for \(n<0\) and \(n>N\), it is necessary and sufficient that the right-hand side of formula (6) be a polynomial in \(z\) of degree \(N\), or that the polynomial in the numerator of (6) be divisible without remainder by the polynomial of the denominator. The latter condition, in turn, is equivalent to the conditions

\[ \left\{ z^k C\!\left(\frac{1}{z}\right) \left[ B\tilde u(z)+\frac{x_0}{z}+\tilde F(z) \right] \right\}^{(q)}_{z=z_p} =0,\qquad q=0,1,\ldots,\alpha_p-1;\quad p=1,2,\ldots,m, \tag{7} \]

where \(z_p\) is a root of multiplicity \(\alpha_p\) of the polynomial \(P(z)=|E-Az|\), and

\[ \sum_{p=1}^{m}\alpha_p=k. \tag{8} \]

The equalities (7) are equivalent to the conditions

\[ \tilde u^{(q)}(z_p)= \frac{\left[z^{k-1}C(1/z)x_0+z^kC(1/z)\widetilde F(z)\right]^{(q)}_{z=z_p}} {\left[z^kC(1/z)B\right]^{(q)}_{z=z_p}} =\beta_p^q, \]

\[ q=0,1,\ldots,\alpha_p-1,\quad p=1,2,\ldots,m. \tag{9} \]

Thus, in order to determine the desired finite control \(u(n)\), \(n=0,1,2,\ldots,N\), it is necessary and sufficient to find a polynomial of degree \(N\) that would be a solution of the interpolation problem (9). Such a polynomial can be constructed using the Lagrange–Hermite formula (7),

\[ \tilde u(z)= \sum_{p=1}^{m}\frac{P(z)}{(z-z_p)^{\alpha_p}} \sum_{q=0}^{\alpha_p-1}\beta_p^q \frac{(z-z_p)^q}{q!} \left[ \frac{(z-z_p)^{\alpha_p}}{P(z)} \right]^{(\alpha_p-1-q)}_{z=z_p} +Q(z), \tag{10} \]

where \(Q(z)\) is an arbitrary polynomial of degree \(N\) having the same roots \(z_p\) with the same multiplicities \(\alpha_p\) as the polynomial \(P(z)\), i.e., \(Q(z)\) must be divisible by \(P(z)\).

In the case when the \(z_p\) are simple roots, \(p=1,2,\ldots,k\), the formula for the desired \(\tilde u(z)\) is simpler:

\[ \tilde u(z)= \sum_{p=1}^{k}\beta_p^0 \frac{P(z)}{P'(z_p)(z-z_p)} +Q(z) = \sum_{n=0}^{N}a_nz^n. \tag{11} \]

As a result, \(\tilde u(z)\) is represented by a polynomial of degree \(N\),

\[ \tilde u(z)=\sum_{n=0}^{N}a_nz^n. \tag{12} \]

Thus, the desired finite control is given by the formula

\[ u(n)=a_n,\quad n=0,1,2,\ldots,N. \tag{13} \]

If \(X_k\) is a real space and \(A\), \(B\), and \(F\) are real matrices, then the real finite control is determined by the formula

\[ u(n)=\operatorname{Re}a_n,\quad n=0,1,2,\ldots,N. \tag{14} \]

Since, for the solution of the interpolation problem (9), it is necessary and sufficient that the quantities \(\beta_p^q\) be finite numbers, \(\beta_p^q\ne\infty\), \(q=0,1,\ldots,\alpha_p-1\), \(p=1,2,\ldots,m\), it follows that for controllability of the system (1) it is necessary and sufficient that the conditions

\[ \left[ z^kC\left(\frac{1}{z}\right)B \right]^{(q)}_{z=z_p} \ne0,\quad q=0,1,\ldots,\alpha_p-1;\quad p=1,2,\ldots,m. \tag{15} \]

be satisfied.

Since the minimal degree of the polynomial solving the interpolation problem (9), generally speaking, for arbitrary \(x_c\), is equal to \(N\), it follows that:

1) for \(N<k-1\) the finite-control problem, generally speaking, for arbitrary \(x_c\), has no solution;

2) for \(N=k-1\) the finite-control problem has a unique solution;

3) for \(N>k-1\), the finite-control problem has an infinite set of solutions, due to arbitrariness in the choice of the polynomial \(Q(z)\).

Formula (10) describes the set of the sought finite controls.

Let us note that in the last case, when \(N>k-1\), the arbitrariness in the choice of the polynomial \(Q(z)\) can be used to optimize the finite control \(u(n)\), \(n=0,1,\ldots,N\), according to a prescribed criterion.

To illustrate the method, let us consider the simplest examples.

Example 1. The controlled system is described by the equations

\[ x(n+1)=x(n)+u(n), \qquad n=0,1,\ldots,N, \]

\[ x(0)=x_0, \qquad x(N+1)=0. \]

To determine the finite control \(u(n)\), \(n=0,1,\ldots,N\), we perform the \(Z\)-transform of the given system. We obtain

\[ \frac{1}{z}\,\widetilde{x}(z)-\frac{1}{z}x_0=\widetilde{x}(z)+\widetilde{u}(z). \]

Hence

\[ \widetilde{x}(z)=\bigl(z\widetilde{u}(z)+x_0\bigr)/(1-z). \]

The root of the polynomial in the denominator of this fraction is

\[ z_1=1. \]

Therefore the condition

\[ \widetilde{u}(1)+x_0=0 \]

must be satisfied.

Hence we obtain the interpolation problem

\[ \widetilde{u}(1)=-x_0. \]

Thus, the system is controllable.

By formula (11) we obtain that the \(Z\)-transform of the sought finite control \(u(n)\) has the form

\[ \widetilde{u}(z)=-x_0+Q(z), \]

where \(Q(z)\) is an arbitrary polynomial of degree \(N\) divisible by \(1-z\), or, in other words, having the root \(z_1=1\), i.e. \(Q(1)=0\).

The last formula can also be written in another form,

\[ \widetilde{u}(z)=-x_0+(1-z)S(z), \]

where \(S(z)\) is an arbitrary polynomial of degree \(N-1\). In particular, for \(N=0\),

\[ \widetilde{u}(z)=-x_0 \]

and, consequently, the finite control is

\[ u(0)=-x_0. \]

Example 2. A system is given:

\[ x_1(n+1)=x_1(n)-x_2(n), \qquad x_2(n+1)=x_1(n)+x_2(n)+u(n), \]

\[ x_1(0)=x_{01}, \qquad x_2(0)=x_{02}, \]

\[ x_1(N+1)=0, \qquad x_2(N+1)=0. \]

We perform the \(Z\)-transform of this system. We obtain

\[ \frac{1}{z}\widetilde{x}_1-\frac{1}{z}x_{01}=\widetilde{x}_1-\widetilde{x}_2, \]

\[ \frac{1}{z}\widetilde{x}_2-\frac{1}{z}x_{02}=\widetilde{x}_1+\widetilde{x}_2+\widetilde{u}. \]

Solving this linear system with respect to \(\tilde{x}_1\) and \(\tilde{x}_2\), we obtain

\[ \tilde{x}_1=\frac{z^2\tilde{u}-x_{01}z+x_{01}+x_{02}z}{2z^2-2z+1}, \]

\[ \tilde{x}_2=\frac{z^2\tilde{u}-z\tilde{u}+x_{01}z+x_{02}z-x_{02}}{2z^2-2z+1}. \]

The polynomial \(2z^2-2z+1\) has two roots

\[ z_1=\frac12+\frac{1}{2}j,\qquad z_2=\frac12-\frac{1}{2}j . \]

The numerators in the expressions for \(\tilde{x}_1\) and \(\tilde{x}_2\) must vanish for \(z=z_1\) and \(z=z_2\). Consequently,

\[ z_i^2\tilde{u}(z_i)-x_{01}z_i+x_{01}+x_{02}z_i=0,\qquad i=1,2, \]

\[ z_i^2\tilde{u}(z_i)-z_i\tilde{u}(z_i)+x_{01}z_i+x_{02}z_i-x_{02}=0,\qquad i=1,2. \]

From the first condition we obtain

\[ \tilde{u}(z_i)=\frac{1-z_i}{z_i^2}x_{01}-\frac{x_{02}}{z_i},\qquad i=1,2. \]

From the second condition we obtain

\[ \tilde{u}(z_i)=-\frac{x_{01}}{z_i-1}-\frac{x_{02}}{z_i},\qquad i=1,2. \]

It is easy to see that the right-hand sides of the last two formulas are finite numbers. Consequently, the system is controllable. Moreover, as was to be expected, the right-hand side of the penultimate formula is equal to the right-hand side of the last formula. Such a situation will always occur in the case of a controllable system. Hence there follows an important practical conclusion: after the \(Z\)-transform it is sufficient to solve the algebraic system with respect to only one component \(\tilde{x}_i\), expressing it in terms of \(x_0\), \(\tilde{u}\), and \(F\).

Thus, we obtain the following interpolation problem:

\[ \tilde{u}\left(\frac12+\frac{1}{2}j\right)=-(1+j)x_{01}-(1-j)x_{02}=\beta_1, \]

\[ \tilde{u}\left(\frac12-\frac{1}{2}j\right)=-(1-j)x_{01}-(1+j)x_{02}=\beta_2. \]

The interpolation polynomial has the form

\[ \tilde{u}(z)=\frac{\beta_1\left(z-\frac12+\frac{1}{2}j\right)}{2j} +\frac{\beta_2\left(z-\frac12-\frac{1}{2}j\right)}{2j} +Q(z), \]

where \(Q(z)\) is an arbitrary polynomial of degree \(N>1\) (since for the given system \(k=2\)), having the roots \(z_1=\frac12+\frac{1}{2}j\) and \(z_2=\frac12-\frac{1}{2}j\).

Hence

\[ \tilde{u}(z)=-2x_{02}+(2x_{02}-2x_{01})z+Q(z). \]

For example, for \(N=1\) the finite control is unique and has the form

\[ u(0)=-2x_{02},\qquad u(1)=2x_{02}-2x_{01}. \]

Moscow Forestry Engineering Institute

Received
18 II 1969

REFERENCES

\[ {}^1 \]
A. G. Butkovskii, DAN, 180, No. 5 (1968).
\[ {}^2 \]
A. G. Butkovskii, L. N. Poltavskii, Avtomatika i telemekh., No. 9 (1967).
\[ {}^3 \]
A. G. Butkovskii, L. N. Poltavskii, ibid., No. 10 (1967).
\[ {}^4 \]
A. G. Butkovskii, L. N. Poltavskii, ibid., No. 2 (1969).
\[ {}^5 \]
Ya. Z. Tsypkin, Theory of Pulse Systems, Moscow, 1958.
\[ {}^6 \]
F. R. Gantmakher, Theory of Matrices, Nauka, 1966.
\[ {}^7 \]
V. L. Goncharov, Theory of Interpolation and Approximation of Functions, Moscow, 1954.