Optimal control of time-varying linear delay systems based on the Bezier curves

Abstract

In this paper, time-delay control systems with quadratic performance are solved by applying the least square method on the Bezier control points. The approximation process is done in two steps. First, the time interval is divided into $2k$ subintervals, then in each subinterval the trajectory and control functions are approximated by the Bezier curves. We have chosen the Bezier curves as piecewise polynomials of degree $n$ and determined the Bezier curves on any subinterval by $n+1$ control points. By considering a least square optimization problem, the control points can be found, then the Bezier curves that approximate the action of control and trajectory can be computed as well. Some numerical examples are given to verify the efficiency of the proposed method.

Appendix

In this section, we specify the derivative of Bezier curve.

By (6), we have

$$\begin{aligned} v_j(t)=\sum _{i=0}^n a_{i}^j B_{i,n}(t),\,t\in [0,1], \end{aligned}$$

where $B_{i,n}(t)=\frac{n!}{i!(n-i)!}t^i(1-t)^{n-i}$.

Now, we have

$$\begin{aligned} \frac{\mathrm{d}B_{i,n}(t)}{\mathrm{d}t}=n(B_{i-1,n-1}(t)-B_{i,n-1}(t)), \end{aligned}$$

(24)

where $B_{-1,n-1}(t)=B_{n,n-1}(t)=0$, and

$$\begin{aligned} B_{i-1,n-1}(t)&= \frac{(n-1)!}{(i-1)!(n-i)!}t^{i-1}(1-t)^{n-i},\\ B_{i,n-1}(t)&= \frac{(n-1)!}{i!(n-i-1)!}t^i(1-t)^{n-i-1}. \end{aligned}$$

Using (24), the first derivative $\mathbf{v}_j(t)$ is shown as

$$\begin{aligned} \frac{\mathrm{d}\mathbf{v}_{j}(t)}{\mathrm{d}t}&= \sum _{i=1}^{n-1}n\mathbf{a}_i^jB_{i-1,n-1}(t)-\sum _{i=0}^{n-1}n\mathbf{a}_i^jB_{i,n-1}(t)\nonumber \\&= \sum _{i=0}^{n-1}n\mathbf{a}_{i+1}^jB_{i,n-1}(t)-\sum _{i=0}^{n-1}n\mathbf{a}_i^jB_{i,n-1}(t)\nonumber \\&= \sum _{i=0}^{n-1}B_{i,n-1}(t)n\{\mathbf{a}_{i+1}^j-\mathbf{a}_{i}^j\}. \end{aligned}$$

(25)

Now, we specify the procedure of derivation (9) from (8).

By (6), we have

$$\begin{aligned} \mathbf{v}_j(t)&= \begin{array}{l}n\\ 0\end{array}\mathbf{a}_0^j\frac{1}{h^n}(t_j-t)^n\nonumber \\&+\ldots +\begin{array}{l}n\\ n\end{array} \mathbf{a}_n^j\frac{1}{h^n}(t-t_{j-1})^n,\end{aligned}$$

(26)

$$\begin{aligned} \mathbf{v}_{j+1}(t)&= \begin{array}{l}n\\ 0\end{array}\mathbf{a}_0^{j+1}\frac{1}{h^n}(t_{j+1}-t)^n\nonumber \\&+\ldots +\begin{array}{l}n\\ n\end{array} \mathbf{a}_n^{j+1}\frac{1}{h^n}(t-t_j)^n, \end{aligned}$$

(27)

by substituting $t=t_j$ into (26) and (27), one has

$$\begin{aligned} \mathbf{v}_j(t_j)&= \mathbf{a}_n^j \frac{1}{h^n} (t_j-t_{j-1})^n, \end{aligned}$$

(28)

$$\begin{aligned} \mathbf{v}_{j+1}(t_j)&= \mathbf{a}_0^{j+1}\frac{1}{h^n}(t_{j+1}-t_j)^n. \end{aligned}$$

(29)

To preserve the continuity of Bezier curves at the nodes, one needs to impose the condition $\mathbf{v}_j(t_j)=\mathbf{v}_{j+1}(t_j)$, so from (28) and (29), we have

$$\begin{aligned}&\mathbf{a}_n^j (t_j-t_{j-1})^n=\mathbf{a}_0^{j+1}(t_{j+1}-t_j)^n. \end{aligned}$$

(30)

From (25), the first derivatives of $\mathbf{v}_j(t)$ and $\mathbf{v}_{j+1}(t)$ are, respectively:

$$\begin{aligned} \frac{\mathrm{d}\mathbf{v}_j(t)}{\mathrm{d}t}&= \sum _{i=0}^{n-1}B_{i,n-1}(t){n(\mathbf{a}_{i+1}^j-\mathbf{a}_i^j)}\nonumber \\&= \sum _{i=0}^{n-1}\left( \begin{array}{c}n-1\\ i\end{array}\right) (t_j-t)^{n-1-i}(t-t_{j-1})^i\nonumber \\&\times \frac{1}{h^n}\{n(\mathbf{a}_{i+1}^j-\mathbf{a}_i^j)\}\nonumber \\&= \left( \begin{array}{c}n-1\\ 0\end{array}\right) \{n(\mathbf{a}_1^j-\mathbf{a}_0^j)\}\frac{1}{h^n}(t_j-t)^{n-1}\nonumber \\&+\ldots +\left( \begin{array}{c}n-1\\ n-1\end{array}\right) \{n(\mathbf{a}_n^j-\mathbf{a}_{n-1}^j)\}\nonumber \\&\times \frac{1}{h^n}(t-t_{j-1})^{n-1}, \end{aligned}$$

(31)

$$\begin{aligned} \frac{d\mathbf{v}_{j+1}(t)}{dt}&= \sum _{i=0}^{n-1}\left( \begin{array}{c}n-1\\ i\end{array}\right) (t_{j+1}-t)^{n-1-i}(t-t_{j})^i\nonumber \\&\times \frac{1}{h^n}\{n(\mathbf{a}_{i+1}^{j+1}-\mathbf{a}_i^{j+1})\}\nonumber \\&= \left( \begin{array}{c}n-1\\ 0\end{array}\right) \{n(\mathbf{a}_1^{j+1}-\mathbf{a}_0^{j+1})\}\frac{1}{h^n}(t_{j+1}-t)^{n-1}\nonumber \\&+\ldots +\left( \begin{array}{c}n-1\\ n-1\end{array}\right) \{n(\mathbf{a}_n^{j+1}-\mathbf{a}_{n-1}^{j+1})\}\nonumber \\&\times \frac{1}{h^n}(t-t_{j})^{n-1}. \end{aligned}$$

(32)

By substituting $t=t_j$ into (31) and (32), we have

$$\begin{aligned} \frac{\mathrm{d}\mathbf{v}_j(t_j)}{\mathrm{d}t}&= {n(\mathbf{a}_n^j-\mathbf{a}_{n-1}^j)} \frac{1}{h^n} (t_j-t_{j-1})^{n-1}, \end{aligned}$$

(33)

$$\begin{aligned} \frac{\mathrm{d}\mathbf{v}_{j+1}(t_j)}{\mathrm{d}t}&= {n(\mathbf{a}_1^{j+1}-\mathbf{a}_{0}^{j+1})}\frac{1}{h^n}(t_{j+1}-t_j)^{n-1}, \end{aligned}$$

(34)

and to preserve the continuity of the first derivative of the Bezier curves at nodes, by equalizing (33) and (34), we have

$$\begin{aligned}&(\mathbf{a}_n^{j}-\mathbf{a}_{n-1}^{j}) (t_j-t_{j-1})^{n-1}=(\mathbf{a}_1^{j+1}-\mathbf{a}_{0}^{j+1})(t_{j+1}-t_j)^{n-1}, \end{aligned}$$

where it shows the equality (9).