Quantized Sampled-Data Control for Exponential Stabilization of Delayed Complex-Valued Neural Networks

Abstract

This paper addresses the problem of quantized sampled-data control for CVNNs with time-varying delay under the assumption that only quantized measurements are transmitted to the controller. Based on the discrete-time Lyapunov stability theory, reciprocally convex approach, a sector bound approach, and some estimation techniques, a reduced conservative stabilization criterion is obtained to guarantee the exponential stabilization of the considered CVNNs. The desired quantized sampled-data controller is designed via converting the complex-valued linear matrix inequality into real-valued ones. The effectiveness of the derived criteria are shown via an illustrative simulation example.

Appendix

Proof

For any $t\in [t_k,t_{k+1})$, according to Eq. (4), we have

$$\begin{aligned} \begin{aligned} ||z(t)||&\le ||z(t_k)||+||\int _{t_k}^tDz(\alpha )d\alpha ||+||Af(z(\alpha ))d\alpha ||\\&\quad +||Bg(z(\alpha _{\tau }))d\alpha ||+||\int _{t_k}^tKz(t_k)d\alpha ||\\&\quad +||\int _{t_k}^tK\varDelta (t_k)z(t_k)d\alpha ||. \end{aligned} \end{aligned}$$

(24)

Based on the Cauchy–Schwarz inequality, it follows from (24) that

$$\begin{aligned} \begin{aligned} ||z(t)||^2&\le 6||z(t_k)||^2+6||\int _{t_k}^tDz(\alpha )d\alpha ||^2\\&\quad +6||\int _{t_k}^tAf(z(\alpha ))d\alpha ||^2+6||\int _{t_k}^tB{g}(z(\alpha _{\tau }))d\alpha ||^2\\&\quad +6||\int _{t_k}^tKz(t_k)d\alpha ||^2+6||\int _{t_k}^tK\varDelta (t_k)z(t_k)d\alpha ||^2. \end{aligned} \end{aligned}$$

(25)

For (25), we utilize the Cauchy–Schwarz inequality again and then we can get

$$\begin{aligned} \begin{aligned} ||z(t)||^2&\le 6||z(t_k)||^2+6h\int _{t_k}^t||Dz(\alpha )||^2d\alpha \\&\quad +6h\int _{t_k}^{t}||Af(z(\alpha ))||^2d\alpha +6h\int _{t_k}^{t}||Bg(z(\alpha _{\tau }))||^2d\alpha \\&\quad +6h\int _{t_k}^t||Kz(t_k)||^2d\alpha +6h\int _{t_k}^t||K\varDelta (t_k)z(t_k)||^2d\alpha . \end{aligned} \end{aligned}$$

According to Assumption 1, we have

$$\begin{aligned} \begin{aligned} ||f(z(\alpha ))||^2\le ||F||^2||z(\alpha )||^2, ~||g(z(\alpha _{\tau }))||^2\le ||G||^2||z(\alpha _{\tau })||^2. \end{aligned} \end{aligned}$$

Thus, we further obtain

$$\begin{aligned} \begin{aligned} ||z(t)||^2&\le 6||z(t_k)||^2+6h(||D||^2\int _{t_k}^t||z(\alpha )||^2d\alpha +||A||^2||F||^2\\&\quad \times \int _{t_k}^t||z(\alpha )||^2d\alpha +||B||^2||G||^2\int _{t_k}^t||z(\alpha _{\tau })||^2d\alpha \\&\quad +\int _{t_k}^t||K||^2||z(t_k)||^2(1+||\varDelta (t_k)||^2)d\alpha )\\&\le 6||z(t_k)||^2+6h||D||^2\int _{t_k}^t||z(\alpha )||^2d\alpha +6h||A||^2||F||^2\\&\quad \times \int _{t_k}^t||z(\alpha )||^2d\alpha +6h||B||^2||G||^2\int _{t_k}^t||z(\alpha _{\tau })||^2d\alpha \\&\quad +6h^2||K||^2||z(t_k)||^2+6h^2\delta ^2||K||^2||z(t_k)||^2\\&=6(1+h^2||K||^2+h^2\delta ^2||K||^2)||z(t_k)||^2+6h||D||^2\\&\quad \times \int _{t_k}^t||z(\alpha )||^2d\alpha +6h||A||^2||F||^2\int _{t_k}^t||z(\alpha )||^2d\alpha \\&\quad +6h||B||^2||G||^2\int _{t_k}^t||z(\alpha _{\tau })||^2d\alpha . \end{aligned} \end{aligned}$$

(26)

By using the Gronwall–Bellman inequality, it follows from (26) that

$$\begin{aligned} \begin{aligned} ||z(t)||^2&\le 6[1+h^2||K||^2(1+\delta ^2)]||z(t_k)||^2\\&\quad \times e^{\int _{t_k}^t6h(||D||^2+||A||^2||F||^2)d\alpha }\\&\quad +6h||B||^2||G||^2\int _{t_k}^t||z(\alpha _{\tau })||^2d\alpha \\&\le 6[1+h^2||K||^2(1+\delta ^2)]||z(t_k)||^2\\&\quad \times e^{6h^2(||D||^2+||A||^2||F||^2)}\\&\quad +6h||B||^2||G||^2\int _{t_k}^t||z(\alpha _{\tau })||^2d\alpha . \end{aligned} \end{aligned}$$

(27)

Then, applying the Lemma 2 to (27), we can obtain Lemma 3 immediately. This completes the proof. $\square$