A Sophisticated Anti-jamming Strategy for a Joint Radar and Communication System

Abstract

In this paper, we consider the problem of determining how a joint (dual) radar and communication system should divide its effort between supporting its radar and communication tasks in the presence of a jammer that wants to obstruct the system’s work by means of jamming. The system, besides the basic objective consisting of two tasks (a) to communicate with a receiver and (b) to track a radar target through the reflections witnessed at the system, also has the secondary objective to achieve the basic objective in a manner that is as unpredictable as possible to the jammer. The signal to interference and noise ratio (SINR) of the radar and communication’s SINR are considered as the metrics that reflect the radar and communication tasks, respectively. The entropy associated with a system’s strategy to switch between two tasks is considered as a metric that reflects unpredictability of its strategy for the jammer. We model this problem by a Bayesian game for a scenario where the system is at a disadvantage to access information about environmental parameters relative to the jammer. The established uniqueness of the equilibrium reflects stability of the designed anti-jamming strategy, even in such a disadvantageous situation for the system.

A Appendix

1.1 A.1 Proof of Proposition 1

By (21), we have that

$$\begin{aligned} \frac{\partial v^2_S(\boldsymbol{x},\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K)}{\partial x^2}=-\frac{w}{x(1-x)}<0. \end{aligned}$$

(48)

Thus, $v_S(\boldsymbol{x},\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K)$ is concave in $\boldsymbol{x}.$ By (22), we have that $v_{J,k}(\boldsymbol{x},\boldsymbol{y}_k)$ is linear on $\boldsymbol{y}_k,$ and the result follows from Nash theorem.   $\square$

1.2 A.2 Proof of Proposition 2

By (21), we have that

$$\begin{aligned} \frac{\partial v_S(\boldsymbol{x},\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K)}{\partial x}&= (1-w)\left( \sum _{k=1}^K\alpha _k(a_k-B_k)-\sum _{k=1}^K\alpha _kD_ky_k\right) \nonumber \\&+w\ln \left( \frac{1-x}{x}\right) \end{aligned}$$

(49)

with $D_k$ given by (28). Thus, for a fixed $w\in (0,1)$ function $\partial v_S(\boldsymbol{x},\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K)/\partial x$ is decreasing on x from infinity for $x\downarrow 0$ to negative infinity for $x\uparrow 1$. Thus, for a fixed $y\in [0,1]$, the best response x is given as the unique root of the following equation:

$$\begin{aligned} (1-w)(\varTheta -\sum _{k=1}^K\alpha _kD_ky_k)+w\ln \left( \frac{1-x}{x}\right) =0. \end{aligned}$$

(50)

Solving this equation by x implies the first row of (25).

For $w=0$, by (49), we have that

$$\begin{aligned} v_S(\boldsymbol{x},\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K)=(\varTheta -\sum _{k=1}^K\alpha _kD_ky_k)x+\sum _{k=1}^K((b_k-B_k)y_k+B_k). \end{aligned}$$

(51)

Thus, $v_S(\boldsymbol{x},\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K)$ is linear in x, and this implies that for a fixed $\boldsymbol{y}_1,\ldots ,\boldsymbol{y}_K$ the best response x is given by the second row of (25)

By (22), we have that

$$\begin{aligned} v_{J,k}(\boldsymbol{x},\boldsymbol{y}_k)=(B_k-b_k+D_kx)y_k+(B_k-a_k)x-B_k. \end{aligned}$$

(52)

Thus, $v_{J,k}(\boldsymbol{x},\boldsymbol{y}_k)$ is linear in $y_k$, and this implies that for a fixed x the best response $y_k$ is given by (29).   $\square$

1.3 A.3 Proof of Theorem 1

Let $(x,\boldsymbol{y}_1,\dots , \boldsymbol{y}_K)$ be an equilibrium. Then, by (29) and (32), we have that there is a t such that

$$\begin{aligned} y_i={\left\{ \begin{array}{ll} 1,&{}i<t,\\ \in [0,1],&{}i=t,\\ 0,&{}i>t. \end{array}\right. } \end{aligned}$$

(53)

Let us consider separately three cases for $\boldsymbol{y}_1,\dots , \boldsymbol{y}_K$: (a) (35) holds, (b) (38) holds and (c) neither (35) nor (38) hold.

Let (35) hold. This corresponds $t=K+1$ in (53). Substituting (35) into (25) implies (36). Then, substituting (36) into (29) implies (34), and (a) follows.

Let (38) hold. This corresponds $t=0$ in (53). Substituting (38) into (25) implies (39). Then, substituting (36) into (29) implies (37), and (b) follows.

(c) Let neither (35) nor (38) hold. Then, by (a) and (b), (40) hold, and, by (29) and (53), two cases arise to consider: (i) $0<y_t<1$ and (ii) $y_t=1$

(i) Let $0<y_t<1.$ Then, by (29)) and (53), we have that

$$\begin{aligned} x=X_{0,t}. \end{aligned}$$

(54)

Substituting (53) into (25), and, then, such obtained x substituting into (54) implies the following equation for $y_t$:

$$\begin{aligned} \frac{1}{\displaystyle 1+\exp \left( \left( \sum _{k=1}^{t-1}\alpha _kD_k+ \alpha _tD_ty_t-\varTheta \right) \delta _w \right) }=X_{t,0}, \end{aligned}$$

(55)

which, by (33), is equivalent to

$$\begin{aligned} \varphi _{t-1}+\alpha _tD_ty_t-\varTheta =\ln (1/X_{t,0}-1)/\delta _w. \end{aligned}$$

(56)

Note that for a fixed t the left side of this equation is increasing on $y_t$ Thus,(33), this equation has the root $y_t$ in [0, 1) if and only if the following relation holds

$$\begin{aligned} (\varphi _{t-1}-\varTheta )\delta _w\le \ln (1/X_{t,0}-1)< (\varphi _t-\varTheta )\delta _w. \end{aligned}$$

(57)

By (33), $(\varphi _t-\varTheta )\delta _w$ is increasing on t. Meanwhile, by (32), $\ln (1/X_{0,t}-1)$ is decreasing on t. Then, since (40) holds, inequalities (57) has the unique solution which we denote by $k_*.$ Finally, solving linear equation (56) on $y_t$ with $t=k_*$ implies (44), and (c-i) follows.

(ii) Let $y_t=1$. Substituting (53) with such $y_t$ into (25) implies that

$$\begin{aligned} x=\frac{1}{\displaystyle 1+\exp \left( \left( \varphi _t-\varTheta \right) \delta _w \right) }. \end{aligned}$$

(58)

Then, by (25) and (53) with $y_t=1,$ we have that

$$\begin{aligned} X_{0,t}\le \frac{1}{\displaystyle 1+\exp \left( \left( \varphi _t-\varTheta \right) \delta _w \right) }<X_{0,t+1}. \end{aligned}$$

(59)

This inequality is equivalent to

$$\begin{aligned} \ln (1/X_{t+1,0}-1)<(\varphi _t-\varTheta )\delta _w\le \ln (1/X_{t,0}-1). \end{aligned}$$

(60)

This implies (45), and (c-ii) follows. Finally, existence of the unique $k_*$ given by (41) and (45) follow from the fact that $\ln (1/X_{t,0}-1)/\delta _w$ is decreasing, meanwhile $\varphi _t-\varTheta$ is increasing on t.   $\square$