Maxmin Strategy for a Dual Radar and Communication OFDM Waveforms System Facing Uncertainty About the Background Noise

Abstract

The paper considers the problem of designing the maxmin strategy for a dual-purpose communication and radar system that employs multicarrier OFDM style waveforms, but faces an uncertain level of background noise. As the payoff for the system, we consider the weighted sum of the communication throughput and the radar’s SINR. The problem is formulated as a zero-sum game between the system and a rival, which may be thought of as the environment or nature. Since the payoff for such a system combines different type of metrics (SINR and throughput), this makes underlying problem associated with jamming such a systems different from the typical jamming problem arising in communication scenarios, where the payoff usually involves only one of these metrics. In this paper, the existence and uniqueness of the equilibrium strategies are proven as well as water-filling equations to design the equilibrium are derived. Finally, using Nash product the optimal value of weights are found to optimize tradeoff of radar and communication objectives.

Appendix: Proof of Theorem 1

Proof

(I) By definition, ${\varvec{P}}$ and ${\varvec{J}}$ are equilibrium strategies if and only if each of them is the best response to the other, i.e., they are solutions of the equations: and By the Karush-Kuhn-Tucker (KKT) theorem, since v is concave on ${\varvec{P}}$, ${\varvec{P}}\in \varPi _S$ is the best response strategy to ${\varvec{J}}$ if and only if there is an $\omega$ (Lagrange multiplier) such that

$$\begin{aligned} w^C\frac{h^C_i}{\sigma ^2+h^C_i P_i+ g^C_i J_i}+w^R\frac{h^R_i}{\sigma ^2+g^R_i J_i} {\left\{ \begin{array}{ll} =\omega ,&{} P_i>0,\\ \le \omega ,&{} P_i=0. \end{array}\right. } \end{aligned}$$

(9)

Similarly, since v is convex on ${\varvec{J}}$, ${\varvec{J}}\in \varPi _R$ is the best response strategy to ${\varvec{P}}$ if and only if there is a $\nu$ (Lagrange multiplier) such that

$$\begin{aligned} \begin{aligned} w^C\frac{h^C_ig^C_i P_i}{(\sigma ^2+h^C_i P_i+ g^C_i J_i)(\sigma ^2+g^C_i J_i)}+w^R\frac{h^R_i g^R_i P_i}{(\sigma ^2+g^R_i J_i)^2} {\left\{ \begin{array}{ll} =\nu ,&{} J_i>0,\\ \le \nu ,&{} J_i=0. \end{array}\right. } \end{aligned} \end{aligned}$$

(10)

Then, (9) and (10) imply that $\omega$ and $\nu$ are positive. By (10) if $P_i=0$ then $J_i=0$. Thus, to find ${\varvec{P}}$ and ${\varvec{J}}$ we have to consider only three cases: (a) $P_i=0, J_i=0$, (b) $P_i=0, J_i>0$, and (c) $P_i>0, J_i>0$.

(a) Let $P_i=0, J_i=0$. Then, by (9) and (10), $w^Ch^C_i/\sigma ^2+w^Rh^R_i/\sigma ^2\le \omega$. Thus, $i\in I_{00}(\omega ,\nu )$, and (I-a) follows.

(b) Let $P_i>0, J_i=0$. Then, by (9) and (10), we have that

$$\begin{aligned} w^Ch^C_i/(\sigma ^2+h^C_i P_i)+w^R h^R_i/\sigma ^2&=\omega , \end{aligned}$$

(11)

$$\begin{aligned} w^Ch^C_ig^C_i P_i/((\sigma ^2+h^C_i P_i)\sigma ^2)+w^Rh^R_i g^R_i P_i/(\sigma ^2)^2&\le \nu . \end{aligned}$$

(12)

By (11), $P_i=P_i(\omega ,\nu )$ is given by (1). Note that, $P_i$ is decreasing with respect to $\omega$. By (1), $P_i>0$ (this holds by assumption of (b)) if and only if:

$$\begin{aligned} w^R h^R_i/\sigma ^2<\omega <w^C h^C_i/\sigma ^2+w^R h^R_i/\sigma ^2. \end{aligned}$$

(13)

Substituting (1) into (12) yields that

$$\begin{aligned} \frac{w^Cg^C_i}{\sigma ^2}+\frac{w^Cw^Rh^R_ig^R_i}{\sigma ^4\left( \omega -w^Rh^R_i/ \sigma ^2\right) } \le \nu +\frac{w^Rh^R_ig^R_i}{\sigma ^2 h^C_i}+\frac{g^C_i}{h^C_i}\left( \omega -\frac{w^Rh^R_i}{\sigma ^2}\right) . \end{aligned}$$

(14)

The left side of (14) is decreasing with respect to $\omega$ from infinity for $\omega =w^Rh^R_i/\sigma ^2$ to $A_L:=w^Cg^C_i/\sigma ^2+w^Rh^R_ig^R_i/(\sigma ^2 h^C_i)$ for $\omega =w^Ch^C_i/\sigma ^2+w^Rh^R_i/\sigma ^2$.

The right side of (14) is increasing with respect to $\omega$ from $\nu +w^Rh^R_ig^R_i/h^C_i$ for $\omega =w^Rh^R_i/\sigma ^2$ to $A_R:=\nu +w^Rh^R_ig^R_i/(\sigma ^2 h^C_i)+w^Chg^C_i/\sigma ^2 =\nu +A_L>A_L$ for $\omega =w^C h^C_i/\sigma ^2+w^R h^R_i/\sigma ^2$. Thus, for any positive $\nu$ there is a unique $\omega =\omega _{+,i}(\nu )$ such that (13) holds, while (14) holds as equality. It is clear that $\omega _{+,i}(\nu )$ is decreasing on $\nu$. Since this is a quadratic equation on $\omega$, $\omega _{+,i}(\nu )$ can be found in closed form, by (2), and (II-b) follows.

(c) Let $P_i>0, J_i>0$. Then, by (9) and (10) we have that

$$\begin{aligned}&w^Ch^C_i/(\sigma ^2+h^C_i P_i+ g^C_i J_i)+w^Rh^R_i/(\sigma ^2+g^R_i J_i)=\omega ,\end{aligned}$$

(15)

$$\begin{aligned}&w^Ch^C_ig^C_i P_i/((\sigma ^2+h^C_i P_i+ g^C_i J_i)(\sigma ^2+g^C_i J_i))+w^Rh^R_i g^R_i P_i/(\sigma ^2+g^R_i J_i)^2=\nu . \end{aligned}$$

(16)

By (15), we have that

$$\begin{aligned} P_i=w^C/(\omega - w^Rh^R_i/(\sigma ^2+g^R_i J_i))-(\sigma ^2+g^C_i J_i)/h^C_i. \end{aligned}$$

(17)

By (17), $P_i$ is decreasing with respect to $J_i$. Substituting (17) into (16) implies (3). The left side of (3) is decreasing with respect to $J_i$ and tends to zero while $J_i$ tends to infinity. Thus, for each $\omega$ and $\nu$, (3) has a root (which is unique) if and only if:

$$\begin{aligned} \frac{w^Cg^C_i}{\sigma ^2}+\frac{w^Cw^Rh^R_ig^R_i}{\sigma ^4\left( \omega - w^Rh^R_i/ \sigma ^2\right) } > \nu +\frac{w^Rh^R_ig^R_i}{\sigma ^2 h^C_i}+\frac{g^C_i}{h^C_i}\left( \omega -\frac{w^Rh^R_i}{\sigma ^2}\right) . \end{aligned}$$

(18)

By (14), the condition (18) is equivalent to $\omega <\omega _{+,i}(\nu )$. Denote this root by $J_i(\omega ,\nu )$. Then, substituting this $J_i(\omega ,\nu )$ into (17) we can uniquely define $P_i$ denoted by $P_i(\omega ,\nu )$, and (I-c) follows.

Note that, by (3), $J_i(\omega ,\nu )$ is decreasing on $\omega$ and $\nu$. The left side of (16) is increasing with respect to $P_i$ and decreasing with respect to $J_i$. Thus, the fact that $J_i(\omega ,\nu )$ is decreasing with respect to $\omega$ implies that $P_i(\omega ,\nu )$ is also decreasing with respect to $\omega$. Also, the left side of (15) is decreasing on $P_i$ and on $J_i$. Thus, the fact that $J_i(\omega ,\nu )$ is decreasing on $\nu$ implies that $P_i(\omega ,\nu )$ is increasing on $\nu$. Thus, $H_J(\omega ,\nu )$ is continuous and decreasing on $\omega$ and $\nu$, while $H_S(\omega ,\nu )$ is continuous and decreasing on $\omega$ and increasing on $\nu$. These monotonous properties yields that solution of (5) is the unique, and (I) follows.

(II) If $w^C=0$ then (9) implies (6). Thus, $J_i(\nu )$ is defined uniquely. Substituting this $J_i(\nu )$ into (9) and taking into account that ${\varvec{P}}\in \varPi _S$ implies the result.