Optimal Beamforming for Multiuser Secure SWIPT Systems (Invited Paper)

Abstract

In this paper, we study the beamforming design for simultaneous wireless information and power transfer (SWIPT) downlink systems. The design is formulated as a non-convex optimization problem which takes into account the quality of service (QoS) requirements of communication security and minimum harvested power. In particular, the proposed design advocates the dual use of energy signal to enable secure communication and efficient WPT. The globally optimal solution of the optimization problem is obtained via the semidefinite programming relaxation (SDR). Our simulation results show that there exists a non-trivial tradeoff between the achievable data rate and the total harvested power in the system. Besides, our proposed optimal scheme provides a substantial performance gain compared to a simple suboptimal scheme based on the maximum ratio transmission (MRT).

6 Appendix – Proof of Theorem 1

It can be verified that the transformed optimization problem in (10) is convex and satisfies the Slater’s constraint qualification, hence, strong duality holds. In other words, solving its dual problem is equivalent to solving the primal problem. In this section, we intend to prove Theorem 1 via first defining the Lagrangian function:

$$\begin{aligned} L= & {} -{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {W}\mathbf {H})-\lambda _{\mathrm {C1}}{{\,\mathrm{\mathrm {Tr}}\,}}((\mathbf {W}+\mathbf {W}_{\mathrm {E}})\mathbf {G})+\lambda _{\mathrm {C2}}{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {W}+\mathbf {W}_{\mathrm {E}})\nonumber \\+ & {} \lambda _{\mathrm {C3}}{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {W}\mathbf {G}_{j})-{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {Y}\mathbf {W})+\varDelta , \end{aligned}$$

(11)

where $\varDelta$ represents the variables and the constants that are independent of $\mathbf {W}$ and therefore irrelevant in the proof. $\mathbf {Y}$ and $\lambda _{\mathrm {C1}}$, $\lambda _{\mathrm {C2}}$, $\lambda _{\mathrm {C3}}$ are dual variables related to the constraints C4 and C1, C2, C3, respectively. Now, we can express the dual problem of (10) as

$$\begin{aligned} \underset{\mathbf {Y},\lambda _{\mathrm {C1}}, \lambda _{\mathrm {C2}}, \lambda _{\mathrm {C3}}}{\max }\,\,\underset{\mathbf {W},\mathbf {W}_{\mathrm {E}}\in \mathbb {H}^{N_{\mathrm {T}}}}{\min } \quad L . \end{aligned}$$

(12)

Then, we study the structure of $\mathbf {W}$ via applying the Karush-Kuhn-Tucker (KKT) conditions:

where (14) is obtained by taking the derivative of the Lagrangian function with respect to $\mathbf {W}$ and $\mathbf {B} = -\lambda _{\mathrm {C1}}\mathbf {G}+\lambda _{\mathrm {C2}}\mathbf {I}+\lambda _{\mathrm {C3}}\mathbf {G}_{j}$. Equation (13) is the complementary slackness property which implies that the columns of matrix $\mathbf {W}$ fall into the null-space spanned by $\mathbf {Y}$ for $\mathbf {W}\ne {{\,\mathrm{\mathbf {0}}\,}}$. Hence, if we can prove that ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {Y}) \ge N_{\mathrm {T}}-1$, the optimal beamforming matrix $\mathbf {W}$ is a rank-one matrix or a zero matrix. Now, we study the structure of $\mathbf {Y}$ via examining (15). First, we prove by contradiction that $\mathbf {B}$ is a positive definite matrix with probability one. Suppose not, $\mathbf {B}$ is a positive semi-definite matrix. Then, there exist at least one zero eigenvalue and we denote the associated eigenvector as $\mathbf {v}$. Without loss of generality, we create a matrix $\mathbf {V} = \mathbf {v}\mathbf {v}^H$ from the eigenvector. By multiplying both sides of (15) with $\mathbf {V}$ and applying the trace operation, we obtain

$$\begin{aligned} {{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {Y} \mathbf {V})= & {} -{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {H}\mathbf {V})+{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {B}\mathbf {V})= -{{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {H}\mathbf {V}). \end{aligned}$$

(16)

Since $\mathbf {H}$ and $\mathbf {G}_j$ are statistically independent, we have ${{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {H}\mathbf {V})>0$. This leads to contradiction as ${{\,\mathrm{\mathrm {Tr}}\,}}(\mathbf {Y} \mathbf {V})\ge 0$. Hence, matrix $\mathbf {B}$ is a positive definite matrix², i.e., ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {B})=N_{\mathrm {T}}$. To further proceed the proof, we introduce the following rank inequality:

Lemma 1

Let $\mathbf {A}$ and $\mathbf {B}$ be two matrices with same dimension. The inequality of matrix ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {A}+\mathbf {B})\ge {{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {A})-{{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {B})$ holds.

Proof:

By basic rule of inequality for the rank of matrix, ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {A})+{{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {B})\ge {{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {A}+\mathbf {B})$ with both matrices of same dimension. Thus we have ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {A}+\mathbf {B})+{{\,\mathrm{\mathrm {Rank}}\,}}(-\mathbf {B})\ge {{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {A})$. Since ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {B})={{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {-B})$, the lemma is proved.

   $\square$

Now, we apply Lemma 1 on (14) which yields:

$$\begin{aligned} {{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {Y})= & {} {{\,\mathrm{\mathrm {Rank}}\,}}(-\mathbf {Y})={{\,\mathrm{\mathrm {Rank}}\,}}(-\mathbf {B}+\mathbf {H})\nonumber \\\ge & {} {{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {-B})-{{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {H})=N_{\mathrm {T}}-1 \end{aligned}$$

(17)

As ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {Y})\ge N_{\mathrm {T}}-1$, we have ${{\,\mathrm{\mathrm {Rank}}\,}}(\mathbf {W})\le 1$ which completes the proof.    $\square$