Energy-based adaptive matching pursuit algorithm for binary sparse signal reconstruction in compressed sensing

Abstract

Compressed sensing gains great attention in the field of signal reconstruction. In order to deal with some practical cases in which the sparsity levels are unknown, this paper proposes an energy-based adaptive matching pursuit (EAMP) algorithm for binary sparse signal reconstruction in the compressed sensing framework. The EAMP algorithm inherits the feature of the sparsity adaptive matching pursuit algorithm, which increases the estimated sparsity level when the energy of the observation residue increases. Meanwhile, the proposed algorithm introduces the measurement vector into the signal reconstruction process. It uses two kinds of step sizes to increase the estimated sparsity level when the energy of the estimated candidate signal is less than half of that of the measurement vector. The experimental results indicate that the proposed EAMP algorithm provides better reconstruction performance than other greedy algorithms.

Appendices

Appendix 1: The fixed step size $s_{fix}< \frac{K}{2}$

Let $T$ denote the support of original signal and $i$ be the iteration index. It is assumed that the measurement matrix satisfies the RIP and the $mth$ iteration is the last one of the stage $\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}$ . Note that $1^{2}+1^{2}+0+0+1^{2}=1+1+0+0+1$. Let $\mathbf {z}$ be a $0-1$ binary signal, we have

$$\begin{aligned} \Vert \mathbf {z}\Vert ^{2}_{2}=\Vert \mathbf {z}\Vert _{1}=\Vert \mathbf {z}\Vert _{0} \end{aligned}$$

(8)

When $\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}$ , the step size is increased by $s^{i}=2^{i}$ and the estimated sparsity level is updated by $k^{i+1}=k^{i}+s^{i}$. For example,

$$\begin{aligned}&k^{1}=2, \quad k^{2}=k^{1}+s^{1}=2+2=4\\&k^{3}=k^{2}+s^{2}=4+2^{2}=8 \quad k^{4}=k^{3}+2^{3}=16. \end{aligned}$$

Therefore,

$$\begin{aligned} k^{i}=2^{i}. \end{aligned}$$

(9)

When $\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}$, by Theorem 1 and (8), one has

$$\begin{aligned} \Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}<\frac{1}{2}\Vert \mathbf {x}_{T}\Vert ^{2}_{2}. \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf {x}_{\Omega }\Vert _{0}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}<\frac{1}{2}\Vert \mathbf {x}_{T}\Vert _{0}. \end{aligned}$$

Furthermore, when $i\le m$, we get

$$\begin{aligned} k^{i}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}<\frac{1}{2}K. \end{aligned}$$

(10)

The step size $s_{fix}$ is fixed at the $mth$ iteration with $s_{fix}=s^{m}$.

Therefore, at the $mth$ iteration

$$\begin{aligned} s_{fix}=s^{m}=k^{m}=2^{m}. \end{aligned}$$

By invoking (10), it follows that

$$\begin{aligned} k^{m}<\frac{1}{2}K. \end{aligned}$$

(11)

Hence,

$$\begin{aligned} s_{fix}<\frac{1}{2}K. \end{aligned}$$

(12)

which completes the proof.

Appendix 2: The procedure will move into the stage of$\frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}<\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{2}\Vert \mathbf {y}\Vert ^{2}_{2}$ as soon as the step size is fixed

The proof is proved by contradiction. It is still assumed that the $m$th iteration is the last one of the stage $\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}$. Suppose that the procedure will move into the stage of $\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}>\frac{1}{2}\Vert \mathbf {y}\Vert ^{2}_{2}$ at the $(m+1)$th iteration.

At the $(m+1)th$ iteration, $\Vert \mathbf {x}_{\Omega }\Vert _{0}=k^{m+1}$ and $\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}>\frac{1}{2}\Vert \mathbf {y}\Vert ^{2}_{2}.$

By invoking (8), it follows that

$$\begin{aligned}&k^{m+1}>\frac{1}{2}\Vert \mathbf {y}\Vert ^{2}_{2}.\end{aligned}$$

(13)

$$\begin{aligned}&k^{m+1}-k^{m}>\frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}. \end{aligned}$$

(14)

This result is a combination of (10) and (13). Therefore,

$$\begin{aligned} s_{fix}> \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}. \end{aligned}$$

(15)

According to “Appendix 1,” $s_{fix}=k^{m}$ and $k^{m}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}$. Therefore,

$$\begin{aligned} s_{fix}\le \frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}. \end{aligned}$$

(16)

It is perceived that (15) clearly contradicts (16). Thus, the process will move into the stage of $\frac{1}{4}\Vert \mathbf {y}\Vert ^{2}_{2}<\Vert \mathbf {x}_{\Omega }\Vert ^{2}_{2}\le \frac{1}{2}\Vert \mathbf {y}\Vert ^{2}_{2}$ as soon as the step size is fixed.