Adventures in Compressive Sensing Based MIMO Radar

Abstract

While radar has been around for many decades, novel developments in recent years have led to significant breakthroughs as well as to exciting new mathematical challenges. In this chapter, we consider a multiple-input-multiple-output (MIMO) radar system. Using sparsity as a key ingredient of our approach and tools from compressive sensing, we derive a mathematical framework for the imaging of targets in the azimuth-range-Doppler domain. Our analysis comprises uniformly spaced linear arrays with random waveforms, as well as random sensor arrays with deterministic waveforms. We also derive results that do not require the “on-the-grid” assumption often used in compressive sensing radar. Algorithmic aspects and numerical simulations are presented as well.

Appendices

Appendix A

In this appendix we collect some auxiliary results.

Lemma 7.

[38, Proposition 34] Let ${\bf x} \in{\mathbb C}^n$ be a vector with $x_k \sim{\cal CN}(0,\sigma^2)$ , then for every $t>0$ one has

$${\mathbb P} \Big( \|{\bf x}\|_2 - {\mathbb E} \|{\bf x}\|_2> t \Big) \le e^{-\frac{t^2}{2\sigma^2}}.$$

(113)

The following lemma is a rescaled version of Lemma 3.1 in [29].

Lemma 8.

Let ${\bf A} \in{\mathbb C}^{n \times m}$ be a Gaussian random matrix with $A_{i,j} \sim{\cal CN}(0,\sigma^2)$ . Then for all ${\bf x}, {\bf y} \in{\mathbb C}^m$ with $\|{\bf x}\|_2 = \|{\bf y}\|_2=\sqrt{m}$ and all $t>0$

$${\mathbb P} \Big\{ |\frac{\sigma^2}{n}\langle{\bf A}{\bf x}, {\bf A}{\bf y} \rangle - \langle{\bf x},{\bf y} \rangle|> tm \Big\} \le 2 {\rm exp} \Big(-n\frac{t^2}{C_1+C_2t}\Big),$$

(114)

with $C_1 = \frac{4e}{\sqrt{6\pi}}$ and $C_2 = \sqrt{8}e$.

For convenience we state the following version of Bernstein’s inequality, which will be used in the proof of Lemma 10.

Lemma 9 (See e.g. [37])

Let $X_1,\dots,X_n$ be independent random varibles with zero mean such that

$${\mathbb E} |X_i|^p \le \frac{1}{2} p! K^{p-2} v_i, \qquad \text{for all $i=1,\dots,n; p\in{\mathbb N}, p\ge 2$},$$

(115)

for some constants $K>0$ and $v_i> 0, i=1,\dots,n$ . Then, for all $t>0$

$${\mathbb P}\Big( \big|\sum_{i=1}^{n} X_i | \ge t \big) \le 2 {\rm exp}\Big( - \frac{t^2}{2v + Kt}\Big),$$

(116)

where $v:=\sum_{i=1}^{n} v_i$.

We also need the following deviation inequality for unbounded random variables. It is a complex-valued and slightly sharpened version of Lemma 6 in [13]. Our proof strategy differs at certain steps from that of Lemma 6 in [13] (and our proof is a bit shorter).

Lemma 10.

Let x _i and y _i , $i=1,\dots,n$ , be sequences of i.i.d. complex Gaussian random variables with variance σ. Then,

$${\mathbb P}\Big( \big|\sum_{i=1}^{n} \bar{x}_i y_i \big|> t \Big) \le 2 {\rm exp} \big(-\frac{t^2}{\sigma^2 (n \sigma^2 + 2t)}\big).$$

(117)

Proof

In order to apply Bernstein’s inequality, we need to compute the moments ${\mathbb E} |x_i y_i|^p$. Since x _i and y _i are independent, there holds

$${\mathbb E} (|x_i y_i|^p) = {\mathbb E} (|x_i|^p) {\mathbb E} (|y_i|^p) =({\mathbb E} (|x_i|^p ))^2.$$

(118)

The moments of x _i are well-known:

$${\mathbb E} |x_i|^{2p} = p! \, \sigma^{2p},$$

(119)

hence

$$({\mathbb E} |x_i|^{2p})^2 = (2p!)^2 (\sigma^{2p})^2 \le \frac{1}{4} (2p)! (\sigma^{2})^{2p} \le \frac{1}{2} (2p)! (\sigma^{2})^{2p-2} \frac{(\sigma^{2})^2}{2}.$$

(120)

We apply Bernstein’s inequality (116) with $K= \sigma^2$ and $v_i = \frac{(\sigma^{2})^2}{2}, i=1,\dots,n$ and obtain (117).

Lemma 11.

Suppose M is an $m\times m$ matrix, α and β are two joint independent random vectors in ${\mathbb C}^m$ with zero means and $|\alpha_k|=|\beta_k|=1$ for $k=1,\dots, m$ . If n is a positive constant, then for any $t>0$ and $s>0$,

• 1. if $|m_{kj}|\le\frac{1}{\sqrt{n}}$ for all $k, j$ , then

  $$\begin{aligned}\mathbb P\Big( |\langle M\alpha, \beta\rangle|\le mt \Big)\ge1-4m{\rm exp}\Big(-\frac{t^2}{4\frac{m}{n}}\Big).\end{aligned}$$

  (121)

  and

  $$\begin{aligned}\mathbb P\Big( |\langle M\alpha, \alpha\rangle|\le 2mt \Big)\ge1-8m{\rm exp}\Big(-\frac{t^2}{2\frac{m}{n}}\Big),\end{aligned}$$

  (122)

• 2. if $|m_{kj}|\le \frac{1}{\sqrt{n}}$ for $k\neq j$ and $m_{jj}=1$ , then

  $$\begin{aligned}\mathbb P\Big( |\langle M\alpha, \beta\rangle|\le s+mt \Big)\ge1-4{\rm exp}\Big(-\frac{s^2}{4m}\Big)-4m{\rm exp}\Big(-\frac{t^2}{4\frac{m}{n}}\Big),\end{aligned}$$

  (123)

  and

  $$\begin{aligned}\mathbb P\Big( m(1-2t)\le|\langle M\alpha, \alpha\rangle|\le m(1+2t) \Big)\ge1-8m{\rm exp}\Big(-\frac{t^2}{2\frac{m}{n}}\Big).\end{aligned}$$

  (124)

Proof

Note that

$$\begin{aligned} \langle M\alpha, \beta\rangle&=\sum_{k,j=1}^mm_{kj}\alpha_j\bar\beta_k\\ &=\sum_{l=1}^m\sum_{j=1}^mm_{j\oplus l,j}\alpha_j\bar\beta_{j\oplus l},\end{aligned}$$

where ⊕ denotes addition modulo m.

Let us first assume that $|m_{kj}|\le\frac{1}{\sqrt{n}}$.

Since α and β are jointly independent, then for any l, the entries in $\sum_{j=1}^mm_{j\oplus l, j}\\\alpha_j\bar\beta_{j\oplus l}$ are all jointly independent and it is easy to check that ${\mathbb E}(m_{j\oplus l, j}\alpha_j\bar\beta_{j\oplus l})=0$ and $|m_{j\oplus l, j}\alpha_j\bar\beta_{j\oplus l}|= |m_{j\oplus l, j}|$, then Theorem 4.5 in [18] will give

$$\begin{aligned} {\mathbb P}\Big( |\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\beta_{j\oplus l}|\le t \Big)&\ge 1-4{\rm exp}\Big(-\frac{t^2}{4\sum_{j}|m_{j\oplus l, j}|^2}\Big)\nonumber\\ &\ge1-4{\rm exp}\Big(-\frac{t^2}{4\frac{m}{n}}\Big).\end{aligned}$$

(125)

We take all m different choices of l, then

$$\begin{aligned}\mathbb P\Big( |\sum_{l=1}^m\sum_{j=1}^mm_{i\oplus l, j}\alpha_j\bar\beta_{j\oplus l}|\le mt \Big)\ge1-4m{\rm exp}\Big(-\frac{t^2}{4\frac{m}{n}}\Big),\end{aligned}$$

(126)

which proves (121).

Now consider

$$ \langle M\alpha, \alpha\rangle=\sum_{l=1}^m\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l},$$

but different from above, the entries in $\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l}$ are no longer all jointly independent. But similar to the proof of Theorem 5.1 in [27] and Lemma 3 in [31], we observe that for any l we can split the index set ${1,\dots,m}$ into two subsets $T_l^1,T_l^2\subset \{1,\dots,m\}$, each of size $m/2$, such that the $m/2$ variables $\alpha_j\bar\alpha_{j\oplus l}$ are jointly independent for $j\in T^1_l$, and analogous for $T^2_l$. (For convenience we assume here that m is even, but with a negligible modification the argument also applies for odd m.) In other words, each of the sums $\sum_{j\in T^r_l} m_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l}, r=1,2$, contains only jointly independent terms.

So for each l,

$${\mathbb P}\Big(|\sum_{j\in T^r_l} m_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l}| \le t \Big) \ge 1-4{\rm exp}\Big(-\frac{t^2}{2\frac{m}{n}}\Big),$$

(127)

which implies that

$$\begin{aligned} {\mathbb P}\Big(|\sum_{j} m_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l}| \le 2t\Big) & \ge 1-8 {\rm exp}\Big(-\frac{t^2}{2\frac{m}{n}}\Big),\end{aligned}$$

(128)

Again, we take all m different choices of l, then

$$\begin{aligned} {\mathbb P}\Big( |\sum_{l=1}^m\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l}|\le 2mt \Big)\ge1-8m{\rm exp}\Big(-\frac{t^2}{2\frac{m}{n}}\Big),\end{aligned}$$

(129)

which proves (122).

Now let us assume that $|m_{kj}|\le \frac{1}{\sqrt{n}}$ for $k\neq j$ and $m_{jj}=1$.

$$\begin{aligned} \langle M\alpha, \beta\rangle&=\sum_{j=1}^mm_{jj}\alpha_j\bar\beta_{j}+\sum_{l=1}^{m-1}\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\beta_{j\oplus l}\\ &=\sum_{j=1}^m\alpha_j\bar\beta_{j}+\sum_{l=1}^{m-1}\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\beta_{j\oplus l}.\end{aligned}$$

Since α and β are joint independent and $|\alpha_j\bar\beta_j|=1$,

$$\begin{aligned}\mathbb P\Big( |\sum_{j=1}^m\alpha_j\bar\beta_{j}|\le s \Big)\ge1-4{\rm exp}\Big(-\frac{s^2}{4m}\Big).\end{aligned}$$

(130)

Similar to the proof of (126) above, we have that

$$\begin{aligned} {\mathbb P}\Big( |\sum_{l=1}^{m-1}\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\beta_{j\oplus l}|\le (m-1)t \Big)\ge1-4(m-1){\rm exp}\Big(-\frac{t^2}{4\frac{m}{n}}\Big),\end{aligned}$$

together with (130), it follows

$$\begin{aligned} {\mathbb P}\Big( |\langle M\alpha, \beta\rangle|\le s+(m-1)t \Big)\ge1-4{\rm exp}\Big(-\frac{s^2}{4m}\Big)-4(m-1){\rm exp}\Big(-\frac{t^2}{4\frac{m}{n}}\Big),\end{aligned}$$

which proves (123).

Finally,

$$ \langle M\alpha, \alpha\rangle=\sum_{j=1}^mm_{jj}+\sum_{l=1}^{m-1}\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l}=m+\sum_{l=1}^{m-1}\sum_{j=1}^mm_{j\oplus l, j}\alpha_j\bar\alpha_{j\oplus l},$$

then (124) results from similar proof as for (122) and the triangle inequality.

Appendix B

We consider a general linear system of equations ${\Psi} {\bf x} = {\bf y}$, where ${\Psi} \in{\mathbb C}^{n \times m}$, ${\bf x} \in{\mathbb C}^m$ and $n \le m$. We introduce the following generic K-sparse model:

• The support $I \subset \{1,\dots,m\}$ of the K nonzero coefficients of ${\bf x}$ is selected uniformly at random.

• The non-zero entries of ${\operatorname{sgn}}({\bf x})$ form a Steinhaus sequence, i.e., ${\operatorname{sgn}}({\bf x}_k):={\bf x}_k/|{\bf x}_k|, k\in I,$ is a complex random variable that is uniformly distributed on the unit circle.

The following theorem is a slightly extended version of Theorem 1.3 in [5], see [31] for its proof.

Theorem 5.

Given ${\bf y} = {\Psi} {\bf x} + {\bf w}$ , where ${\Psi}$ has all unit-ℓ ₂ -norm columns, ${\bf x}$ is drawn from the generic K-sparse model and ${\bf w}_i \sim{\cal CN}(0,\sigma^2)$ . Assume that

$$\mu({\Psi}) \le \frac{C_0}{\log m},$$

(131)

where $C_0>0$ is a constant independent of $n,m$ . Furthermore, suppose

$$K \le \frac{c_0 m}{\|{\Psi} \|_{\text{op}}^2 \log m}$$

(132)

for some constant $c_0> 0$ and that

$$\underset{k\in I}{\min}\, |{\bf x}_k|> 8 \sigma \sqrt{2 \log m}.$$

(133)

Then the solution $\hat{\bf x}$ to the debiased lasso computed with $\lambda = 2 \sigma \sqrt{2 \log m}$ obeys

$$\operatorname{supp} (\hat{\bf x}) = {\operatorname{supp}} ({\bf x}),$$

(134)

and

$$\frac{\|\hat{\bf x} - {\bf x} \|_2}{\|{\bf x}\|_2} \le \frac{\sigma \sqrt{3 n}}{\|{\bf y}\|_2}$$

(135)

with probability at least

$$1 - 2m^{-1}(2\pi \log m + Km^{-1}) - {\cal O}(m^{-2 \log 2}).$$

(136)