Experimental investigations on the applicability of LASSO based sparse recovery for compressively sensed OFDM channel estimation

Abstract

Owing to its success in 4G, Orthogonal Frequency Division Multiplexing (OFDM) made its contenting presence in 5G communications. To accommodate massive 5G data, it would be quite opportune to maintain significant reduction in overhead, while also ensuring reliable symbol detection at the receiver. For instance, reducing pilot symbol transmissions required for OFDM channel estimation, still retaining the efficacy of symbol detection is propitious. In this regard, contemporary compressive sensing framework with sparse recovery approaches such as Orthogonal Matching Pursuit (OMP) and Sparse Bayesian Learning (SBL) have been widely investigated in the literature. However study on approaches such as Least Absolute Shrinkage and Selection Operator (LASSO), Split Augmented Lagrangian Shrinkage Algorithm (SALSA) and corresponding investigations with experimentally acquired data frames are very less explored. In this work, by modifying the group LASSO optimization, a sparse recovery approach is proposed for OFDM channel estimation, particularly for a severely less pilot scenario. A non-coherent OFDM wireless link is developed using two Universal Software Radio Peripheral (USRP) devices, and several complex baseband OFDM frames are extracted in an indoor wireless lab environment for different receiver gains. Sparse recovery approaches are incorporated in two scenarios, namely number of pilots less than and severely less than channel length. The results of Mean Square Error (MSE) and Bit Error Rate (BER) versus receiver gain, show that the proposed approach outperforms all the other considered approaches. Particularly, while estimating CIR of length 20 with the proposed approach, employing 17 pilots resulted in BER of 0.0050, which degraded to only 0.0099 in a 5 pilot scenario, thus accommodating severe reduction in pilot overhead.

Proposed solution of group LASSO

Substituting $F({\textbf {h}})$ from (20) and the chosen $G({\textbf {h}})$ in (21), the modified Lagrangian $J({\textbf {h}})$ becomes

$$\begin{aligned} J({\textbf {h}})&=\frac{1}{2}\left\| {\textbf {y}}_{f,p} - \mathbf {\Phi }{} {\textbf {h}} \right\| _2 ^2 + \lambda \sum _{k=1}^{G} {\textbf {d}}_k \left\| {\textbf {h}}_k \right\| _2\nonumber \\ {}&\quad + \sum _{k=1}^{G} \frac{1}{2} ({\textbf {h}}_k -{\textbf {h}}_k ^{i})^T (\alpha _k {\textbf {I}}_L-\mathbf {\Phi }_k^H\mathbf {\Phi }_k) ({\textbf {h}}_k -{\textbf {h}}_k ^{i}) \end{aligned}$$

(A1)

,where the first term is split using $\mathbf {\Phi }_{-k}$, ${\textbf {h}}_{-k}$ and $\mathbf {\Phi }_{k}$, ${\textbf {h}}_k$ as

$$\begin{aligned}&\sum _{k=1}^{G} \left\{ \frac{1}{2}\left\| \textbf{y}_{f,p}-\mathbf {\Phi }_{-k} \textbf{h}_{-k}-\mathbf {\Phi }_{k} \textbf{h}_{k}\right\| _2^2+ \lambda {\textbf {d}}_k \left\| {\textbf {h}}_k \right\| _2 \right. \nonumber \\ {}&\quad + \left. \frac{1}{2} ({\textbf {h}}_k -{\textbf {h}}_k ^{i})^T (\alpha _k {\textbf {I}}_L-\mathbf {\Phi }_k^H\mathbf {\Phi }_k)({\textbf {h}}_k -{\textbf {h}}_k ^{i})\right\} \end{aligned}$$

(A2)

Vector differentiating (A2) for ${\textbf {h}}_k \ne {\textbf {0}}$ we get $\frac{\partial J(\textbf{h})}{\partial \textbf{h}_k}$ as

$$\begin{aligned} \frac{\lambda d_k \textbf{h}_k}{\left\| {\textbf {h}}_k \right\| _2} -\mathbf {\Phi }_k^H (\textbf{r}_k-\mathbf {\Phi }_k\textbf{h}_k) +(\alpha _k {\textbf {I}}_L - \mathbf {\Phi }_k^H \mathbf {\Phi }_k) (\textbf{h}_k- {\textbf{h}_k}^{i}) \end{aligned}$$

(A3)

Equating (A3) to zero, and simplifying it further, we get

$$\begin{aligned} \left( 1+ \frac{\lambda d_k }{\alpha _k \left\| {\textbf {h}}_k \right\| _2} \right) {\textbf {h}}_k= \frac{1}{\alpha _k} \left[ \mathbf {\Phi }_k^H \left( \textbf{r}_k - \mathbf {\Phi }_k \textbf{h}_k^i \right) \right] +\textbf{h}_k^i \end{aligned}$$

(A4)

Notating the right hand side of (A4) as $\textbf{s}_k$, the solution is obtained as

$$\begin{aligned} {\textbf {h}}_k = \left( 1+ \frac{\lambda d_k }{\alpha _k \left\| {\textbf {h}}_k \right\| _2} \right) ^{-1} \textbf{s}_k \end{aligned}$$

(A5)

However, (A5) has to be simplified to remove the dependency of $\left\| {\textbf {h}}_k \right\| _2$ in it. Taking the norm of (A5), we get

$$\begin{aligned} \left\| {\textbf {h}}_k \right\| _2 = \left( 1+\frac{\lambda d_k}{\alpha _k \left\| {\textbf {h}}_k \right\| _2 } \right) ^{-1} \left\| \textbf{s}_k \right\| _2 \end{aligned}$$

(A6)

where $\lambda , d_k, \alpha _k$ are all positive constants and hence the inverse term, which is a scalar is written outside the norm operation. Simplifying (A6), we obtain

$$\begin{aligned} \left\| \textbf{h}_k \right\| _2 = \left\| \textbf{s}_k \right\| _2 - \frac{\lambda d_k}{\alpha _k} \end{aligned}$$

(A7)

Substituting (A7) in the right hand side term of (A5) and simplifying it, the solution becomes

$$\begin{aligned} \textbf{h}_k = \left( 1-\frac{\lambda d_k}{\alpha _k \left\| {\textbf {s}}_k \right\| _2} \right) {\textbf {s}}_k \end{aligned}$$

(A8)

Note that (A8) is valid when ${\textbf {h}}_k \ne {\textbf {0}}$. If ${\textbf {h}}_k = {\textbf {0}}$, $\left\| \textbf{h}_k \right\| _2$ becomes non-differentiable. In this case, the subdifferential of $\left\| \textbf{h}_k \right\| _2$ is considered. Correspondingly, for the subgradient vector ${\textbf {v}}_k$ of $\left\| \textbf{h}_k \right\| _2$ should satisfy

$$\begin{aligned} \lambda d_k {\textbf {v}}_k -\mathbf {\Phi }_k^H (\textbf{r}_k-\mathbf {\Phi }_k\textbf{h}_k) +(\alpha _k {\textbf {I}}_L - \mathbf {\Phi }_k^H \mathbf {\Phi }_k) (\textbf{h}_k- {\textbf{h}_k}^{i}) = {\textbf {0}} \end{aligned}$$

(A9)

and further using the condition that $\left\| {\textbf {v}}_k \right\| _2 \le 1$, when ${\textbf {h}}_k={\textbf {0}}$ [49]. Note that (A9) is similar to (A3) except that the sub differentiation of $\left\| \textbf{h}_k \right\| _2$ is denoted as ${\textbf {v}}_k$, in place of differentiation of $\left\| \textbf{h}_k \right\| _2$ in (A3). Simplifying (A9), we get ${\textbf {v}}_k=\frac{\alpha _k}{\lambda d_k}{} {\textbf {s}}_k$ and simplifying it with $\left\| {\textbf {v}}_k \right\| _2 \le 1$, we get

$$\begin{aligned} \frac{\lambda d_k}{\alpha _k \left\| {\textbf {s}}_k \right\| } \ge 1, ~ \textbf{h}_k= \textbf{0} \end{aligned}$$

(A10)

Combining step (A8) to include the condition (A10), the closed from expression for the solution of group LASSO optimization in (20) is

$$\begin{aligned} \textbf{h}_k = \left( 1-\frac{\lambda d_k}{\alpha _k \left\| {\textbf {s}}_k \right\| _2 } \right) _\mathbf {+}{} {\textbf {s}}_k \end{aligned}$$

(A11)

where $\left( a\right) _\mathbf {+}$ indicates soft thresholding operation.