Joint Atomic Norm Based Estimation of Sparse Time Dispersive SIMO Channels with Common Support in Pilot Aided OFDM Systems

Abstract

We consider the problem of estimation of sparse time dispersive Single Input Multiple Output (SIMO) channels, using a single transmit and multiple receive antennas in pilot aided OFDM systems. The channels we consider are with a continuous time delays and sparse, and we assume a common support of the channel coefficients of the SIMO channels associated with different antennas, resulting from the same scatterer. To exploit these properties, we propose a new channel estimation algorithm based on the atomic norm minimization for the Multiple Measurement Vector (MMV) model. A joint estimation of the delays corresponding to the same scatterer is obtained using the combination of the atomic norm regularized minimization for the MMV model and the MUSIC method. Then, based on the availability of the channel correlation information, the path gains are estimated using the LS or the MMSE method. Additionally, we derive a theoretical estimate of the channel estimate Mean Square Error for the asymptotically increasing number of receive antennas. To evaluate the proposed algorithm, we compare its performance with other state of the art algorithms.

Appendices

Appendix A

The value of μ _X in Eq. 10 is obtained by estimating the expected dual atomic norm of the noise Z ₁. Namely, in order to guarantee that the expected MSE of the estimate obtained by the atomic norm minimization is upperbounded, and that the solution of the atomic norm is asymptotically consistent (when P→∞) in [20, 21, 26] it is proposed to use the expected dual atomic norm of the noise as an upper bound of μ _X . Using the inequality (36) from [26], modified by the inequality (69) from [21], and applying the inequality below (43) in [26] we obtain an estimate of the expected dual atomic norm of the noise:

$$\begin{array}{@{}rcl@{}} \mathbb{E}[||\mathbf{Z}_{1}||_{\mathcal{A}}^{*}]\leq && \sqrt{\frac{{\sigma_{n}^{2}}}{2}}(1-\frac{2\pi\overline{P}N_{r}}{R})^{-1/2}(2N_{r}+2\ln R\\ && +2\sqrt{2N_{r}\ln R} +\sqrt{2\pi N_{r}}+2)^{1/2} \end{array} $$

where R is a value that should be optimized to obtain a tight estimate of the expected dual atomic norm. We set $\phantom {\dot {i}\!}R = \rho 2\pi \overline {P}N_{r}$ where ρ>1 is the variable to be optimized, and, after the substitution, we obtain:

$$\begin{array}{@{}rcl@{}} \mathbb{E}[||\mathbf{Z}_{1}||_{\mathcal{A}}^{*}]&\leq& \sqrt{\frac{{\sigma_{n}^{2}}}{2}}(1-\frac{1}{\rho})^{-1/2} (2N_{r}+2\ln 2\pi\overline{P}N_{r}\\ &&+ 2\ln\rho+2\sqrt{2N_{r}}\sqrt{\ln 2\pi\overline{P}N_{r}+ \ln\rho}\\ &&+\sqrt{2\pi N_{r}}+2)^{1/2} \end{array} $$

(17)

Using the substitutions $\phantom {\dot {i}\!}a = 2N_{r} +2\ln 2\pi \overline {P}N_{r}+2\sqrt {\frac {\pi N_{r}}{2}}+2$, $\phantom {\dot {i}\!}b=2\sqrt {2 N_{r}}$ and $\phantom {\dot {i}\!}c = \ln 2\pi \overline {P}N_{r}$, we further obtain:

$$\begin{array}{@{}rcl@{}} &&\mathbb{E}[||\mathbf{Z}_{1}||_{\mathcal{A}}^{*}]\leq\\ &&\ \ \ \sqrt{\frac{{\sigma_{n}^{2}}}{2}}(1-\frac{1}{\rho})^{-1/2} (a + 2\ln\rho +b\sqrt{c + \ln\rho})^{1/2} \end{array} $$

(18)

Since the right-hand side of Eq. 18 is always positive, we minimize its square by finding its derivative with respect to ρ and setting it to zero, resulting in:

$$\begin{array}{@{}rcl@{}} \rho = && \frac{1}{2\sqrt{c+\ln\rho}}((a+2)\sqrt{c+\ln\rho}\\ &&+b(c+\ln\rho)+\frac{b}{2}+2\sqrt{c+\ln\rho}\ln\rho-\frac{b\rho}{2}) \end{array} $$

(19)

For values of the parameters that correspond to real communication systems, and assuming ρ>1, the function on the right-hand side of Eq. 19 has a maximum in ρ = ρ _m ( ρ _m ≈1), with a value much greater than ρ _m , and it is a strictly monotone decreasing function when ρ>ρ _m . Thus, there is a single solution of Eq. 19 in terms of ρ in the range of interest of ρ. The derivative of the right-hand side of Eq. 19 is:

$$\begin{array}{@{}rcl@{}} &&\frac{1}{2\rho(\sqrt{c+\ln\rho})^{3}}(b(c+\ln\rho)+\frac{b\rho}{2}\\ &&+4(c+\ln\rho)\sqrt{c+\ln\rho}-\frac{b}{2}-b\rho(c+\ln\rho)) \end{array} $$

(20)

For the right-hand side of Eq. 19 to be strictly monotone decreasing, it needs to have a negative derivative:

$$ b(c+\ln\rho)+\frac{b\rho}{2}+4(c+\ln\rho)\sqrt{c+\ln\rho}-\frac{b}{2}<b\rho(c+\ln\rho) $$

(21)

Neglecting the last term in the left-hand side of Eq. 21 and dividing (21) by c+ lnρ, we obtain:

$$ b+\frac{b\rho}{2(c+\ln\rho)}+4\sqrt{c+\ln\rho}<b\rho $$

(22)

Using the fact that $\phantom {\dot {i}\!}\frac {1}{c+\ln \rho }<\frac {1}{c}$ and $\phantom {\dot {i}\!}\sqrt {c+\ln \rho }<\sqrt {c+\rho }<\sqrt {c}+\sqrt {\rho }$ we obtain:

$$ \frac{b+4\sqrt{c}}{b(1-\frac{1}{2c})}+\frac{4}{b(1-\frac{1}{2c})}\sqrt{\rho}<\rho $$

(23)

From Eq. 23 for

$$\begin{array}{@{}rcl@{}} \rho > &&\frac{b+4\sqrt{c}}{b(1-\frac{1}{2c})}+\frac{8}{(b(1-\frac{1}{2c}))^{2}}\\ &&+\frac{2}{b(1-\frac{1}{2c})}\sqrt{(\frac{4}{b(1-\frac{1}{2c})})^{2}+4\frac{b+4\sqrt{c}}{b(1-\frac{1}{2c})}} \end{array} $$

(24)

the right-hand side of Eq. 19 is a monotone decreasing function. Thus, using a starting value of ρ satisfying (24), any gradient based optimization algorithm will find the solution of Eq. 19. After finding the optimal value of ρ, it is substituted in the right-hand side of Eq. 17 to obtain a tight estimate of μ _X .

Appendix B

Here we show the derivation of the MSE estimate (14). Assuming that Δτ _r,i ≪T _s , and that $\phantom {\dot {i}\!}\hat {\tau }_{i}$’s are sufficiently close to the τ _i ’s we can approximate D _r using its first order approximation as $\phantom {\dot {i}\!}\mathbf {D}_{r} = \hat {\mathbf {D}}_{r}+\mathbf {G}\hat {\mathbf {D}}_{r} \mathbf {\Delta f}_{r}$ where G is a diagonal matrix with nonzero elements $\phantom {\dot {i}\!}[\mathbf {G}]_{i,i} = -j\frac {2\pi }{L}n_{i}^{,}$ and # #Δ# # f _r is a diagonal matrix with nonzero elements $\phantom {\dot {i}\!}[\mathbf {\Delta f}_{r}]_{i,i} = \frac {\Delta \hat {\tau }_{r,i}}{T_{s}}=\frac {\tau _{r,i}-\hat {\tau }_{i}}{T_{s}}$. Using matrix notation, Eq. 3 can be rewritten as:

$$ \mathbf{H}_{r} =\mathbf{A}_{r}\mathbf{h}_{I,r} \ r = 1,...,N_{r} $$

(25)

where $\phantom {\dot {i}\!}[\mathbf {A}_{r}]_{n,i}= e^{-j2\pi \frac {\tau _{r,i}}{NT_{s}}n}$, i=0,...,I−1, n=0,...,N−1. Using the first order approximation of A _r we can write:

$$ \mathbf{A}_{r} = \hat{\mathbf{A}}_{r}+\mathbf{G}_{g}\hat{\mathbf{A}}_{r}\mathbf{\Delta f}_{r} $$

(26)

where $\phantom {\dot {i}\!}[\hat {\mathbf {A}}_{r}]_{n,i}=e^{-j2\pi \frac {\hat {\tau }_{i}}{NT_{s}}n}$ and G _g is a diagonal matrix with nonzero elements $\phantom {\dot {i}\!}[\mathbf {G}_{g}]_{i,i} = -j\frac {2\pi }{N}i$. Using the approximations introduced above, we can rewrite $\phantom {\dot {i}\!}{\Delta }\mathbf {H}_{r}=\mathbf {H}_{r}-\hat {\mathbf {H}}_{r}$ as:

$$\begin{array}{@{}rcl@{}} {\Delta}\mathbf{H}_{r} = && \mathbf{G}_{g}\hat{\mathbf{A}}_{r}\mathbf{\Delta f}_{r}\mathbf{h}_{I,r} - \hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{D}}_{r}^{H} \mathbf{G}\hat{\mathbf{D}}_{r} \mathbf{\Delta f}_{r}\mathbf{h}_{I,r}\\ && -\hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{D}}_{r}^{H}\mathbf{Z}_{1_{r}} \end{array} $$

(27)

where represents the measurement noise i.e. $\phantom {\dot {i}\!}\mathbf {Z}_{1_{r}} =$ $\phantom {\dot {i}\!}[Z_{1_{r}}(n_{0}^{,}\frac {N}{L}),$ $\phantom {\dot {i}\!} \ldots , Z_{1_{r}}(n_{P-1}^{,}\frac {N}{L})]^{T}$.

Using the assumption that h _r,i ’s are asymptotically i.i.d. (when N _r →∞), which allows for the independence of the entries in # #Δ# # f _r and h _I,r , and averaging over the noise, # #Δ# # f _r and h _I,r , the total mean square channel estimation error per antenna is:

$$\begin{array}{@{}rcl@{}} &&\mathbb{E}_{\mathbf{Z}_{1_{r}},\mathbf{\Delta f}_{r},\mathbf{h}_{i,r}}[{\Delta}\mathbf{H}_{r}^{H}{\Delta}\mathbf{H}_{r}] = \\ &&{\sigma_{h}^{2}}\sigma_{\frac{\Delta\tau}{T_{s}}}^{2}[\text{trace}(\mathbf{G}_{g}\hat{\mathbf{A}}_{r}\hat{\mathbf{A}}_{r}^{H}\mathbf{G}_{g}^{H}) \\ &&- \text{trace}(\mathbf{G}_{g}\hat{\mathbf{A}}_{r}\hat{\mathbf{D}}_{r}^{H}\mathbf{G}^{H}\hat{\mathbf{D}}_{r}\left( (\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\right)^{H}\hat{\mathbf{A}}_{r}^{H})\\ &&- \text{trace}(\hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{D}}_{r}^{H}\mathbf{G}\hat{\mathbf{D}}_{r}\hat{\mathbf{A}}_{r}^{H} \mathbf{G}_{g}^{H})\\ &&+\text{trace}\left( \hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{D}}_{r}^{H}\mathbf{G}\hat{\mathbf{D}}_{r}\right.\\ &&\hspace*{3pc}\left.\hat{\mathbf{D}}_{r}^{H}\mathbf{G}^{H}\hat{\mathbf{D}}_{r}\left( (\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\right)^{H}\hat{\mathbf{A}}_{r}^{H})\right)\\ &&+{\sigma_{n}^{2}}\text{trace}(\hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{A}}_{r}^{H}) \end{array} $$

(28)

where $\phantom {\dot {i}\!}\mathbb {E}[\mathbf {\Delta f}_{r}\mathbf {\Delta f}_{r}^{H}] = \sigma _{\frac {\Delta \tau }{T_{s}}}^{2}\mathbf {I}$. When I=1, averaging over all the possible combinations of the $\phantom {\dot {i}\!}n^{,}_{p}$’s, the expressions in Eq. 28 can be calculated as:

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{n_{p}^{,}}\left[\text{trace}(\mathbf{G}_{g}\hat{\mathbf{A}}_{r}\hat{\mathbf{A}}_{r}^{H}\mathbf{G}_{g}^{H})\right] &=&\frac{4\pi^{2}}{N^{2}}\sum\limits_{n=1}^{N-1}n^{2}\\ &=& 4\pi^{2}\frac{(N-1)(2N-1)}{6N} \end{array} $$

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{n_{p}^{,}}\left[\text{trace}(\mathbf{G}_{g}\hat{\mathbf{A}}_{r}\hat{\mathbf{D}}_{r}^{H}\mathbf{G}^{H}\hat{\mathbf{D}}_{r}\left( (\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\right)^{H}\hat{\mathbf{A}}_{r}^{H})\right]\\ =\frac{4\pi^{2}}{NL}\frac{1}{P}\left( \sum\limits_{n=1}^{N-1}n\right)\mathbb{E}_{n_{p}^{,}}[\sum\limits_{p=1}^{P}n_{p}^{,}]=\frac{\pi^{2}(N-1)(L-1)}{L} \end{array} $$

$$\begin{array}{@{}rcl@{}} \mathbb{E}_{n_{p}^{,}}\left[\text{trace}(\hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{D}}_{r}^{H}\mathbf{G}\hat{\mathbf{D}}_{r}\hat{\mathbf{A}}_{r}^{H}\mathbf{G}_{g}^{H})\right] \\ =\frac{4\pi^{2}}{NL}\frac{1}{P}\left( \sum\limits_{n=1}^{N-1}n\right)\mathbb{E}_{n_{p}^{,}}[\sum\limits_{p=1}^{P}n_{p}^{,}]=\frac{\pi^{2}(N-1)(L-1)}{L} \end{array} $$

$$\begin{array}{@{}rcl@{}} &&\mathbb{E}_{n_{p}^{,}}\left[\text{trace}(\hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{D}}_{r}^{H}\mathbf{G}\hat{\mathbf{D}}_{r}\right.\\ &&\ \ \ \ \ \ \ \ \ \ \left.\hat{\mathbf{D}}_{r}^{H}\mathbf{G}^{H}\hat{\mathbf{D}}_{r}\left( (\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\right)^{H}\hat{\mathbf{A}}_{r}^{H})\right]\\ &&= \frac{4\pi^{2}N}{L^{2}P^{2}}\mathbb{E}_{n_{p}^{,}}\left[\left( \sum\limits_{p=1}^{P}n_{p}^{,}\right)^{2}\right]\\ &&=\pi^{2}N(\frac{(P-1)(L-1)}{PL}+\frac{2(L-P)(2L-1)}{3PL^{2}}) \end{array} $$

$$ \ \ \mathbb{E}_{n_{p}^{,}}\left[\text{trace}(\hat{\mathbf{A}}_{r}(\hat{\mathbf{D}}_{r}^{H}\hat{\mathbf{D}}_{r})^{-1}\hat{\mathbf{A}}_{r}^{H})\right]=N/P $$

where $\phantom {\dot {i}\!}\mathbb {E}_{n_{p}^{,}}[\cdot ]$ shows the expectation over all the possible combinations of $\phantom {\dot {i}\!}n_{p}^{,}$ for p=0,...,P−1. To derive the equations above, we used the trace property trace(A B ^H) = trace(B ^H A) and the equalities $\phantom {\dot {i}\!}\hat {\mathbf {D}}_{r}^{H}\hat {\mathbf {D}}_{r}$ = P, $\phantom {\dot {i}\!}\hat {\mathbf {A}}_{r}^{H}\hat {\mathbf {A}}_{r}$ = N, $\phantom {\dot {i}\!}{\sum }_{n=1}^{N-1}n=\frac {(N-1)N}{2}$, $\phantom {\dot {i}\!}{\sum }_{n=1}^{N-1}n^{2}$ =$\phantom {\dot {i}\!}\frac {(N-1)N(2N-1)}{6}$, $\phantom {\dot {i}\!}\mathbb {E}_{n_{p}^{,}}[{\sum }_{p=1}^{P}n_{p}^{,}]$ =$\phantom {\dot {i}\!}\frac {P}{L}\frac {(L-1)L}{2}$ and $\phantom {\dot {i}\!}\mathbb {E}_{n_{p}^{,}}\left [\left ({\sum }_{p=1}^{P}n_{p}^{,}\right )^{2}\right ]=$ $\phantom {\dot {i}\!}\frac {P(P-1)}{L(L-1)}$ $\phantom {\dot {i}\!}\left (\frac {(L-1)L^{2}}{2})\right )^{2}$ $\phantom {\dot {i}\!}+\frac {P(L-P)}{L(L-1)}\frac {(L-1)L(2L-1)}{6}$. Thus, (28) becomes:

$$\mathbb{E}[{\Delta}\mathbf{H}_{r}^{H}{\Delta}\mathbf{H}_{r}] \approx N{\sigma_{h}^{2}}\sigma_{\frac{\Delta\tau}{T_{s}}}^{2}\frac{\pi^{2}}{3}+\frac{N}{P}{\sigma_{n}^{2}} $$

When I>1, as long as τ _i ’s are sufficiently separated and uncorrelated for different i’s, Eq. 28 can be approximated as:

$$ \mathbb{E}[{\Delta}\mathbf{H}_{r}^{H}{\Delta}\mathbf{H}_{r}] \approx IN{\sigma_{h}^{2}}\sigma_{\frac{\Delta\tau}{T_{s}}}^{2}\frac{\pi^{2}}{3}+\frac{IN}{P}{\sigma_{n}^{2}} $$

(29)

Dividing (29) by the number of subcarriers N, we obtain the MSE estimate per subcarrier given by Eq. 14.