RBSE

Real Binary Spectral Estimator(RBSE): A modified spectral estimator of phase retrieval for 0-1 measurement matrix.

The Phase Retrieval Problem: Recovering an unknown signal vector $x$ from observations $y_m=|\mathbf{a}_m^H x|^2$ where $\mathbf{a}$ are given measurement vectors.

It is meaningful to consider that the elements of $\mathbf{a}_m$ belong to the set {0,1}, for example, when $\mathbf{a}_m$ is a matrix generated by DMD.

Let $ \mathbf{a}_{m} $ are $M$ dimensional i.i.d. random vectors, the probability that the element has a value of 1 is $p$ (i.e., the duty cycle). Using $a$ to denote any element of $\mathbf{a}_{m}$, and the vector $\mathbf{a}$ to denote any $\mathbf{a}_{m}$ . Define

$$a^{'} = \frac{1}{p}(a - p),\frac{a}{p} = a^{'} + 1$$

$$ \begin{matrix} B_{ij} = & \\ & 0 & i = j \\ & \frac{1}{Mp^{2}} \sum_{m}{\left\lbrack \mathbf{a}_{m}^{'} \right\rbrack_{i}\left\lbrack \mathbf{a}_{m}^{'} \right\rbrack_{j}y_{m}} & i \neq j \\ C_{i} = & \\ & \frac{1}{Mp^{2}}\sum_{m}{\left\lbrack \mathbf{a}_{m}^{'} \right\rbrack_{i} y_{m}} \\ D_{ij} = &\\ &\frac{2}{\left( \mathbb{E}\left\lbrack a^{'3} \right\rbrack + 2\mathbb{E}\left\lbrack a^{'2} \right\rbrack \right)}\left( C_{i} - \frac{1}{\mathbb{E}\left\lbrack a^{'2} \right\rbrack}\sum_{n} B_{in} \right) & i = j \\ &\frac{1}{\mathbb{E}\left\lbrack a^{'2} \right\rbrack^{2}}B_{ij} & i \neq j \\ \end{matrix} $$

By the large number theorem, we have $D_{ij} \approx x^{*}_{i} x_{j} + x_{j}^{*}x_{i}$ , or $\mathbf{D} \approx \mathbf{x}\mathbf{x}^{H} + \mathbf{x}^{*}\mathbf{x}^{T}$

In the case that $\mathbf{x}^{T}\mathbf{x}$ is not zero, the following two real vectors are the eigenvectors of $\mathbf{x}\mathbf{x}^{H} + \mathbf{x}^{*}\mathbf{x}^{T}$

$$\mathbf{z}_{1} = \frac{1}{2}\sqrt{\mathbf{x}^{T}\mathbf{x}}\mathbf{x}^{*} + \frac{1}{2}\sqrt{\mathbf{x}^{H}\mathbf{x}^{*}}\mathbf{x}$$

$$\mathbf{z}_{2} = \frac{1}{2\text{i}}\sqrt{\mathbf{x}^{T}\mathbf{x}}\mathbf{x}^{*} - \frac{1}{2\text{i}}\sqrt{\mathbf{x}^{H}\mathbf{x}^{*}}\mathbf{x}$$

The corresponding eigenvalues are

$$\lambda_{1} = \left| \mathbf{x} \right|^{2} + \left| \mathbf{x}^{T}\mathbf{x} \right|,\lambda_{2} = \left| \mathbf{x} \right|^{2} - \left| \mathbf{x}^{T}\mathbf{x} \right|$$

Calculating the eigenvectors of the matrices yields $\mathbf{z}_{1}$ and $\mathbf{z}_{2}$ after normalization, and then $\left| \mathbf{z}_{1} \right|$ and $\left| \mathbf{z}_{2} \right|$ can be determined by eigenvalues $\lambda_{1}$ and $\lambda_{2}$, and so we get ±$\mathbf{z}_{1}$ and ±$\mathbf{z}_{2}$, then $\mathbf{x}$ or $\mathbf{x}^{*}$.

When $\mathbf{x}^{T}\mathbf{x}=0$ , $\lambda_{1} =\lambda_{2}$, and any linear superposition of $\mathbf{x}$ and $\mathbf{x}^{*}$ is an eigenvector