# Stationarity, discrete-translation operator, and the power spectral density matrix

Let $$\mathbf{T}$$ be the translation operator/matrix in discrete-time domain which can be written as $$\mathbf{T} = \mathbf{\Phi} \mathbf{P} \mathbf{\Phi}^*$$ where $$\mathbf{P} = \exp(-i Diag([w_0, w_1, \ldots, w_M]))$$ is a diagonal matrix where $$w_k= 2\pi(k-1)/M$$. We know that the stochastic signal $$\mathbf{x}$$ is called wide-sense stationary (WSS) w.r.t. the translation operator on in discrete-time domain $$\mathbf{T}$$ if and only if for all $$r$$:

$${\mathbf R}_{\mathbf{x}} = \mathbb{E}[\mathbf{x}\mathbf{x}^{*}] = \mathbb{E}\big[(\mathbf{T}^{r} \mathbf{x})\big(\mathbf{T}^{r} \mathbf{x}\big)^{*}\big].$$

We need to show that: a process in second-order moment is WSS if and only if \begin{align} {\mathbf R}_{{\mathbf{x}}} = \mathbf{\Phi} {\mathbf S}_{{\mathbf{x}}} \mathbf{\Phi}^{*}, \end{align} where $$\mathbf{S}_{\mathbf x}$$ is a diagonal matrix with non-negative entries on its main diagonal and $$\mathbf{\Phi}^{*}$$ is the DFT matrix.

Proof: We can write for the power spectral density matrix as $$$${\mathbf S}_{{\mathbf{x}}} = \mathbb{E}[\widehat{\mathbf{x}}\widehat{\mathbf{x}}^{*}] = \mathbf{\Phi}^{*}\mathbb{E}[{\mathbf{x}}{\mathbf{x}}^{*}]\mathbf{\Phi} = \mathbf{\Phi}^{*}{\mathbf R}_{{\mathbf{x}}}\mathbf{\Phi},$$$$ and hence $${\mathbf R}_{{\mathbf{x}}} = \mathbf{\Phi} {\mathbf S}_{{\mathbf{x}}} \mathbf{\Phi}^{*}.$$ Via the condition above for WSS signal, one can obtain $$$$\begin{split} {\mathbf R}_{\mathbf{x}} = \mathbb{E}[{\mathbf{x}}{\mathbf{x}}^{*}] &= \mathbb{E}\big[\big({\mathbf{T}^{r}\mathbf{x}}\big) \big(\mathbf{T}^{r}{\mathbf{x}}\big)^{*}\big] \\ &= \mathbf{T}^{r} \mathbb{E}[{\mathbf{x}}{\mathbf{x}}^{*}] \mathbf{T}^{{-r}}. \end{split}$$$$ On can easily obtain \begin{align} \begin{split} {\mathbf S}_{{\mathbf{x}}} &= \big(\mathbf{\Phi}^{*} \mathbf{T}^{r} \mathbf{\Phi}\big) {\mathbf S}_{{\mathbf{x}}} \big(\mathbf{\Phi}^{*} \mathbf{T}^{r}\mathbf{\Phi}\big)^{-1} \\ &= \mathbf{P}^{r} {\mathbf S}_{{\mathbf{x}}} \mathbf{P}^{-r}, \end{split} \end{align} I just get confused here. How we can say from this that $${\mathbf S}_{{\mathbf{x}}}$$ is diagonal which is a well-known fact in signal processing?

• I am certain that the definition of P is incorrect here. The translation operator T should be unitary, i.e. $\mathbf{T}^{-1} = \mathbf{T}^{\dagger}$, where $\dagger$ indicates complex conjugate transpose (* in your notation). T is unitary if you change the definition of P to $\textrm{diag}(1,e^{2\pi i/M},\ldots,e^{2\pi i (M-1)/M})$, where $i = \sqrt{-1}$. With that definition, $\mathbf{P}\mathbf{P}^{\dagger} = \mathbf{I}$. – Joe Mack Jun 12 '20 at 18:34
• @JoeMac: Sorry. That was typo. Corrected. – Amin Jun 12 '20 at 22:56

Given the definition of the correlation matrix $$\mathbf{R}_{\mathbf{x}}$$ here, I am assuming that $$\mathsf{E}[\mathbf{x}] = \mathbf{0}$$. I do this because the correlation matrix is usually defined as $$\mathsf{E}[(\mathbf{x} - \mathsf{E}[\mathbf{x}])(\mathbf{x} - \mathsf{E}[\mathbf{x}])^{\dagger}]$$, where $$\dagger$$ indicates complex conjugate tranpose.

Note that since $$\mathbf{R}_{\mathbf{x}}$$ is a correlation matrix, it is Hermitian positive semi-definite, which means that it is unitarily diagonalizable and has all non-negative eigenvalues: $$$$\mathbf{R}_{\mathbf{x}} = \mathbf{U}\mathbf{\Lambda}\mathbf{U}^{\dagger},$$$$ where $$\mathbf{U}$$ is a unitary matrix and $$\mathbf{\Lambda}$$ is a diagonal matrix with non-negative numbers on its diagonal.
Let's consider the translation operator $$\mathbf{T}$$: $$$$\mathbf{T} = \mathbf{\Phi}\mathbf{P}\mathbf{\Phi}^{\dagger},$$$$ where $$\mathbf{\Phi}^{\dagger}$$ is the DFT matrix and $$\mathbf{P}$$ is diagonal. As I mentioned in a comment, $$\mathbf{T}$$ should be unitary, and that requires that the diagonal matrix $$\mathbf{P}$$ have complex exponentials on its diagonal. In fact, to make $$\mathbf{\Phi}\mathbf{P}\mathbf{\Phi}^{\dagger}$$ a translation operator, $$\mathbf{P}$$ should be $$$$\mathbf{P} = \textrm{diag}(1,e^{2\pi i/M},\ldots,e^{2\pi i(M-1)/M}).$$$$

I do not think this will have an impact on the proof, but it is important for applications.
Now let's assume that $$\mathbf{R}_{\mathbf{T}\mathbf{x}} = \mathbf{R}_{\mathbf{x}}$$: $$\begin{eqnarray} \mathsf{E}[(\mathbf{T}\mathbf{x})(\mathbf{T}\mathbf{x})^{\dagger}] &=& \mathbf{R}_{\mathbf{x}}\\ \mathbf{T}\mathsf{E}[\mathbf{x}\mathbf{x}^{\dagger}]\mathbf{T}^{\dagger} &=& \mathbf{R}_{\mathbf{x}}\\ \mathbf{T}\mathbf{R}_{\mathbf{x}}\mathbf{T}^{\dagger} &=& \mathbf{R}_{\mathbf{x}}\\ \mathbf{T}\mathbf{R}_{\mathbf{x}} &=& \mathbf{R}_{\mathbf{x}}\mathbf{T} \end{eqnarray}$$ Since

• $$\mathbf{R}_{\mathbf{x}}$$ and $$\mathbf{T}$$ are both diagonalizable
• and they commute with each other,

they are simultaneously diagonalizable. This means that there is a single matrix that diagonalizes both.

We have been told from the beginning that $$\mathbf{\Phi}$$ diagonalizes $$\mathbf{T}$$, so now we know that $$\mathbf{\Phi}$$ diagonalizes $$\mathbf{R}_{\mathbf{x}}$$, too. This means that the unitary matrix $$\mathbf{U}$$ that diagonalizes $$\mathbf{R}_{\mathbf{x}}$$ must be $$\mathbf{\Phi}$$: $$$$\mathbf{R}_{\mathbf{x}} = \mathbf{\Phi}\mathbf{\Lambda}\mathbf{\Phi}^{\dagger}.$$$$ We have already established that $$\mathbf{\Lambda}$$ is diagonal with non-negative real numbers on its diagonal. The $$\mathbf{S}_{\mathbf{x}}$$ that we have sought, is $$\mathbf{\Lambda}$$.

• Thanks @JoeMac. My confusion is resolved. – Amin Jun 12 '20 at 23:08