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 \begin{equation} {\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}, \end{equation} 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{equation} \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} \end{equation} 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?

  • 1
    $\begingroup$ 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, $\mathbf{P}\mathbf{P}^{\dagger} = \mathbf{I}$. $\endgroup$ – Joe Mack Jun 12 at 18:34
  • $\begingroup$ @JoeMac: Sorry. That was typo. Corrected. $\endgroup$ – Amin Jun 12 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: \begin{equation} \mathbf{R}_{\mathbf{x}} = \mathbf{U}\mathbf{\Lambda}\mathbf{U}^{\dagger}, \end{equation} 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 $\mathbf{T}$: \begin{equation} \mathbf{T} = \mathbf{\Phi}\mathbf{P}\mathbf{\Phi}^{\dagger}, \end{equation} 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 \begin{equation} \mathbf{P} = \textrm{diag}(1,e^{2\pi i/M},\ldots,e^{2\pi i(M-1)/M}). \end{equation}

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 $\mathbf{\Phi}$: \begin{equation} \mathbf{R}_{\mathbf{x}} = \mathbf{\Phi}\mathbf{\Lambda}\mathbf{\Phi}^{\dagger}. \end{equation} 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 $\mathbf{\Lambda}$.

| improve this answer | |
  • $\begingroup$ Thanks @JoeMac. My confusion is resolved. $\endgroup$ – Amin Jun 12 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.