Given a Wide Sense Stationary Multi Variate (Vector) Random Process $ \boldsymbol{x} \left[ n \right] $ it Auto Covariance Matrix Function is given by:

$$ {R}_{x, x} \left[ m \right] = \mathbb{E} \left[ \boldsymbol{x} \left[ n \right] \boldsymbol{x}^{T} \left[ n - m \right] \right] $$

Prove that the Power Spectrum Density Matrix is Positive Semi Definite (PSD) Matrix where it is given by:

$$ {S}_{x, x} \left( f \right) = \sum_{m = -\infty}^{\infty} {R}_{x, x} \left[ m \right] {e}^{-j 2 \pi f m} $$


Pay attention that $ {R}_{x, x} \left[ m \right] $ isn't necessarily Positive Semi Definite matrix (It is for $ m = 0 $, yet nothing can be said in general for other time indices). It is even not necessarily symmetric.

  • $\begingroup$ In what way are $R_{x,x}(m)$ and $S_{x,x}(f)$ considered to be matrices? It looks like they are sequences of numbers: OK, at best matrices in which one dimension is 1 but how is positive definiteness defined in this case? $\endgroup$ Oct 4 '17 at 11:20
  • $\begingroup$ @DilipSarwate, If we're dealing with a Process of Random Vector those are matrices. Not $ 1 \times 1 $ but actual matrices. For example if $ \boldsymbol{x} \left[ n \right] \in \mathbb{R}^{k} $ then $ {R}_{x, x} \left[ m \right] \in \mathbb{R}^{k \times k} \; \forall m $. $\endgroup$
    – Royi
    Oct 4 '17 at 12:07

Pay attention that for a Scalar Random Process the Power Spectrum Density is non negative.

Namely, let $ y \left[ n \right] \in \mathbb{R} $ be a WSS Random process with its Auto Correlation function given by:

$$ {R}_{y, y} \left[ m \right] = \mathbb{E} \left[ y \left[ n \right] y \left[ n - m \right] \right] $$

Then the Power Spectrum Density is:

$$ {S}_{y, y} \left( f \right) = \sum_{m = -\infty}^{\infty} {R}_{y, y} \left[ m \right] {e}^{-j 2 \pi f m} \geq 0 $$

Then, by defining $ z \left[ n \right] = \boldsymbol{v}^{T} \boldsymbol{x} \left[ n \right] $ one would get:

$$ \begin{align*} 0 \leq {S}_{z, z} \left( f \right) & = \sum_{m = -\infty}^{\infty} \mathbb{E} \left[ z \left[ n \right] {z}^{T} \left[ n - m \right] \right] {e}^{-j 2 \pi f m} \\ & = \sum_{m = -\infty}^{\infty} \mathbb{E} \left[ \boldsymbol{v}^{T} \boldsymbol{x} \left[ n \right] \boldsymbol{x}^{T} \left[ m - m \right] \boldsymbol{v} \right] {e}^{-j 2 \pi f m} \\ & = \boldsymbol{v}^{T} \left( \sum_{m = -\infty}^{\infty} \mathbb{E} \left[ \boldsymbol{x} \left[ n \right] \boldsymbol{x}^{T} \left[ n - m \right] \right] {e}^{-j 2 \pi f m} \right) \boldsymbol{v} \\ & = \boldsymbol{v}^{T} {S}_{x, x} \left( f \right) \boldsymbol{v} \\ & \Rightarrow {S}_{x, x} \left( f \right) \succeq 0 \end{align*} $$

Some remarks regarding simulating it in MATLAB:

  • If one use MATLAB's xcorr() to calculate the Auto Correlation one should use ifftshift() to shift the function to be "Symmetric" to MATLAB in order to have the DFT Real and Non Negative. This is due to the fast MATLAB's fft() expects the first sample to be of index $ 0 $ (See the lags output of xcorr()).

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.