Given a Wide Sense Stationary Multi Variate (Vector) Random Process $ \boldsymbol{x} \left[ n \right] $ its 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, 2017 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, 2017 at 12:07

1 Answer 1


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()).
  • $\begingroup$ Why is this true in the scalar case? $\endgroup$
    – Exodd
    Jun 28, 2022 at 14:29
  • $\begingroup$ @Exodd, what do you mean? $\endgroup$
    – Royi
    Jun 30, 2022 at 5:44
  • $\begingroup$ why is the PSD nonnegative in the scalar case, if you define it starting from the autocorrelation? $\endgroup$
    – Exodd
    Jun 30, 2022 at 8:39
  • $\begingroup$ Because the cross correlation basically equals to the DFT of the signal multiplied by its conjugate. So it is norm, non negative. $\endgroup$
    – Royi
    Jun 30, 2022 at 8:43

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