Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.14

Author: Panagiotis

4 Mrz

In [1, p. 62 exercise 3.14] it is requested to proof prove that the autocorrelation matrix given by [1, eq. 3.46] is also positive semidefinite. This shall be done by usage of the definition of the semidefinite property of the ACF in [1, eq. 3.45].
Solution: A matrix $\mathbf{A}$ is positive semidefinite if $\mathbf{x}^{H}\mathbf{A}\mathbf{x}\geq 0$ . And the autocorrelation matrix given by ([1, eq. 3.46]) is:

$\mathbf{R}_{xx}$	$=$	$E\left\{\mathbf{x}\mathbf{x}^{H}\right\}$	(1)
	$=$	$\left[\begin{array}{cccc}r_{xx}[0] & r_{xx}[-1]& \cdots&r_{xx}[-(M-1)] \\r_{xx}[1] &r_{xx}[0] &\cdots&r_{xx}[-(M-2)] \\ \vdots &\vdots & \ddots& \vdots \\ r_{xx}[M-1]&r_{xx}[M-2]&... &r_{xx}[0]\end{array}\right]$	(2)

Let $\mathbf{\alpha}=\left[\begin{array}{ccc} \alpha[0]&\cdots & \alpha[M-1]\end{array}\right]^{T}$ and $\mathbf{x}=\left[\begin{array}{ccc} x[0]&\cdots & x[M-1]\end{array}\right]^{T}$ then from [1, eq. 3.45] we obtain:

$E\left\{\left\| \sum\limits_{n=0}^{M-1}\alpha^{\ast}[n]x[n] \right\|^{2}\right\}$	$\geq$	$0$
$E\left\{\left\| \mathbf{\alpha}^{H}\mathbf{x} \right\|^{2}\right\}$	$\geq$	$0$
$E\left\{ \mathbf{\alpha}^{H}\mathbf{x}\mathbf{x}^{H}\mathbf{\alpha} \right\}$	$\geq$	$0$
$\mathbf{\alpha}^{H}E\left\{\mathbf{x}\mathbf{x}^{H}\right\}\mathbf{\alpha}$	$\geq$	$0$
$\mathbf{\alpha}^{H}\mathbf{R}_{xx}\mathbf{\alpha}$	$\geq$	$0$