Example 7.14 from Kay Estimation Theory

A common WSS random process has the ACF

$$r_{xx}[k]= \begin{cases} 1+b[1]^{2}+b[2]^{2} & \text{k = 0}\\ b[1] +b[1]b[2] & \text{k=1}\\ b[2] & \text{k=2}\\ 0 & k \geq 3 \end{cases} $$

The PSD is easily shown to be $$P_{xx}(f) = | 1+ b[1]\exp(-j2 \pi f) + b[2] \exp(-j4 \pi f)|^{2}$$

I don't understand how the PSD was calculated. The PSD is the Fourier Transform of the ACF. When I take the Fourier Transform of the ACF I get the following:

$$1+b^{2}[1]+b^2[2]+(b[1] +b[1]b[2])e^{-j2 \pi f}+b[2] e^{-j4 \pi f}$$

How do I convert the PSD I calculated into the magnitude squared from in the text?


1 Answer 1


You're forgetting an important property of the autocorrelation sequence:


I.e., it is conjugate symmetric, which is necessary for the power spectrum to be real-valued. Because of this symmetry, the autocorrelation sequence is only given for non-negative values of $k$. If you use $(1)$, you obtain

$$S_{xx}(f)=1+b_1^2+b_2^2+2b_1(1+b_2)\cos(2\pi f)+2b_2\cos(4\pi f)\tag{2}$$

It's straightforward to show the equivalence of $(2)$ and the given expression for $S_{xx}(f)$.

Alternatively, it can be shown that

$$r_{xx}[k]=g[k]\star g[-k]\tag{3}$$



where $\star$ denotes convolution.

From $(3)$ it follows that


where $G(f)$ is the DTFT of $g[k]$. Eq. $(5)$ is identical to the given expression for $S_{xx}(f)$ because we have

$$G(f)=1+b_1e^{-j2\pi f}+b_2e^{-j4\pi f}\tag{6}$$


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