# Kay Statistical Signal Processing Estimation Theory Example 7.14

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?

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

$$r_{xx}[k]=r_{xx}^*[-k]\tag{1}$$

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

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

with

$$g[k]=\delta[k]+b_1\delta[k-1]+b_2\delta[k-2]\tag{4}$$

where $$\star$$ denotes convolution.

From $$(3)$$ it follows that

$$S_{xx}(f)=\big|G(f)\big|^2\tag{5}$$

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

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