# Explanation of Spectral Line Splitting when AR process is overmodeled

In the book Statistical Digital Signal Processing and Modeling by Monson Hayes, it is shown (in section 8.5.1) that when an AR(2) process described by the following difference equation $$x(n) = 0.9x(n-2) + w(n)$$

is modeled with two different values of model orders $$p = 4$$ and $$p = 12$$, a single peak appears in the power spectrum for the former case and two peaks occur for the latter one. How do I explain this? What is the relation between the model order and the number of peaks. Please forgive my lack of proper fundamentals, but I also wish to know why I am supposed to expect only a single peak to occur and not more than one?

The equations involved in estimating model parameters $$a_{p}(k)$$ and $$b(0)$$ and power spectrum of the AR process are given below. $$r_{x}(k)$$ refers to the autocorrelation sequence.

$$$$\hat{P}_{AR}(e^{j\omega}) = \frac{|\hat{b}(0)|^{2}}{\left|1 + \sum_{k=1}^{p}\hat{a}_{p}(k)e^{-jk\omega}\right|^{2}}$$$$

$$$$\begin{bmatrix} r_{x}(0) & r_{x}(1) & r_{x}(2) & \dots & r_{x}(p)\\ r_{x}(1) & r_{x}(0) & r_{x}(1) & \dots & r_{x}(p-1)\\ r_{x}(1) & r_{x}(0) & r_{x}(1) & \dots & r_{x}(p-2)\\ \vdots & \vdots & \vdots & \vdots & \vdots\\ r_{x}(p) & r_{x}(p-1) & r_{x}(p-2) & \dots & r_{x}(0)\\ \end{bmatrix} \begin{bmatrix} 1\\ a_{p}(1) \\ a_{p}(2) \\ \vdots \\ a_{p}(p) \end{bmatrix} = \epsilon_{p} \begin{bmatrix} 1\\ 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$$$

$$$$|b(0)|^{2} = \epsilon_{p} = r_{x}(0) + \sum_{k=1}^{p}a_{p}(k)r_{x}^{*}(k)$$$$

• could you also add the formula that contains $p$? – Marcus Müller Nov 14 '18 at 8:31
• hm, what do the $a, b$ and $r$ mean? the answer, I feel, really lies in defining what you see in these formulas... – Marcus Müller Nov 14 '18 at 8:59

One way of doing power spectrum estimation of random processes actually requires finding a set of ARMA model parameters that best approximates the given observed random process $$x[n]$$.

In your case, you selected an all-pole signal model with $$p$$ poles as the roots of the denominator polynomial $$A_p(z)$$ with coefficients $$a_p[k]$$, and a $$b(0)$$ related with power scaling of the process. Each pole represents a peak in the spectrum for $$0 \leq \omega \leq \pi$$.

So, PSD estimation is reduced to finding (estimating) all-pole model parameters $$a_p[k]$$ and $$b[0]$$, from the given data by first estimating the auto-correlation sequence values $$r_x[k]$$ for $$k=0,1,2,...,p$$ , and then solving a set of linear equations (Prony or Yule-Walker etc) to find those $$a_p[k]$$ and $$\epsilon_p$$.

Note that there are a number of approaches in estimating the ACS $$r_x[k]$$ of $$x[n]$$, such as Auto-Correlation, Covariance, Burg's, Modified Covariance, etc...

Certain of these techniques (such as the autocorrelation and Burg's method) produce an artifact known as spectral splitting when they are forced to fit a high order model to an actually low order data; aka a model-data mismatch problem. There is not much you can do to prevent it. I don't know an analytic derivation of the relation between the chosen model order $$p$$ and the number of splitted peaks you will observe. That's also dependent on the observed noise instance being processed. So that number of peaks is also random.

However, it's stated that modified covariance method of ACS estimation can overcome the spectral splitting problem.

Furthermore, if you continue reading, you will see in the section named Selecting the Model Order (p. 445, Eq.8.23) that by trying to make a good guess of model order you may reduce the chances of generating that line splitting problem.

• But if each pole contributed a peak in the spectrum, why do we observe only one peak for p = 4? Why not 4 peaks? Am I missing out on any concept? Please correct me – MaxFrost Nov 15 '18 at 1:48
• I'm sorry for the confusion. Number of peaks in the DTFT PSD spectrum depends on the distinct roots of the denominator polynomial. However if some of those roots are small or large (compared to |z|=1) then they will not create sharp peaks. Furthermore, if your assumed model order $p$ is larger than the true model order, then high order coefficients tends to go zero hence you will not observe that many peaks. – Fat32 Nov 15 '18 at 10:53
• That means the transfer function for p=4 will contain only one pole near |z| = 1? Because I have not checked for it – MaxFrost Nov 15 '18 at 11:24
• Yes; indeed the signal has a singel pole close to unit circle, then your model will also have a single pole close to unit circle creating a spike in PSD (DTFT) (other than artefacts as defined to be line splitting) – Fat32 Nov 15 '18 at 14:16