Skip to main content

Spectrum of ARMA processes

Given the $ARMA$ process $y(t)=\frac{C(z)}{A(z)}e(t)$ with $e(t) \sim WN(0,\lambda^2)$ the spectrum is given by: $$\Gamma_y(\omega)=\mid\frac{C(e^{j\omega})}{A(e^{j\omega})}\mid^2\times\lambda^2$$ Since both $C(z)$ and $A(z)$ are polynomial the result is a rational spectrum (ratio of polynomials of $e^{j\omega}$):

$$\Gamma_y(\omega)=\frac{\beta_0+\beta_1e^{j\omega}+\beta_2e^{2j\omega}+...}{\alpha_0+\alpha_1e^{j\omega}+\alpha_2e^{2j\omega}+...}$$

THEOREM

If the spectrum of a stationary stochastic process is rational then the process is an $ARMA$ process.

Let $y(t)$ be a SSP with rational spectrum. There exists a digital filter $W(z)$ and a white noise $e(t)$ with suitable mean $\mu$ and variance $\lambda^2$ such that $y(t)=W(z)e(t)$.

The representation is not unique: if the process $y(t)$ is an $ARMA$ process then there are an infinite moltitude of transfer function $W(z)$ and white noises $e(t)$ such that $y(t)=W(z)e(t)$.

Canonical representation of ARMA processes (Spectral Factorization Theorem)

Let $y(t)$ be a SSP with rational spectrum. There exists a unique $ARMA$ representation, that is a unique pair of transfer function $W(z)=\frac{C(z)}{A(z)}$ and white noise $e(t)$, such that:

  1. $C(z)$ and $A(z)$ are monic: the coefficient of the term with the highest power is 1
  2. $C(z)$ and $A(z)$ have null relative degree: the highest power in the numerator is the same of the one in the denominator
  3. $C(z)$ and $A(z)$ are coprime: they do not have common factors
  4. Poles have absolute values < 1 and zeros have absolute values less or equal than 1. $$ |poles|<1\newline |zeros|\le1 $$

How to find the canonical representation

Given an $ARMA$ process $y(t)=W(z)e(t)$ with $e(t) \sim WN(\mu,\lambda^2)$ the following operations are applicable in order to find the canonical representation:

  1. With $\alpha \in \mathbb{R}$: $$ \bar{W(z)}=\alpha W(z)\newline \bar{e(t)}=\frac{1}{\alpha}e(t) \sim WN(\frac{\mu}{\alpha}, \frac{\lambda^2}{\alpha^2}) $$
  2. With $k=\pm1,\pm2,...$: $$ \bar{W(z)}=z^{-k} W(z)\newline \bar{e(t)}=z^{k}e(t) \sim WN(\mu, \lambda^2) $$
  3. With $p$ a pole of $W(z)$: $$ \bar{W(z)}=W(z)\frac{z-p}{z-p}\newline \bar{e(t)}=e(t) \sim WN(\mu, \lambda^2) $$
  4. We can use the all pass filter. From: $$ W(z)=\frac{C(z)}{A(z)}=\frac{(z-p_1)(z-p_2)...(z-p_m)}{(z-q_1)(z-q_2)...(z-q_n)}=W_1(z)(z-q) $$ Then: $$ y(t)=W_1(z)(z-q)\frac{z-\frac{1}{q}}{z-\frac{1}{q}}e(t) $$ Which results in: $$ \bar{W(z)}=W_1(z)(z-\frac{1}{q})\newline \bar{e(t)}=\frac{z-q}{z-\frac{1}{q}}e(t) \sim WN(-q\mu, q^2\lambda^2) $$

Questions from past exams

By means of an example, show that the ARMA representation of a stationary stochastic process with rational spectral density is not unique. Enunciate then the Spectral Factorization Theorem and give some comments.

See above.

Back to top