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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 to move a zero/pole which is outside the immaginary unit disk to inside it. From: $$ W(z)=\frac{C(z)}{A(z)}=\frac{(z-q_1)(z-q_2)...(z-q_n)}{(z-p_1)(z-p_2)...(z-p_m)}=W_1(z)(z-q) $$ Then: $$ y(t)=W_1(z)(z-\frac{1}{q})\frac{z-q}{z-\frac{1}{q}}e(t) $$ 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.

Prove that the steady state output of an all-pass filter feb by a white noise Is a white noise itself. Illustrate why all-pass filters are useful to obtain canonical forms of ARMA.

The all pass filter is $\frac{z-q}{z-\frac{1}{q}}$. See above on why they are useful to obtain the canonical form.

Given the white noise $e(t) \sim WN(\mu,\lambda^2)$, if we applu an all pass filter to it: $$\bar{e}(t)=\frac{z-q}{z-\frac{1}{q}}e(t)$$ It is well defined for $|q|>1$ and it is still a white noise since: $$ \Gamma_{\bar{e}(t)}(\omega)=\mid \frac{e^{j\omega}-q}{e^{j\omega}-\frac{1}{q}} \mid^2 \Gamma_{e(t)}(\omega)=\frac{e^{j\omega}-q}{e^{j\omega}-\frac{1}{q}} \frac{e^{-j\omega}-q}{e^{-j\omega}-\frac{1}{q}} \lambda^2= $$ $$ =\frac{1+q^2-q(e^{j\omega}+e^{-j\omega})}{1+\frac{1}{q^2}-\frac{1}{q}(e^{j\omega}+e^{-j\omega})} \lambda^2=q^2\frac{1+\frac{1}{q^2}-\frac{1}{q}(e^{j\omega}+e^{-j\omega})}{1+\frac{1}{q^2}-\frac{1}{q}(e^{j\omega}+e^{-j\omega})} \lambda^2= $$ $$ =q^2\lambda^2 $$ Which is constant $\forall \omega$ (as the white noise).

For the mean: $$\bar{e}(t)=\frac{z-q}{z-\frac{1}{q}}e(t)$$ $$\bar{e}(t+1)-\frac{1}{q}\bar{e}(t)=e(t+1)-qe(t)$$ $$\mathbb{E}[\bar{e}(t+1)-\frac{1}{q}\bar{e}(t)]=\mathbb{E}[e(t+1)-qe(t)]$$ $$m_{\bar{e}}-\frac{1}{q}m_{\bar{e}}=\mu-q\mu$$ $$m_{\bar{e}}=\frac{1-q}{1-\frac{1}{q}}\mu=-q\frac{1-\frac{1}{q}}{1-\frac{1}{q}}\mu=-q\mu$$

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