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**ELG5377 Adaptive Signal Processing**

Lecture 7: Stochastic Models: Moving Average (MA), Autoregressive (AR) and ARMA

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Stochastic Models The term model is used for any hypothesis that may be applied to describe the hidden laws that govern the generation of physical data. A time series u(k) consisting of highly correlated observations may be generated by applying a series of statistically independent “shocks” to a linear filter. The shocks are drawn from a fixed distribution (usually Gaussian) with zero mean and constant variance. This time series of shocks is v(k). E[v(k)v*(k-i)]=sv2 for i = 0 and 0 for i ≠ 0. Discrete-time Linear Filter v(k) u(k)

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**Types of Models u(k) = Sbi*v(k-i) Moving Average Model**

FIR filter implementation Sai*u(k-i) = v(k) (usually a0 = 1) u(k) = v(k) – Sai*u(k-i) (i ≠ 0) Autoregressive Model IIR filter implementation ARMA Sai*u(k-i) = Sbi*v(k-i) Cascade of FIR and IIR filters

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**Autoregressive Models**

The time series u(k), u(k-1), …, u(k-M) represents the realization of an autoregressive model of order M if it satisfies the difference equation u(k) + a1*u(k-1)+…+aM*u(k-M) = v(k). Or u(k) = w1*u(k-1)+…+wM*u(k-M) = v(k). Where wi = -ai. It is called an autoregressive model since u(k) is regressed on previous values of itself.

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**Correlation function of an asymptotically stationary AR process**

Starting with Sai*u(k-i) = v(k) Multiply both sides by u*(k-l) and take the expectation. E[Sai*u(k-i)u*(k-l)] = E[v(k)u*(k-l)] The right side of the equation is 0 for l > 0. The left side is Sai*r(l-i) Therefore for l > 0, Sai*r(l-i) = 0 (for i = 0, 1, … M) (a0 = 1) Or we can write this as r(l) = Swi*r(l-i) = 0 (for i = 1,2, … M) Want to find w1, w2, … wM.

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**Coefficients of AR model**

Recall that r(-x) = r*(x). If we take the complex conjugate of both sides, we get or r = Rw

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**Coefficients of AR model 2**

r = [r*(1) r*(2) … r*(M)]T and w=[w1 w2 … wM]T. R is the M×M correlation matrix of u(k). Therefore w = R-1r. a0 = 1 and ai = -wi. Next, let l = 0. E[v(k)u*(k)] = E[v(k)(Swi*u(k-i)+v(k))*] = E[v(k)v*(k)] = sv2. The right side of the equation becomes Sair(i). Therefore:

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AR model example Find a third order AR model that produces a process with the following correlation function r(i) = sinc(i/2) Solution M = 3. r(0) = 1, r(1) = 0.637, r(2) = 0, r(3) =

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**AR model example continued**

w= R-1r. w = [1.552, , 0.703]T. a0 = 1, a1 = , a2 = and a3 = sv2 = Sair(i) = 0.16 v(k) u(k) + + w1 + w2 w3

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**Applying autoregressive models to nonstationary systems**

Let us consider a first order autoregressive model. w(k) = b1w(k-1)+v(k) We need r(0) and r(1). b1 = r(1)/r(0). sv2 = r(0)-r2(1)/r(0). Next let us consider the relationship between x(k) and d(k) in a stationary system. d(k) = yo(k)+eo(k) = woHx(k)+eo(k). Suppose that wo is time varying. Then the cross correlation between d(k) and x(k) would also be time-varying. Nonstationary system. Can represent as d(k) = woH(k)x(k)+eo(k). where wo(k)=awo(k-1)+w(k)

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