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# ELG5377 Adaptive Signal Processing

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ELG5377 Adaptive Signal Processing
Lecture 7: Stochastic Models: Moving Average (MA), Autoregressive (AR) and ARMA

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)

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

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.

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.

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

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:

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) =

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

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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