Linear Filtering of Random ProcessesLecture 13Spring 2002Wide-Sense StationaryA stochastic process X(t) is wss if its mean is constantE[X(t)] = µand its autocorrelation depends only on τ = t1− t2Rxx(t1,t2)=E[X(t1)X∗(t2)]E[X(t + τ)X∗(t)] = Rxx(τ)Note that Rxx(−τ)=R∗xx(τ)andRxx(0) = E[|X(t)|2]Lecture 13 1ExampleWe found that the random telegraph signal has the autocorrelationfunctionRxx(τ)=e−c|τ|We can use the autocorrelation function to find the second momentof linear combinations such as Y (t)=aX(t)+bX(t − t0).Ryy(0) = E[Y2(t)] = E[(aX(t)+bX(t − t0))2]= a2E[X2(t)] + 2abE[X(t)X(t − t0)] + b2E[X2(t − t0)]= a2Rxx(0) + 2abRxx(t0)+b2Rxx(0)=(a2+ b2)Rxx(0) + 2abRxx(t0)=(a2+ b2)Rxx(0) + 2abe−ct0Lecture 13 2Example (continued)We can also compute the autocorrelation Ryy(τ)forτ =0.Ryy(τ)=E[ Y (t + τ)Y∗(t)]= E[(aX(t + τ)+bX(t + τ − t0))(aX(t)+bX(t − t0))]= a2E[X(t + τ)X(t)] + abE[X(t + τ)X(t − t0)]+ abE[X(t + τ − t0)X(t)] + b2E[X(t + τ − t0)X(t − t0)]= a2Rxx(τ)+abRxx(τ + t0)+abRxx(τ − t0)+b2Rxx(τ)=(a2+ b2)Rxx(τ)+abRxx(τ + t0)+abRxx(τ − t0)Lecture 13 3Linear Filtering of Random ProcessesThe above example combines weighted values of X(t)andX(t − t0)to form Y (t). Statistical parameters E[Y ], E[Y2], var(Y )andRyy(τ)are readily computed from knowledge of E[X]andRxx(τ).The techniques can be extended to linear combinations of more thantwo s amples of X(t).Y (t)=n−1k=0hkX(t − tk)This is an example of linear filtering with a discrete filter with weightsh =[h0,h1,...,hn−1]The corresponding relationship for continuous time processing isY (t)=∞−∞h(s)X(t − s)ds =∞−∞X(s)h(t − s)dsLecture 13 4Filtering Random ProcessesLet X(t, e) be a random process. For the moment we show theoutcome e of the underlying random experiment.Let Y (t, e)=L[X(t, e)]be the output of a linear system when X(t, e)is the input. Clearly, Y (t, e) is an ensemble of functions selected bye, and is a random process.What can we say about Y when we have a statistical description ofX and a description of the system?Note that L does not need to exhibit random behavior for Y to berandom.Lecture 13 5Time InvarianceWe will work with time-invariant (or shift-invariant) systems. Thesystem is time-invariant if the response to a time-shifted input isjust the time-shifted output.Y (t + τ)=L[X(t + τ)]The output of a time-invariant linear system can be represented byconvolution of the input with the impulse response, h(t).Y (t)=∞−∞X(t − s)h(s)dsLecture 13 6Mean ValueThe following result holds for any linear s ystem, whether or not itis time invariant or the input is stationary.E[LX(t)] = LE[X(t)] = L[µ(t)]When the process is stationary we find µy= L[µx], which is just theresponse to a constant of value µx.Lecture 13 7Output AutocorrelationThe autocorrelation function of the output isRyy(t1,t2)=E[y(t1)y∗(t2)]We are particularly interested in the autocorrelation function Ryy(τ)of the output of a linear system when its input is a wss randomprocess.When the input is wss and the system is time invariant the outputis also wss.The autocorrelation function can be found for a process that isnot wss and then specialized to the wss case without doing muchadditional work. We will follow that path.Lecture 13 8Crosscorrelation TheoremLet x(t)andy(t) be random processes that are related byy(t)=∞−∞x(t − s)h(s)dsThenRxy(t1,t2)=∞−∞Rxx(t1,t2− β)h(β)dβandRyy(t1,t2)=∞−∞Rxy(t1− α, t2)h(α)dαTherefore,Ryy(t1,t2)=∞−∞Rxx(t1− α, t2− β)h(α)h(β)dαdβLecture 13 9Crosscorrelation TheoremProof Multiply the first equation by x(t1) and take the expectedvalue.E[x(t1)y(t)] =∞−∞E[x(t1)x(t − s)]h(s)ds =∞−∞Rxx(t1,t− s)h(s)dsThis proves the first result. To prove the second, multiply the firstequation by y(t2) and take the expected value.E[y(t)y(t2)] =∞−∞E[x(t − s)y(t2)]h(s)ds =∞−∞Rxy(t − s, t2)h(s)dsThis proves the second and third equations. Now substitute thesecond equation into the third to prove the last.Lecture 13 10Example: Autocorrelation for Photon ArrivalsAssume that each photon that arrivesat a detector generates an impulse ofcurrent. We want to model this pro-cess so that we can use it as the exci-tation X(t) to detection systems. As-sume that the photons arrive at a rateλ photons/second and that each photongenerates a pulse of height h and width.To compute the autocorrelation function we must findRxx(τ)=E[ X(t + τ)X(t)]Lecture 13 11Photon Pulses (continued)Let us first assume that τ>. Then it is impossible for the instantst and t + τ to fall within the same pulse.E[X(t + τ)X(t)] =x1x2x1x2P (X1= x1,X2= x2)=0· 0P (0, 0) + 0 · hP (0,h)+h · 0P (h, 0) + h2P (h, h)= h2P (X1= h)P (x2= h)The probability that the pulse waveform will be at level h at anyinstant is λ, which is the fraction of the time occupied by pulses.Hence,E[X(t + τ)X(t)] = (hλ)2for |τ| >Lecture 13 12Photon Pulses (continued)Now consider the case |τ| <. Then, by the Poisson assumption,there cannot be two pulses so close together so that X(t)=h andX(t + τ)=h only if t and t + |τ| fall within the same pulse.P (X1= h, X2= h)=P (X1= h)P (X2= h|X1= h)=λP (X2= h|X1= h)The probability that t + |τ| also hits the pulse isP (X2= h|X1= h)=1−|τ|/Hence,E[X(t + τ)X(t)] = h2α 1 −|τ|for |τ|≤If we now let  → 0 and keeph = 1 the triangle becomes animpulse of area h and we haveRxx(τ)=λδ(τ )+λ2Lecture 13 13Detector Response to Poisson PulsesIt is common for a physical de-tector to have internal resis-tance and capacitance. A se-ries RC circuit has impulse re-sponseh(t)=1RCe−t/RCstep(t)The autocorrelation function of the detector output isRyy(t1,t2)=∞−∞Rxx(t1− α, t2− β)h(α)h(β)dαdβ=1(RC)2∞0λδ(τ + α − β)+λ2e−(α+β)/RCdαdβ=λ(RC)2∞0e−(τ+2α)/RCdα + λ21RCe−u/RCdu2=λ2RCe−τ/RC+ λ2with τ ≥ 0Lecture 13 14White NoiseWe will say that a random process w(t) is white noise if its valuesw(ti)andw(tj) are uncorrelated for every tiand tj= ti. That is,Cw(ti,tj)=E[w(ti)w∗(tj)] = 0 for ti= tjThe autocovariance must be of the formCw(ti,tj)=q(ti)δ(ti− tj) where q(ti)=E[|w(ti)|2] ≥ 0is the mean-squared value at time ti. Unless specifically stated tobe otherwise, it is assumed that the mean value of white noise iszero. In that case,