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ENEE630 ADSP Part II w/ solution1. Determine if each of the following are valid autocorrelation matrices of WSS processes.(Correlation Matrix)Ra=4 1 1−1 4 1−1 −1 4, Rb=2 1 11 2 01 0 2, Rc=2j 0 j0 2j 0−j 0 2j, Rd=1 0 20 1 02 0 1.Solution:Recall that the properties of an autocorrelation matrix for a WSS process is that (1) R isToeplitz; (2) RH= R; (3) R is non-negative definite.Rais NOT Hermitian; Rbis NOT Toeplitz; Rcis NOT Hermitian; Rdis NOT non-negativedefinite (λ = 1, −1, 3).2. Consider the random process y(n) = x (n) + v(n), where x(n) = Aej(ωn+φ)and v(n) is zeromean white Gaussian noise with a variance σ2v. We also assume the noise and the complex sinusoidare independent. Under the following conditions, determine if y(n) is WSS. Justify your answers.(WSS Process)(a) ω and A are constants, and φ is a uniformly distributed over the interval [0, 2π].(b) ω and φ are constants, and A is a Gaussian random variable ∼ N (0, σ2A).(c) φ and A are constants, and ω is a uniformly distributed over the interval [ω0− ∆, ω0+ ∆]for some fixed ∆.Solution:(a)E[y(n)] = AejωnEφ[ejφ] + Ev[v(n)] = 0E[y(n)y∗(n − k)] = Eφ[(Aej(ωn+φ)+ v(n))(A∗e−j(ω(n−k)+φ)+ v∗(n − k))]= |A|2Eφ[ejωk] + σ2vδ(k)= |A|2ejωk+ σ2vδ(k)1st and 2nd moments are independent of n. Thus, the process is WSS.(b)E[y(n)] = EA[A]ej(ωn+φ)+ Ev[v(n)] = 0E[y(n)y∗(n − k)] = EA[(Aej(ωn+φ)+ v(n))(A∗e−j(ω(n−k)+φ)+ v∗(n − k))]= EA[AA∗]ejωk+ σ2vδ(k)= σ2Aejωk+ σ2vδ(k)11st and 2nd moments are independent of n. Thus, the process is WSS.(c)E[y(n)] = Eω[x(n)] + Ev[v(n)] = A · Eω[ejωn] · ejφ=Aejφ2jn∆ejωnω0+∆ω0−∆⇒ |E[y(n)]| ≤ |Aejφ2jn∆| · 2 → 0 as n → ∞E[y(n)y∗(n − k)] = Eω[(Aej(ωn+φ)+ v(n))(A∗e−j(ω(n−k)+φ)+ v∗(n − k))]= |A|2Eω[ejωk] + σ2vδ(k)= |A|2ejω0ksin(k∆)k∆+ σ2vδ(k)The sequence defined here is actually NOT a WSS process, but its 1st and 2nd moment statisticsare approximately independent of n as n → ∞.3. [Rec.II P2(a) revisited] Determine the PSD of the WSS process y(n) = Aej(ω0n+φ)+ v(n),where v(n) is zero mean white Gaussian n oise with a variance σ2v, and φ is uniformly distributedover the interval [0, 2π]. (Power Spectral Density)Solution:In the autocorrelation function in P2(a) isry(k) = A2ejωk+ σ2vδ(k)By taking discrete time Fourier tr an sform on ry(k), we getPy(ω) = 2πA2δ(ω − ω0) + σ2v4. Assume v(n) is a white Gaussian random process with zero mean and variance 1. Th e twofilters in Fig. RII.4 are G(z) =11−0.4z−1and H(z) =21−0.5z−1. (Auto-Regressive Process)v(n)-G(z)-H(z)-u(n)Figure RII.4:(a) Is u(n) an AR process? If so, find th e parameters.(b) Find the autocorrelation coefficients ru(0), ru(1), and ru(2) of the process u(n ).Solution:(a) U(z) =21−0.9z−1+0.2z−2V (z), u(n) = 0.9u(n − 1) − 0.2u(n − 2) + 2v(n ), a1= −0.9, a2= 0.2.2(b) Apply the Yule-Walker equation,ru(0) ru(1)ru(1) ru(0)−0.90.2= −ru(1)ru(2),from which we getru(1) = −a11+a2ru(0) =34ru(0)ru(2) =a211+a2− a2ru(0) =1940ru(0)Moreover, since ru(0) + a1ru(1) + a2ru(2) = 4σ2v(Here, ‘4’ because in this model it is ‘2v(n)’ ratherthan ‘v(n)’), we have ru(0) =1+a21−a24σ2v(1+a2)2−a21=20021. T hen, ru(1) =507, and ru(2) =9521.Note:1. In general, for a p-order AR model, given {σ2v, a1, a2, . . . , ap}, we can find {r (0), r(1), r(2), . . .};and vice versa. T hey are related by Yule-Walker Equations.2. r(−k) = r∗(k) in general (and hence matrix R is Hermitian), and r(−k) = r(k) for real-valuedsignals. r(0) is the power of s equence u(n), and hence r(0) > 0 from physical point of view.3. For an AR model, u(n) =Ppk=1−aku(n−k)+v(n) has NO correlation with future v(m), m =n + 1, n + 2, . . . (convince yourself). Simply multiply both sides by u∗(n) and take expectation, weget r(0) =Ppk=1−akr(−k) + E(v(n)u∗(n)). Note that E(v(n)u∗(n)) = E(v(n)(Ppk=1−a∗ku∗(n −k) + v∗(n))) but E(v(n)u∗(n − k)) = 0 for k ≥ 1. Then, r(0) =Ppk=1−akr(−k) + σ2v, which wehave used to find the relation of r(0) (signal power) and σ2v(model parameter) in part (b). Wecould multiply u∗(n − k) instead of u∗(n) and take the expectation, and this is h ow the Yule-Walkerequations are derived.5. Let a real-valued AR(2) process be described byu(n) = x(n) + a1x(n − 1) + a2x(n − 2)where u(n ) is a white noise of zero-mean and variance σ2, and u(n) and past values x(n−1), x(n−2)are uncorrelated. (Yule-Walker Equ ation)(a) Determine and solve the Yule-Walker Equations for the AR process.(b) Find the variance of the process x(n).Solution: (a) Solve the Yule-Walker equation, we haverx(0) = −a1rx(−1) − a2rx(−2) + σ2rx(1) = −a1rx(0) − a2rx(−1)rx(2) = −a1rx(1) − a2rx(0)Use the relation that rx(k) = rx(−k) and solve this we get3rx(0) =σ21 −a211 + a2+ a2(a211 + a2− a2)rx(1) = −a11 + a2rx(0)rx(2) = (a211 + a2− a2)rx(0)(b) The process is zero m ean, so the variance is rx(0).6. [Problem II.4 continued ] Assume v(n) and w(n) are white Gaussian random processes withzero mean and variance 1. The two filters in Fig. RII.6 are G(z) =11−0.4z−1and H(z) =21−0.5z−1.(Wiener Filter)Figure RII.6:(a) Design a 1-order Wiener fi lter such that the desired output is u(n). What is the MSE?(b) Design a 2-order Wiener fi lter. What is the MS E?Solution:(a) Rx="ru(0) + 1 ru(1)ru(1) ru(0) + 1#, and pxd="ru(0)ru(1)#. The filter is w = R−1xp with MSEru(0) − pHxdR−1xpxd.(b) Similar to (a), except Rx=ru(0) + 1 ru(1) ru(2)ru(1) ru(0) + 1 ru(1)ru(2) ru(1) ru(0) + 1, and pxd=ru(0)ru(1)ru(2).MSE is still the s ame expression, i.e. ru(0) − pHxdR−1xpxd.Note:1. In general, for a p-order AR model, given {σ2v, a1, a2, . . . , ap}, we can find {r (0), r(1), r(2), . . .};and vice versa. T hey are related by Yule-Walker Equations.2. r(−k) = r∗(k) in general (and hence matrix R is Hermitian), and r(−k) = r(k) for real-valuedsignals. r(0) is the power of s equence u(n), and hence r(0) > 0 from physical point of view.3. For an AR model, u(n) =Ppk=1−aku(n−k)+v(n) has NO correlation with future v(m), m =n + 1, n + 2, . . . (convince yourself). Simply multiply both sides by u∗(n) and take expectation, weget r(0) =Ppk=1−akr(−k) + E(v(n)u∗(n)). Note that E(v(n)u∗(n)) =


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UMD ENEE 630 - Part II w/ solution

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