Time : 3 Hours Max. Marks: 60. Instructions : Answer any two questions from each Module. All questions carry equal marks. Module I

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1 Reg. No.:... Name:... First Semester M.Tech. Degree Examination, March 4 ( Scheme) Branch: ELECTRONICS AND COMMUNICATION Signal Processing( Admission) TSC : Random Processes and Applications (Solutions to Model Question Paper) Time : Hours Max. Marks: 6 Instructions : Answer any two questions from each Module. All questions carry equal marks. Module I. A computer chip manufacturer finds that, historically, for every chips produced, 85 meet specification, needs reworking and 5 needs to be discarded. Ten chips are chosen for inspection. (a) What is the probability that all meet specifications? (b) What is the probability that two or more need to be discarded? (c) What is the probability that 8 meet specification, one needs reworking and one will be discarded? It is the case of generalized Bernoulli trials. No. of trials, n =. Sp be the event of a chip meeting specification, Re be the event of a chip requiring rework and, D be the event of a chip being defective and discarded. Probability of a chip meeting specification, P Sp= 85 =.85 Probability of a chip requiring rework, P Re= =. Probability of a chip discarded, P D= 5 =.5 (a) Here sample space can be considered as S={Sp, Sp} with P Sp=.85 and P Sp=.5. P All chips meeting specifications=c P Sp P Sp =.969 (b) Here sample space can be considered as S={D, D} with P D=.5 and P D=.95. P chips discarded & chip discarded=c P D P D +C P D P D 9 =.99 = P or more chips need to be discarded = -.99 =.86. (c) Here sample space is the original S={Sp, Re, D}. P 8 chips meet specification, needs reworking, discarded=! 8!!! P Sp 8 P Re P D =.6. Let X be a Gaussian random variable with pdf f X (x) = π e x ; < x <

2 Let Y = g(x) where g(.) is the nonlinear function given as (a) Sketch g(x). for x < Y = g(x) = x <, x <, x > (b) Find F Y (y), the probability distribution of Y. (c) Find and sketch f Y (y), the probability density function of Y. (a) (b) F Y (y) = f Y (y) (c) (i) for P Y = = P x <, = ( ) erf() =.6 (ii)for <, x <,, Y = X and hence f Y (y) = f X (x) (iii)p Y = = P x > = P x <, =.6 Hence f y (y) is given by Page

3 . X and Y are two independent random variables both uniform in -, and let Z = X + Y, Find (a) the probability density function of Z. (b) the characteristic function of Z. Z = X + Y and X and Y are independent and uniform in -, = f Z (z) = f X (z) * f Y (z). i.e., z 4 + ; z z f Z (z) = 4 + ; < z ; elsewhere (a) Let W = Z. w = The two solutions are z = - w and z = w. f W (w) = f Z(z ) dw dz atz = z 4 + = - w + f Z(z ) dw dz atz + z 4 + for w. { z ; z z ; < z { dw ; z dz = ; < z Page

4 The required density function is f W (w) where W = Z. (b) Characteristic function of Z = φ Z (ω) = Ee jωz = ejωz z 4 + dz + ejωz z 4 + dz = ejωz f Z (z)dz = ωsin(ω)+cos(ω) ω + Sin(ω) ω. Module II 4. If X and Y are two zero mean jointly Gaussian random variables with equal variance, find (a) the conditional density of the random variable X given that Y = y. (b) the conditional expectation of the random variable X given that Y = y. (c) the co-variance matrix of the random vector containing X and Y. (a) X, Y N(, σ ) f XY (x, y) = πσ e (x rxy+y ) r σ ( r ) f Y (y) = e y σ πσ ; < y < ; < x, y < f X/Y (x/y) = f XY (x,y) f Y (y) = e (x ry) σ ( r ) ; < x, y < πσ ( r ) i.e., X/Y = y N(ry, σ ( r )). (b) EX/Y =y=ry (from above Gaussian pdf). (c) C XY = C Y X = EXY EX.EY = rσ C XX = EX (EX) = σ C Y Y = EY (EY ) = σ σ rσ C = rσ σ. 5. If X is a random vector with zero mean and co-variance matrix given by R X =. 4 Find a linear transformation Y = AX such that the co-variance matrix of Y is an identity matrix. Page 4

5 Given the co-variance matrix of X we have to find a transformation matrix A such that the co-variance matrix of Y is an identity matrix where Y = AX. Then A must be A = Z {U T } where U T is transpose of the matrix formed from unit eigenvectors and λ Z = λ where λ and λ are the eigenvalues of the co-variace matrix of X If λ is an eigenvalue of R X then λ 4 λ ( λ)(4 λ) = 7λ + λ = λ 7λ + = solving these equations we get λ = and λ = 5 Now, (R X λ I)V λ = v v = where V λ is eigenvector of λ solving for v and v we get V λ = normalised eigenvector is as follows: Page 5

6 similarly for λ (R X λ I)V λ = v v = where V λ is eigenvector of λ solving for v and v we get V λ = normalised eigenvector is as follows: U matrix is U= Z= 5 now the transformation A is A = Z U T and A= 5 A= where A is the transformation matrix 6. For a Poisson process N(t) with parameter λ, find the following. (a) The distribution t n, the time to the n th arrival. (b) The autocorrelation function of N(t). (a) The time between the arrivals, τn = t n t n, where t n is time to n th arrival. Its distribution is exponential and is given by, f τ (t) = λte λt u(t) Page 6

7 Let t = τ + τ If t = t f t (t) = λ te λt u(t) = λ(λt)e λt u(t) Now let t = τ + τ + τ f t (t) = λ e λt t (λt) u(t) = λ e λt u(t) Generalising, let t n = τ + τ τ n f tn (t) = λ (λt)n (n )! e( λt) u(t) Hence the distribution of time to the n th arrival is Erlang distribution. (b) Taking t < t Autocorrelation = EN(t )N(t ) =E(N(t ) + N(t ) N(t ))N(t ) =EN (t ) + EN(t ) N(t )EN(t ) by independent increment property. =λt + λ t + λ(t t )λt =λt + λ t t. Similarly if t < t we merely interchange t and t. Generally Autocorrelation is given by R NN (t, t ) =EN(t )N(t ) =λmin(t, t ) + λ t t. Module III 7. The resistors r, r, r and r 4 are independent random variables and each of them are uniform in the interval(45,55). Let R = r + r + r + r 4 (a) Using the central limit theorem,find P {9 r + r + r + r 4 }. (b) Find the Chebyshev bound on P {9 r + r + r + r 4 }. (a) The resistors r, r, r and r 4 are independent random variables and each of them are uniform in the interval(45,55). Er i = = 5 varr i = (55 45) = 8. Let R = r + r + r + r 4 ER = 4 5 = varr = 4 8. =. By using the central limit theorem, P {9 r + r + r + r 4 } = erf(. ) erf( 9. ) = erf( ) erf( ) = erf( ) (b) By Chebyshev s inequality, P X µ > a σ a ie. P µ a < X < µ + a σ a Page 7

8 Here µ =, a = P 9 < X <. = A computer device can be either in a busy mode (state ) processing a task, or in an idle mode (state ), when there are no tasks to process. Being in a busy mode, it can finish a task and enter an idle mode any minute with the probability.. Thus, with the probability.8 it stays another minute in a busy mode. Being in an idle mode, it receives a new task any minute with the probability. and enters a busy mode. Thus, it stays another minute in an idle mode with the probability.9. The initial state is idle. Let X n be the state of the device after n minutes. (a) Find the distribution of X, (b) Find the steady-state distribution of X n. (a) The one-step transition matrix P is given by, p p P = p p The distribution of X is given by, = p () = p ().P where p () is given as, We can find P as, P = p () = = Therefore, we obtain the distribution of X as, p ().66.4 =.7.8 =.7.8 (b) We can find the steady-state distribution Π using the relation, ΠP = Π Substituting the values, we get.8. π π..9 = π π Equating the values, we get the equation,.8π +.π = π which is equivalent to.π =.π Page 8

9 Also, we know that, π + π = Therefore, solving these two equations, we get,.π =.( π ).π =. Therefore the steady state probabilities are π = π = 9. (a) Find the autocorrelation function of a Wiener process and then find its K-L expansion. (b) Show how a white noise process can be obtained from a Wiener process. (a) Autocorrelation: Let W (t) be a Wiener process. Assume that t t E{W (t )W (t ) W (t )} = EW (t )EW (t ) W (t ) () Since Wiener process has independent increment property () changes to E{W (t )W (t ) W (t )} = Since W (t) is a Wiener noise EW (t ) =. EW (t )W (t ) W (t ) = EW (t )W (t ) = EW (t ) For Wiener noise EW (t ) = αt because individual RVs are Gaussian Similarly by taking t t we can prove that From () & () R W W (t, t ) = αt () R W W (t, t ) = αt () R W W (t, t ) = αmin(t, t ) (4) KL Expansion According to KL Expansion any random process can be expanded as X(t) = Σ n=x n φ n (t) The basis functions are obtained by solving the integral equation T R XX (t, s)φ n (s) dx = λ n φ(t) Page 9

10 From (4) for a Wiener process R W W (t, s) = αmin(t, s). Let us assume a standard Wiener process so α =. Let us assume that tɛ(, T ) t t min(t, s)φ(s) ds = λφ(t) (5) T sφ(s) ds + t tφ(s) ds = λφ(t) (6) Differentiating (6) w.r.t t tφ(t) + d T dt t φ(s) ds = λφ (t) (7) Differentiating (9) w.r.t t tφ(t) + t( φ(t)) + t T t T t φ(s) ds = λφ (t) (8) φ(s) ds = λφ (t) (9) CP of () is given by.φ(t) = λφ (t) () φ(t) = λφ (t) () λφ (t) + φ(t) = () λ + = λ = ±j φ(t) = A sin( t ) + B cos( t ) λ λ Evaluate A and B using boundary conditions At t = + φ(t) = B = At t = T φ (t) = A cos( T ) = λ λ T λ = (n ) π λ = (n ) π T Choose A such that T φ (t) ds = = A = T Page

11 Now put λ = λ n and φ(t) = φ n (t) φ n (t) = T sin(n )πt, t T, n T So KL expansion of Wiener Process is given by X(t) = T Σ n=x n sin(n )πt, t T, n T where T X n = X(t) sin(n T )πt T dt (b) As we know Autocorrelation of Wiener Process (W (t)) is given by EW (t )W (t ) = αmin(t, t ) Now consider the process W (t) = W (t + t) W (t) t We are going to prove that as a limiting case t the process W (t) will be White process { } { } E W (t)w W (t + t) W (t)w (t τ + t) W (t τ) (t τ) = E t { } { } W (t + t)w (t τ + t) W (t)w (t τ + t) = E ( t) E ( t) { } { } W (t + t)w (t τ) W (t)w (t τ) E ( t) + E ( t) Above four terms and their sums are listed in the table. Condition Term Term Term Term 4 Result τ = α(t+ t) αt αt αt α < τ < t τ t ( t) α(t+ t) ( t) α(t τ+ t) ( t) ( t) αt ( t) α(t τ+ t) ( t) ( t) α(t τ) ( t) α(t τ) ( t) Therefore the autocorrelation of W (t) is given by ( t) αt ( t) α(t τ) ( t) t R W (τ) = α τ ( ) if < τ < t t t Otherwise α t ( τ t ) This represents equation of a triangle wave. As t we not that R W (τ) = αδ(τ). Hence differentiating a Wiener Process w.r.t time gives White Noise. Page