PROBABILITY AND RANDOM PROCESSESS

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1 PROBABILITY AND RANDOM PROCESSESS SOLUTIONS TO UNIVERSITY QUESTION PAPER YEAR : JUNE 2014 CODE NO : 6074 /M PREPARED BY: D.B.V.RAVISANKAR ASSOCIATE PROFESSOR IT DEPARTMENT MVSR ENGINEERING COLLEGE, NADERGUL

2 PROBABILITY & RAMDOM PROCESSESS SOLUTIONS OF JUNE-2014, CODE NO /M (B.E. 2/4 IT II-SEM MAIN EXAMINATION) PART-A (25M) 1.Define properties of Joint Cumulative distribution function. (2M) Sol. The following are the properties of Joint cdf I.F(-,y) = 0, F(x,-) =0, F(,) =1 II. P{x 1 <X x 2,Y y} = F(x 2,y) F(x 1,y) P{X x,y 1 Y y 2 } = F(x,y 2 ) F(x,y 1) III. P{x 1 X x 2, y 1 Y y 2 } = F(x 2,y 2 ) F(x 1,y 2 ) F(x 2,y 1) +F(x 1,y 1) 2. If A and B are independent events, prove that A and B are also independent (2M) Sol. We know that if A and B are independent P(AB) = P(A). P(B) P A = 1 P A, P B = 1 P B P A P B = 1 P A 1 P B = 1 P A P B + P A P B = 1 P A + P B P AB = 1 P A B = P A B = P(A B) (The above two identities are from addition theorem and Demorgan s theorem) 3.State any three properties of characteristic function (3M) Sol. I) The n th order moment m n is the coefficient of j n ω n in the expansion of Ф(ω) =E{e jωx } II) m n = 1 dn (ω at ω = 0 j n dω n III) If X and Y are independent RVs, then Ф X+Y (ω) = Ф X (ω) Ф Y (ω) 4.Define Axiomatic definition of probability (2M) Sol. Let S be the sample space and A be an event associated with a random experiment. Then the probability of the event A denoted by P(A) is defined as a real number satisfying the following axioms. 1.0 P(A) 1 2.P(S)=1 3. If A and B are mutually exclusive events, P(AՍB) = P(A) + P(B) 5.What is the difference between random variable and random processes (3M) Sol. A random variable is a rule that assigns a real number to every outcome of a random experiment, while a random process is a rule that assigns a time function to every outcome of a random experiment. n!

3 6. Find the mean and variance of poisson distribution (3M) Sol. For a random variable X with parameter λ the probability function of poisson distribution λ λk is given by P X = k = e k = 0,1,2,. k! The mean of poisson distribution E X = k P X = k k=0 = k=0 k λ λk e k! = λ E X 2 = k 2 k=0 λ k 1 ke λ k k 1! =λ e λ e λ k=1 = λ P X = k, k = 0,1,.., = k=2 k(k 1) λ 2 e λ λ k 2 k k 1 k 2! + k=0 k λ λk e k! = λ 2 + λ The variance of poisson distribution = E{X 2 } [ E{x}] 2 = λ 2 + λ - λ 2 = λ Hence the mean and variance of poisson distribution is λ 7. Write any three properties of auto-correlation (3M) Sol. I. R xx (τ) = R xx (-τ) II. R xx (0)= E{X 2 (t)} III. X 2 = lim τ R xx (τ), if X(t) is a stationary process with no periodic component 8. Define Gaussian process (3M) Sol. A real valued random process X(t) is called a Gaussian process or normal process if the r random variables X(t 1 ), X(t 2 ), X(t 3 ),., X(t n ) at the n time instants t 1,t 2,., t n will have a n th order density function given by f x x 1, x 2,., x N ; t 1, t 2,, t N = exp { 1 2 [x X]t [C x ] 1 x X } (2π) N C x Where matrix[x X] t = [x X 1 x X 2.. x X N ] C 11 C 1N Co-Variance matrix [C x ] = is a N N matrix C N1 C NN

4 9.State any three properties of power spectral density function (2M) Sol. I. S xx (ω)=s xx (-ω) II.S xx (ω) = d2 dω 2 S xx (ω) III. S xx (ω) State Wiener-Khinchin thorem Sol. For a random process X(t) the power spectral density and auto correlation function will form a Fourier transform pair. S ω = R τ = 1 2π R τ e jωτ dτ S(ω)ejωτ dω (OR) The area of the power spectrum of any random process is positive. i.ee{x 2 t } = R 0 = 1 S ω dω 0 2π PART-B (50 M) 11. (a) Suppose box 1 contains a white balls and b black balls, and box 2 contain c white balls and b black balls. One ball of unknown color is transferred from the first box into the second one and then a ball is drawn from the later. What is the probability that it will be white ball. (5M) (Papoulis Example 2.14) Sol. Let us define the following events W={ transferred ball is white} B={transferred ball is black} We know that W U B = S P(W) = a a+b and P(B) = b a+b The event of interest A ={ white ball is drawn from the second box} The probability of A P A = P A W B = P{ A W A B } =P A W + P(A B) ( Using Axiom 3) = P(A W) P(W) + P(A B) P(B)

5 But P(A W) = c+1 c +d+1 and P(A B) = c Hence P A = a c+1 + a+b c+d+1 which is the required probability c+d+1 bc a+b c+d+1 = ac +bc +a a+b c+d+1 (b) Show that P A 1 A 2 A 3 A n = P A 1 + P A P A n P(A 1 A 2 A 3.. A n ) (5M) Sol. Consider two events A 1 and A 2 which can be represented on a venn Diagram From the venn diagram event A 1 = (A 1 A 2 ) U (A 1 A 2 ) andevent A 2 = (A 1 A 2 ) U (A 1 A 2 ) The events on the R.H.S of the above two are mutually exclusive. Hence by applying the probability on both the sides and by using axiom 3 P(A 1 )= P(A 1 A 2 ) + P(A 1 A 2 ) and P(A 2 ) = P(A 1 A 2 ) +P(A 1 A 2 ) Adding both of them we have P(A 1 )+P(A 2 )= P(A 1 A 2 ) + P(A 1 A 2 ) +P(A 1 A 2 ) +P(A 1 A 2 ) The first 3 terms on the R.H.S of above represents P(A 1 U A 2 ) Therefore P(A 1 )+P(A 2 )= P(A 1 U A 2 ) +P(A 1 A 2 ) Or P(A 1 U A 2 ) = P(A 1 )+P(A 2 )-P(A 1 A 2 ), which is the desired result and A 1 A 2 is the intersection of A 1 and A 2 The above proof can be extended to n events A 1,A 2,.,A n to obtain the desired result P A 1 A 2 A 3 A n = P A 1 + P A P A n P(A 1 A 2 A 3.. A n ) This is also called as addition theorem of probability 12.Three switches S 1,S 2,and S 3 connected in parallel operate independently and each switch remains closed with probability p. (10 M) ( Papoulis Example 2.21) (a) Find the probability of receiving an input signal at the output (b) Find the probability that the switch S 1 is open given that an input signal is received at the output. Sol. (a) Let A i = Switch S i is closed Then P(A i )=p, i=1,2,3 Since switches operate independently P(A 1 A 2 A 3 ) = P(A 1 ) P(A 2 ) P(A 3 ) The event R ={ Input signal is received at the output } Because the 3 switches S 1,S 2,and S 3 are connected in parallel R =A 1 U A 2 U A 3

6 P(R) = 1-P(R ) = 1-P(A 1 A 2 A 3 ) = 1-P(A 1 ) P(A 2 ) P(A 3 ) = 1-(1-p) 3 = 3p-3p 2 +p 3 (b) We have to determine P(A 1 R) From Baye s theorem P(A 1 R) = P R A 1 ) P(A 1 ) P(R) 13.Given f xy (x,y) = Cx(x-y), 0<x<2, y and 0 elsewhere, a)evaluate C, b) find f x (x) c) find f y/x (y/x) = 2p p2 (1 p) 3 3p+p 2 solution: For the given problem the limits are x < y <x, o< x<2 a) we know that f xy x, y dxdy = 1 x x 0 2 cx x y dxdy = 1 8c=1, c=1/8 b) we know that f x x = f xy x, y dy = 1 x x y dy y= x 8 c) f y x = f(x,y) f(x) = 1 2x 2 x = x 3 4, o < x < (a) If Y=X 2 where X is a Gausssian random variable with zero mean and variance σ 2, find the Pdf of the random variable Y. (b) If the continuous random variable X has pdf f x (x) = 2/9 (x+1), 1<x<2 and 0 elsewhere. Find the pdf of Y=X 2 Sol. (a) Given Y=g(X)= X 2 Define function y=g(x) =x 2, there are two solutions x 1 = y and x 2 = y dy dx = 2x dx dy = 1 2x = 1 2 y Therefore the density of Y f y (y) = dx dy [f x x 1 + f x x 2 ] f y (y) = 1 2 y [f x( y) +f x ( y)] (b)since f x (x) = 2/9 (x+1), 1<x<2 The density of Y f y (y) = f x (x) dx dy f y (y)= 2 9 x x = 1 9 (1 + 1 y )

7 15 (a) Show that the random X(t) = a cos(ω 0 +θ) is wide-sense stationary (WSS), if A and ω 0 are constants and θ is a uniformly distributed random variable in (0,2ᴨ) (5M) Sol. The density of θ f θ (θ) = 1 2π To prove X(t) as a WSS process we have to show that mean of X(t) is constant and Auto-correlation function (ACF) of X(t) must depend on difference in the time intervals 2π θ=0 Mean of X(t) E{X(t)} = X t f θ θ dθ = acos ω 0 t + θ 1 2π 0 By solving the above integral E{X(t)} = 0 2π dθ ACF of X(t),R xx (т) = E{X(t) X(t+т)} = E{acos ω 0 t + θ acos (ω 0 t + τ + θ)} 2π = a 2 cos ω 0 t + θ cos ω θ=0 0 t + τ + θ 1 dθ 2π By applying trigonometric identity and solving the above integral we will get R xx (т) = a 2 2 cos (ω 0τ). Since the mean is constant and ACF is dependent on the difference in the time intervals the given process X(t) is a WSS process. (b) Find the power spectral density (PSD) of a WSS process with auto correlation function (ACF) R(t) = a 2 e -2β t Sol. We know that the PSD and ACF form a fourier transform pair. Therefore the PSD S ω = R t e jωt dt = a 2 e 2β t e jωt dt The above integral can be splitted into two parts containing real and imaginary terms as following S ω = a 2 e 2β t cos ωt dt j a 2 e 2β t sin ωt dt The frirst integral is an even function i.e f(x) = f(-x) and second integral is a odd function i.e f(x) = -f(-x). Hence the second integral is equated to zero. S ω = 2 0 a 2 e 2βt cos ωt dt By solving the above integral using the following identity ʃe ax cos bx =e ax / a 2 +b 2 [ a cos bx +bsinbx] and subsituiting the limits o and The PSD of the given problem is given by S ω = 4a 2 β 4β 2 +ω 2

8 16. (a) If the power spectral density (PSD) of a WSS process is given by S ω = b a ω, ω a a = 0 ω a Find the auto correlation function (ACF) of the process. (5M) 2π Sol. The ACF of X(t) R τ = 1 S(ω) e jωτ dω = 1 b a ω 2π a ejωτ dω a The integral on the RHS can be splitted into real and imaginary parts as R τ = 1 a a b 2π a a ω cos ωτ dω + j 1 b a ω sin ωτ dω a 2π a a By applying the concept of even and odd functions the second integral will become zero. a R τ = 1.2 b 2π 0 a a a ω cos ωτ dω The integral on the RHS is simplified as b πa { a ω sin ωτ the imits 0 and a. By substituting the limits the ACF is given by b R τ = (1 cos aτ) πaτ2 τ cos ωτ τ 2 } between (b) Write short notes on (i) Thermal noise (ii) Filters (5M) Sol. (i) Thermal Noise : is the noise because of the random motion of electrons in conducting media such as a resistor. Thermal noise generated in resistors and semiconductors is assumed to be a Zero mean, stationary Gaussian random process {N(t)}with a PSD S NN (ɷ) that is flat over range of frequencies. Therefore Thermal noise is also called as White noise.psd S NN ω = N 0 2 (ii) Filters : Filters are used in electrical systems to reject undesirable signals and noise and to select the desired signal. Filtering actually means selecting carefully the transfer function H(ɷ) in a stable, linear, time-invariant system, so as to modify the spectral components of the input signal. The system function H(ɷ) is referred to as filter. The 4 types of filters are low-pass, high-pass, band-pass and band-stop filters. 17. Consider a white Gaussian noise of zero mean and power spectral density N 0 /2 applied to low-pass RC filter whose transfer function is H(f) = of the output random process. (10M) 1 1+j 2π f RC. Find the auto correlation Sol. Given a low-pass RC circuit with transfer function H(f). This is a liner time-invariant system. If X(t) and Y(t) are the input & output random process of the RC filter with PSDs S xx (ɷ) and S yy (ɷ) respectively.

9 The PSDs are defined by the following relation. S yy (ɷ)= H(ɷ) 2 S xx (ɷ) (1) For the given problem the transfer function is given in terms of frequency. Hence (1) can be written as S yy (f) = H(f) 2 S xx (f) (2) Given the applied input is having a PSD S xx (f) = N 0 /2, (2) can be written as the following S yy f = 1 N 0 1+4π 2 f 2 R 2 C 2 2 The autocorrelation function of the output random process Y(t) is given by R yy τ = S yy f e j2πfτ from 3 & 4 R yy τ = N 0 = (3) df(4) ( Where dɷ = 2π df) e j 2πτf 2 N 0 e j 2πτ f df 8π 2 R 2 C 2 1 ( 2πRC )2 + f 2 e jmx a 2 +x 2 1+4π 2 f 2 R 2 C 2 df By comparing the integral on the RHS of (6) with dx, which can be evaluated by contuour integration technique which results in the following. (5) (6) e jmx a 2 +x 2 dx = π a e m a (7) Using (7) in (6) with m = 2πτ and a = 1 2πRC R yy τ = 2πτ N 0 π 2π RC 8π 2 R 2 e 2πRC (8) C 2 Therefore the ACF of the output random process R yy τ = N 0 2 e τ /RC

for valid PSD. PART B (Answer all five units, 5 X 10 = 50 Marks) UNIT I

for valid PSD. PART B (Answer all five units, 5 X 10 = 50 Marks) UNIT I Code: 15A04304 R15 B.Tech II Year I Semester (R15) Regular Examinations November/December 016 PROBABILITY THEY & STOCHASTIC PROCESSES (Electronics and Communication Engineering) Time: 3 hours Max. Marks: