MA6451 – RANDOM PROCESSES PART-A 1.The mean and variance of the binomial distribution are 4 and 3 respectively. Find p(X=0) Solution: Let x be a binomial random variable with parameters n and p. P(X=x)=n𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 𝑥 = 0,1,2, …n Mean=np, variance=npq Given mean=4 and variance=3 np=4→ (1), 𝑛𝑝𝑞 = 3 → (2) (2) 𝑛𝑝𝑞 3 → = (1) 𝑛𝑝 4 3 3 1 𝑞 = ,𝑝 = 1 − 𝑞 = 1 − = 4 4 4 4 4 n =𝑝 = 1 = 16 4

1 𝑥

3 16−𝑥,

P(X=x)=16𝐶𝑥 (4) (4)

𝑥 = 0,1,2, … 16

3 16

P(X=0)=𝑞16 = (4)

______________________________________________________________________________ 2.If the probability that a target isdestroyed on any one short is0.5.What is the probability that it would be destroyed on the 6th attempt Solution: Here destruction is success. First Success is on the 6th attempt. Let the random variable X denote the number of trails required for the first success. Given p=0.5 so q=0.5 and x=6 1

So, X follows geometric distribution with parameter p=0.5=2

Therefore the geometric distribution is P(X=x)=𝑞 𝑥−1 𝑝, 𝑥 = 1,2, … 1 𝑥−1 1

=(2)

1 𝑥

.2

=(2) , 𝑥 = 1,2, … 1 6

Therefore P(X=6) =(2)

=0.0156 ______________________________________________________________________________

3.The CDF of a continuous random variable is given by F(x)= {

𝟎,

𝒙<𝟎

Find the 𝟏 − 𝒆 ,𝟎 ≤ 𝒙 < ∞ −𝒙 𝟓

PDF and mean of x Solution: Given the CDF of the Continuous random variable X is −𝑥

F(x)={1 − 𝑒 5 , 𝑥 ≥ 0 0, 𝑥 < 0 The PDF of X is f(x)=𝐹1 (𝑥) 1

={

−𝑥

𝑒 5 ,𝑥 ≥ 0 5

0 , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 ∞

∞𝑥

Mean of X =E(X)=∫−∞ 𝑥𝑓(𝑥)𝑑𝑥 = ∫0 −𝑥

1

= 5 [𝑥

𝑒5

−1 5

−𝑥 5

−𝑥

− 1.

𝑒5 (

=-[𝑒 (𝑥 + 5)]

−1 2 ) 5

∞

∞

−𝑥

𝑒 5 𝑑𝑥 5

] 0

0

=-[𝑒 −∞ − 𝑒 0 . 5] =5

______________________________________________________________________________ 4. Establish the memory less property of the exponential distribution If X is exponentially distributed with parameter  , then for any two positive integers ‘s’ and ‘t’P[X>s+t/X>s] = P[x>t] − x ,x ≥ 0 Proof: The PDF of X is f(x)={  e 0, 𝑂𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Now P[X>k] =∫𝑘  e−  x ∞



 e  x  =      k

=e−  k Consider for s,t≥0, P[X>s+t/X>s] = = =

𝑃[𝑋>𝑠+𝑡∩𝑋>𝑠] 𝑃[𝑋>𝑠] 𝑃[𝑋>𝑠+𝑡] 𝑃[𝑋>𝑠] 𝑒 −  (s+t) e−  s

= e−  t =P[X>t] Thus P[X>s+t/X>s] = P[x>t] ∀ 𝑠, 𝑡 ≥ 0 ______________________________________________________________________________ 5. If y=x2,where X is a Normal random variable with zero mean and variance 𝝈𝟐 ,find the PDF of therandom variable Y. −𝑥2

1

Solution:Given fx(x)=𝜎√2𝜋 𝑒 2𝜎2 − ∞ < 𝑥 < ∞ Fy(y) = P[Y≤y]=P[X2≤y] = p[−√𝑦 ≤ 𝑥 ≤ √𝑦 ] =Fx(√𝑦 )- Fx(-√𝑦 ) fy(y)=2

1

√𝑦 1

=2

[𝑓𝑥 √𝑦 + 𝑓𝑥 (−√𝑦)] −𝑦

1

−𝑦

1

[ 𝑒 2𝜎2 + 𝜎√2𝜋 𝑒 2𝜎2 ] 𝑦 𝜎√2𝜋

√ 1

−𝑦

fy(y) =𝜎√2𝜋𝑦 . 𝑒 2𝜎2 ,y>0 ______________________________________________________________________________ 6.Find the acute angle between two lines of regression. Solution : 1− 𝑟 2

If 𝜃 is the angle between two regression lines , then 𝑡𝑎𝑛𝜃 = (

𝑟

)

𝜎𝑥 𝜎𝑦 𝜎𝑥 2 + 𝜎𝑦 2

.

______________________________________________________________________________ 7.The joint p.df of the two dimensional random variable (X,Y) is given by 8𝑥𝑦

𝑓(𝑥, 𝑦) = {

(i) (ii)

9 1 ≤ 𝑥 ≤ 𝑦 ≤ 2. Find 0 , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 Marginal densities of X and Y The conditional density functions 𝑓(𝑥/𝑦) and 𝑓(𝑦/𝑥)

Solution : (i)

The Marginal density functions of X and Y are given by 2

𝑓𝑋 (𝑥) = ∫𝑥 𝑓(𝑥, 𝑦)𝑑𝑦 2 8𝑥𝑦

∫𝑥

9

𝑑𝑦

8 𝑥𝑦 2

= ⌊ 9

2

⌋

2 𝑥

= =

8𝑥

⌊

9

4𝑥

4 − 𝑥2 2

⌋

⌊4 − 𝑥 2 ⌋ ,

9

1≤𝑥≤2

𝑦

𝑓𝑌 (𝑦) = ∫1 𝑓(𝑥, 𝑦)𝑑𝑥 𝑦 8𝑥𝑦

∫1

9

𝑑𝑥

8 𝑦𝑥 2

= ⌊ = = (ii)

⌋

9

2

8𝑦

𝑦2

⌊

9

4𝑦 9

𝑦

1 −1

2

⌋

(𝑦 2 − 1) ,

1≤𝑦≤2

The conditional density functions are given by 𝑓(𝑥/𝑦) = =

𝑓(𝑥,𝑦) 𝑓𝑋 (𝑥) 4𝑥 9

8𝑥𝑦 9

⌊4− 𝑥2 ⌋ 2𝑦

1≤𝑥≤𝑦≤2

= (𝑦 2 − 1) 𝑓(𝑦/𝑥) = =

𝑓(𝑥,𝑦) 𝑓𝑌 (𝑦)

8𝑥𝑦 9

4𝑦 2 (𝑦 − 1) 9

2𝑥

1≤𝑥≤𝑦≤2

= ⌊4− 𝑥 2 ⌋

______________________________________________________________________________ 8.If X has mean 4 and variance 9 , while Y has mean -2 and variance 5 , and the two are independent, find (a) E(XY) (b) E(X𝑌 2 ) Solution: Given E(X) = 4 , V(X) = 9, E(Y) = -2, V(Y) = 5 X and Y are independent ⇒ 𝐸(𝑋𝑌) = 𝐸(𝑋). 𝐸(𝑌) = 4 × −2 = E(X𝑌 2 ) = 𝐸(𝑋).E(𝑌 2 ) =4×9

8

= 36 ______________________________________________________________________________ 9.. Let X and Y be two discrete random variables with joint probability mass function 𝑃(𝑋 = 1

𝑥, 𝑌 = 𝑦) = {8

(𝑥 + 𝑦), 𝑥 = 1,2 𝑎𝑛𝑑 𝑦 = 0,1 0 , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Find the marginal probability mass functions of X and Y.

Sol:

The PMF is X Y 1 2 P(y)

0

1 0.125 0.25 0.375

0.25 0.375 0.625

P(x) 0.375 0.625 1

______________________________________________________________________________ 10.Define Wide sense stationary process A random process {X(t)} is called Wide sense stationary if its mean is constant and autocorrelation function depends only on the time difference i.e (i) E(X(t)) is always a constant (ii)E(X(t)X(t+𝜏))= Rxx(𝜏) ______________________________________________________________________________ 11.If {X(t)} is a normal process with µ(t)=10 and C(t1,t2)=16 𝒆−|𝒕𝟏 −𝒕𝟐| find the variance of X(10)X(6) Solution: Since {X(t)} is a Normal process, any member of {X(t)} is a normal random variable By definition C(t1,t2)= R(t1,t2)-E[X(t1)].E[X(t2)] C(t1,t2)= Var(X(t1)) C(t1,t2)= Var {X(t)} Now X (10) is a normal random variable with mean µ(10)=10 and variance C(10,10) =16 Var(U)= Var[X(10)-X(6)] =Var[X(10)]+Var[X(6)]-2Cov[X910,X(6)] =Cov(10,10)+Cov(6,6)-2Cov(10,6) =16 𝑒 −|10−10| + 16𝑒 −|6−6| − 2 × 16𝑒 −|10−6| =16+16-2 16e-4

= 31.4139 𝜎𝑢=√31.4139 =5.6048 ______________________________________________________________________________ 12.Prove that a first order stationary process has a constant mean Solution: Let X(t) be a first orderStationary process Therefore the first order density function is unaffected by shift in time origin fx(x,t)= fx(x,t+𝛿) for any → (1) ∞

Now E[X(t)]=∫−∞ 𝑥𝑓𝑥 (𝑥, 𝑡)𝑑𝑥 ∞

And E[X(t+ 𝛿)] = ∫−∞ 𝑥𝑓𝑥 (𝑥, 𝑡 + 𝛿)𝑑𝑥 ∞

=∫−∞ 𝑥𝑓𝑥 (𝑥, 𝑡)𝑑𝑥 = 𝐸[𝑋(𝑡)]𝑢𝑠𝑖𝑛𝑔 (1) Therefore E[X(t)]=E[X(t+ 𝛿)] 𝑓𝑜𝑟 𝑎𝑛𝑦 𝛿 Hence E[X(t)] is aconstant. Hence mean is constant ______________________________________________________________________________ 13.State the postulates of a Poisson process If X(t) represents the number of occurrences of a certain event in (0,t) then the discrete random process {X(t)} is called the Poisson Process,Provided the following Posulates are satisfied. (i) P(1 Occurrence in (t,t+∆𝑡))=  ∆𝑡 + 0(∆𝑡) (ii)P (0 Occurrence in (t,t+∆𝑡))=1-  ∆𝑡 + 0(∆𝑡) (iii)P(2 or more Occurrence in (t,t+∆𝑡)) = 0(∆𝑡) (iv) X(t) is independent of the number of occurrences of the event in any interval before and after interval (0,t) ______________________________________________________________________________ 14.Define ergodic random process A random process X(t) is said to be ergodic if its ensemble averages are equal to its appropriate time averages

15. Find the autocorrelation function of a stationary process, whose power spectral densityfunction is 2   ,   1   0,   1.

given by S ( )   Sol:

R( )   

1 2 1 2 1 2





S ( ) ei d  

 1



2

1

1  2 ei d  2 1

(cos   sin  ) d

1 1

2   cos  d 

1

1

1

 

2

cos  d

0

1   sin  2 cos       2 1  sin  2 cos  2sin        2 3 2

1

2sin      3 0  . 

_____________________________________________________________________________________ 15. Given that the autocorrelation function for a stationary ergodic process with no periodic components is Rxx    25 

4 . Find the mean and the variance of the process  X (t ) . 1  6 2

Sol: By the property of autocorrelation function,

 x2  lim Rxx ( )  

 25



x  5 . E  X 2 (t )  Rxx (0)  25  4  29



Var  X (t )  E  X 2 (t )   E  X (t )  29  25  4. 2

_____________________________________________________________________________________ 16. Show that Spectral density function of a real random process is even function. Sol : By definition S ( ) 



 R( ) e



 i

d



S ( ) 



 R( ) e

i

d



Putting   u ,

S ( ) 



 R(u) e

 iu

du

 



 R(u) e

 iu

du

[since R( ) is an even function]



 S ( ). Therefore, S ( ) is an even function of  _____________________________________________________________________________________ 17. Define Cross Correlation function and state any two of its properties. Sol: The cross correlation of two processes X(t) and Y(t) is denoted by

RXY (t1 , t 2 )  E[ X (t1 )Y (t 2 )] for any t1 and t 2 If X(t) and Y(t) are jointly WSS, then RXY (t , t   )  R XY ( ) Properties (i) RXY ( )  RYX ( ) (ii) R XY ( ) 

R XX (0) RYY (0)

_____________________________________________________________________________________ 18. Define Cross-Spectral density. Sol: Let X(t) any Y(t) be two jointly stationary processes with cross correlation function R XX ( ) . Then the cross power spectrum of X(t) and Y(t) is S XY ( ) 



R

XY

( )e i  d



_____________________________________________________________________________________ 19. Define a System. When is it called a linear system? Sol: A system is a functional relation between input x(t ) and output y (t ). A System y(t )  f [ x(t )] given by the function f is said to be linear if for any two inputs x1 (t ) and x 2 (t ) the output

f [a1 x1 (t )  a2 x2 (t )] = a1 f [ x1 (t )]  a2 f [ x2 (t )] for any constants a1 and a 2 . i.e., the system is linear if the principle of superposition holds good. Otherwise, the system is said to be non-linear.

_____________________________________________________________________________________

20. Define Band-Limited white noise. Sol: Noise having a non-zero and constant power spectral density over a finite frequency band and zero elsewhere is called band-limited white noise.  The PSD of the band-limited white noise is given by

 N0  S NN ( )   2   0

, for   WB , elsewhere

_____________________________________________________________________________________ 21. Define White noise process. Sol: A sample function X(t) of a WSS noise random process {X(t)} is called white noise if the power spectral density of {X(t)} is a constant at all frequencies. We denote the power spectral density of white noise w(t) as S w ( f ) 

N0 2

_____________________________________________________________________________________ 22. Define Linear time invariant system. Sol: A linear system given by y(t )  f [ x(t )] is said to be time-invariant if y(t  h)  f [ x(t  h)] for any

h  (, ) i.e., any time shift h in the input results in the same shift in time of the output. _____________________________________________________________________________________

23. .State the Convolution form of the output of a linear time invariant system. Sol: If X (t ) is the input and h(t ) be the system weighting function and Y (t ) is the output, then 

Y (t )  X (t ) * h(t )  h(u ) X (t  u )du 

_____________________________________________________________________________________ 24.

PART B 1. (A) The density function of random variable X is given by f(x)={

𝑲𝒙(𝟐 − 𝒙), 𝟎 ≤ 𝒙 ≤ 𝟐 𝟎, 𝒆𝒍𝒔𝒆𝒘𝒉𝒆𝒓𝒆

Find the value of K, the mean, variance and rth moment 𝐾𝑥(2 − 𝑥), 0 ≤ 𝑥 ≤ 2 Solution: Given the pdf of Continuous random variable X is f(x)={ 0, 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒 ∞

∫ f(x)dx = 1 −∞ 2

∫ 𝑘𝑥(2 − 𝑥)𝑑𝑥 = 1 0 2

𝑘 ∫(2𝑥 − 𝑥 2 )𝑑𝑥 = 1 0 2

2𝑥 2 𝑥 3 𝑘[ − ] =1 2 3 0 23 𝑘 [2 − ] = 1 3 2

4𝑘 =1 3 3 =𝐾 4 3/4𝑥(2 − 𝑥), 0 ≤ 𝑥 ≤ 2 f(x)={ 0, 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒 ∞

𝜇𝑟1

= 𝐸(𝑋

𝑟)

= ∫ 𝑥 𝑟 𝑓(𝑥)𝑑𝑥 −∞

2

3 = ∫ 𝑥 𝑟 𝑥(2 − 𝑥)𝑑𝑥 4 0

2

3

= 4 ∫0 (2𝑥 𝑟+1 − 𝑥 𝑟+2 )𝑑𝑥 2

3 𝑥 𝑟+2 𝑥 𝑟+3 = [2 − ] 4 𝑟+2 𝑟+3 0 2 2.2𝑟+2

= 3[ =

𝑟+2

−

2𝑟+3 𝑟+3

]

3 𝑟+3 𝑟 + 3 − (𝑟 + 2) 2 [ ] (𝑟 + 2)(𝑟 + 3) 4

3 2𝑟+3 → (1) 4 (𝑟 + 2)(𝑟 + 3)

𝜇𝑟1 = Substituting r=1, 2 in (1) 𝜇11

3 24 3 25 6 1 = = 1, 𝜇2 = = 4 3.4 4 4.5 5 6

1

Mean of X is E(X) =𝜇11 = 1 and Var(X)=𝜇21 − (𝜇11 )2= 5 − 1 = 5 1. (B) X is distributing normally, with mean 16 and standard deviation 3. Find(i)P(X≥19),(ii)P(12.5
Put Z= (i)

𝜎

=

𝑋−16 3

19−16

When X=19, Z=

3

=1

P(X≥19)=P(Z≥1) =05-P(0
P((12.5
When X=12.5,Z=

3 19−16

When X=19 ,Z=

(iii)

3

= −1.16

=1

Therefore P(12.5
When X=10,Z=

3 25−16

When X=10,Z=

3

= −2 =3

P(10X<25)=P(-21/2. Solution: (i)∑7𝑥=0 𝑝(𝑥) = 1 K+2K+2K+3K+K2+2k2+7k2+K=1 10k2+9K-1=0

4 3k

5 K2

6 2k2

7 7k2+k

5 1 100

6 2 100

7 17 100

1

K=10 or K=-1 But since P(x) cannot de negative ,K=-1 is rejected 1

Hence K=10 X

0 0

1 2 1 2 P(x) 10 10 (ii) P(X<6)=P(X=0)+P(X=1)+….+P(X=5) 1

2

2

3

1

3 2 10

4 3 10

81

=10+10 + 10 + 10 + 100 = 100 81

19

P(X≥6)=1-P(X<6)=1-100=100 P(0
(iii) P(X≤3)= 2

8

8 10

1

P(X≤4) =10 > 2 ______________________________________________________________________________ 3. If the p.d.f of a two dimensional random variable (X,Y) is given by 𝑓(𝑥, 𝑦) = 𝑥 + 𝑦 , 0 ≤ 𝑥, 𝑦 ≤ 1. Find the p.d.f of U=XY. 𝑢

Solution: 𝑢 = 𝑥𝑦 , 𝑣 = 𝑦. Hence 𝑥 = 𝑣 and 𝑦 = 𝑣. 𝐽=

𝜕𝑥

𝜕𝑧

|𝜕𝑢 𝜕𝑦

𝜕𝑣 | 𝜕𝑦

𝜕𝑢

𝜕𝑣

1 𝑣 = |−𝑢 𝑣2

0 |= 1

1 𝑣

The joint p.d.f. of 𝑢 and 𝑣 is given by 𝑔(𝑢, 𝑣) = 𝑓(𝑥, 𝑦)|𝐽| = (𝑥 + 𝑦)

1 𝑢 1 𝑢 𝑢 = ( + 𝑣) = 2 + 1 = 1 + 2 𝑣 𝑣 𝑣 𝑣 𝑣

Since , 0 ≤ 𝑦 ≤ 1 , 0 ≤ 𝑣 ≤ 1, , 0 ≤ 𝑥 ≤ 1 ⇒ , 0 ≤ 𝑢 ≤ 𝑣 𝑣 varies from 𝑣 = 𝑢 to 𝑣 = 1. Hence the p.d.f of U is given by ∞

𝑓(𝑢) = ∫ 𝑔(𝑢, 𝑣)𝑑𝑣 1

−∞

𝑢 ] 𝑑𝑣 𝑣2 𝑢 𝑢1 = [𝑣]1𝑢 − [ ] = 1 − 𝑢 − 𝑢 + 1 𝑣 𝑢 = ∫ [1 +

𝑓(𝑢) = 2(1 − 𝑢), 0 ≤ 𝑥 ≤ 1. ______________________________________________________________________________

4.

The Probability distribution of the process{X(t)} is given by (𝒂𝒕)𝒏−𝟏 , 𝒏 = 𝟏, 𝟐, … 𝒏+𝟏 𝑷(𝑿(𝒕) = 𝒏) (𝟏 + 𝒂𝒕) 𝒂𝒕 𝒏=𝟎 { 𝟏 + 𝒂𝒕 , Show that {X(t)} is not stationary. Solution: Given

X(t)=n P{X(t)}=p(xn)

0

1

𝑎𝑡 1 + 𝑎𝑡

1 (1 + 𝑎𝑡)2

2

𝑎𝑡 (1 + 𝑎𝑡)3

Mean E{X(t)}=∑∞ 𝑛=0 𝑛𝑝(𝑥𝑛 ) 𝑎𝑡

1

(𝑎𝑡)2

𝑎𝑡

=0.1+𝑎𝑡 +1.(1+𝑎𝑡)2 +2.(1+𝑎𝑡)3 + 3. (1+𝑎𝑡)4 + ⋯ 1

𝑎𝑡

𝑎𝑡

2

=(1+𝑎𝑡)2 [1 + 2. 1+𝑎𝑡 + 3. (1+𝑎𝑡) + ⋯ ]

3

(𝑎𝑡)2 (1 + 𝑎𝑡)4

... ...

1

𝑎𝑡

= (1+𝑎𝑡)2 [1 − 1+𝑎𝑡] 1

= (1+𝑎𝑡)2 × [ 1

−2

1+𝑎𝑡−𝑎𝑡 −2 1+𝑎𝑡 1

]

=(1+𝑎𝑡)2 × (1+𝑎𝑡)−2 1

= (1+𝑎𝑡)2 (1 + 𝑎𝑡)2 =1, which is a constant Now 2 E(X2(t)) =∑∞ 𝑛=0 𝑛 𝑝(𝑥𝑛 ) = ∑∞ 𝑛=0[𝑛(𝑛 + 1) − 𝑛]𝑝(𝑥𝑛 ) ∞ =∑∞ 𝑛=0 𝑛(𝑛 + 1)𝑝(𝑥𝑛 ) − ∑𝑛=0 𝑛𝑝(𝑥𝑛 ) 1

(𝑎𝑡)2

𝑎𝑡

={0 + 1.2 (1+𝑎𝑡)2 + 2.3 (1+𝑎𝑡)3 + 3.4 (1+𝑎𝑡)4 + ⋯ }-E{X(t)} 2

𝑎𝑡

𝑎𝑡

2

=(1+𝑎𝑡)2 {1 + 3. (1+𝑎𝑡) + 6. (1+𝑎𝑡) + ⋯ } − 1 2

𝑎𝑡

=(1+𝑎𝑡)2 [1 − 1+𝑎𝑡] =

2 (1+𝑎𝑡)2

−3

−1

. (1 + 𝑎𝑡)3 − 1

=2(1+at)-1 = 2+2at-1 =1+2at Var{X(t)}= E(X2(t))-[E(x(t))]2 =1+2at-1 =2at Which is dependent on t {x(t)} is not stationary ______________________________________________________________________________ 5. (A) If the Wss process {X(t)} is given by X(t)=10cos(100t+𝜽),where 𝜽 is uniformly distributed over (-π,π).Prove that {X(t)} is correlation ergodic Solution: We Know that Rxx(𝜏)= E(X(t).X(t+ 𝜏)) =E(100cos(100t+𝜃)cos(100t+100 𝜏 + 𝜃)) =100E(cos(100t+𝜃+100t+100 𝜏 + 𝜃)+cos(100t+𝜃-100t-100 𝜏 − 𝜃)) 100

=

2

E(cos(200t+2𝜃+100 𝜏) +cos(-100 𝜏))

50 E(cos(200t+2𝜃+100 𝜏) +cos100 𝜏) = 50cos100 𝜏+50E(cos(200t+2𝜃+100 𝜏)) → (1) 1

𝜋

Now E(cos(200t+2𝜃+100 𝜏)) =2𝜋 ∫−𝜋 cos(200t + 2𝜃 + 100 𝜏)𝑑𝜃 1

𝜋

= 𝜋 ∫0 cos(200t + 2𝜃 + 100 𝜏)𝑑𝜃

𝑠𝑖𝑛(200t + 2𝜃 + 100 𝜏) 𝜋 =[ ] 2 0 1

= 2𝜋 [ 𝑠𝑖𝑛(200t + 2𝜋 + 100 𝜏)- 𝑠𝑖𝑛(200t + 100 𝜏)] =0 Substituting in Equation (1),we have E(X(t).X(t+ 𝜏))=50cos100 𝜏 → (2) 1

𝑇

Therefore ZT= 2𝑇 ∫−𝑇 𝑋(𝑡). 𝑋(𝑡 + 𝜏)𝑑𝑡 1

𝑇

50

𝑇

25

𝑇

=2𝑇 ∫−𝑇 100 cos(100𝑡 + 𝜃) cos(100𝑡 + 100𝜏 + 𝜃) 𝑑𝑡 =2𝑇 ∫−𝑇[𝐶𝑐𝑜𝑠(200𝑡 + 2𝜃 + 100𝜏) + cos(−100𝜏) 𝑑𝑡 = 𝑇 ∫−𝑇 cos(100𝜏) 𝑑𝑡 + 25

𝑇

25 𝑇

𝑇

∫𝑇 𝐶𝑜𝑠(200𝑡 + 100𝜏 + 2𝜃)𝑑𝑡

=50cos(100 𝜏)+ 𝑇 ∫𝑇 𝐶𝑜𝑠(200𝑡 + 100𝜏 + 2𝜃)𝑑𝑡 25

1

= 50cos(100 𝜏)+ 𝑇 [200 (sin(200𝑡 + 2𝜃 + 100𝜏) − (𝑠𝑖𝑛200𝑡 + 2𝜃))] 1

=50cos(100 𝜏)+4𝑇 [sin(200𝑡 + 2𝜃 + 100𝜏) − (𝑠𝑖𝑛200𝑡 + 2𝜃)] Now lim (𝑍𝑇 ) = 50cos(100 𝜏) 𝑇→∞

=R(𝜏) Therefore {X(t)} is correlation ergodic ______________________________________________________________________________ 5. (B) If customers arrive at a counter in accordance with poisson process with a mean rate of 2 per minute, find the probability that the interval between 2 consecutive arrivals is (a)more than 1 minute (b) between 1 minute and 2 minute and (c) 4 minute or less Solution: Let T be the random variable denoting inter arrival time. By Property (3) fT(t)= e  t and P(T>t)= e  t Here  =2/minute Therefore fT(t)=2 e

2 t

∞

(a) P(T>1)=∫1 2𝑒 −2𝑡 𝑑𝑡 = 0.1353 2

(b) P(1≤T≤2)=∫1 2𝑒 −2𝑡 𝑑𝑡 = 0.117 4

(c) P(T≤4) =∫0 2𝑒 −2𝑡 𝑑𝑡 = 0.99967

6. State and prove Wiener-Khintchine theorem Statement: Let X(t) be a real WSS process with power density spectrum S XX ( ) . Let X T (t ) be a portion of the

 X (t ) ,T  t  T X T (t )   , elsewhere  0 process X(t) in time interval –T to T. i.e.,

Let X T ( ) be the Fourier

transform of X T (t ) , then

S XX ( ) 

lim

1

T   2T



E X T ( )

2



Proof: Given X T ( ) is the Fourier transform of X T (t )

 X T ( ) 



X

T

(t )e it dt

 T



X

T

(t )e it dt

T T



 X (t )e

it

dt

T

X T ( )  X T* ( ) X T ( ) [where * denotes complex conjugate] 2

T

T

T

T

it it  X (t )e dt.  X (t )e dt



[ X (t ) is real]

T T it  it 2 dt   X (t )e 1 dt .  X (t )e 1 1 2 2 T T T T



  X (t ) X (t 1

2

)e i ( t2 t2 ) dt1dt 2

T T

T T



lim T 



E X T ( )

2

  T   2T lim

1

  X (t ) X (t 1

2

)e i ( t2 t2 ) dt1dt 2

T T

But EX ((t1 )(t 2 )  RXX (t1 , t 2 ) if  T  t1 , t 2  T T T



lim T 



E X T ( )

2

  T   2T lim

1

 R

T T

XX

(t1 , t 2 )e i ( t2 t2 ) dt1dt 2

We shall now make a change of variables as below Put t1  t and t 2  t1    t 2    t  the jacobian of transformation is

t1 J  t t 2 t

1 0 t1  1 1 1  t 2 

 dt1dt 2  J dtd The limits of t and –T and T When t 2  T ,  T  t and t 2  T ,  T  t



lim

1

T   2T



E X T ( )

2

1   T lim   2T

T t

T

 R

XX

(t , t   )e i  dtd

XX

( )e i  dtd

T t T

Since X(t) is WSS Process, RXX (t , t   )  RXX ( )



lim

1

T   2T



E X T ( )

2

1   T lim   2T

T t

T

 R

T t T





lim

1

T t

T

T t

T

T   2T

lim

i   RXX ( )e d .  dt

1

T t

T   2T

R

XX

( )e i  d .2T

T t



lim T 

T t

R

XX

( )e i  d

T t





R

XX

( )e i  d  S XX ( )





 S XX ( )  lim E X T ( ) T  2T

2



, by definition.

Hence proved. _____________________________________________________________________________________

7. If X(t) is the input voltage to a circuit (system) and Y(t) is the output voltage.{X(t)} is a stationary process with  x  0 and Rxx ( )  e function is H ( ) 

 

. Find  y , S yy and Ryy ( ) , if the power transfer

R . R  iL 

 h(u) X (t  u)du

Y (t )  Sol: (i) We know that

 EY (t ) 





 h(u) E[ X (t  u)]du



Since X(t) is stationary with mean 0, E[ X (t )]  0 for all t  E[ X (t  u)]  0  E[Y (t )]  0

S XX ( ) 



R

XX

( )e i  d



(ii)We know that





e

 

e i  d





0

  e e i  d   e  e i  d 

0 

0

  e ( i ) d   e ( i ) d 

0 0



 e ( i )   e  ( i )          i     (  i )  0 1 1  [1  0]  [0  1]   i   i 1 1 2   i    i    2    i   i (  i )(  i )    2 Given H ( ) 

R . R  iL

We know that S yy ( )  H ( ) S xx ( ) 2

2

R2 2  2 R  iL     2



R2 2 . 2 2 2 2 R  L   2

(iii) The autocorrelation function of Y(t) is

RYY ( ) 

1 2

1  2 



S

YY

( )e i  d

 

2R 2 e i   2 2 2 . 2 2 d  ( R  L  )   

R 2 



1 e i  .  ( R 2  L2 2 )  2   2 d

1 2 2 First we shall write ( R  L  )(   ) as partial fraction, treating 2

2

2

ω2 as u. We shall write the special partial fraction as

1 A B  2  2 2 2 2 2 ( R  L  )(   ) R  L   2 2

2

2

R2 L2 we get A  L2  2 L2  R 2 1 Put u   2 we get B  2 R  L2 2 L2 1 2 2 2 2 1  L  R  R  L2 2  2  ( R  L2 2 )( 2   2 ) R 2  L2 2  2 2 Put u  

L2 1 1 1  2 2 2 2 2 2 2 2 2 2  L R R L R  L   2



  R 2 L2 1 R 2 1 i  RYY ( )  .e d  .e i d 2 2 2 2 2 2 2 2 2 2    ( L  R )  ( R  L  )  ( L  R )    2



R 2 L2 1 2 2 2  ( L  R ) L2



1





2

(

R 2) 2 L

By Contour integration, we know that 

e imz  ma  z 2  a 2 dz  a e , m  0

.e i d 



R 2  ( L  R 2

2

2

)  

2

1 .e i d 2 

 RYY ( ) 

R 2 L2   R 2    e  e 2 2 2 2 2 2  ( L  R )  R   ( L  R )    L





LR  2 L2  R 2

e



LR R2 L2 ( 2  2 ) L

e

R2 R   e   2 2 2 L  L  R   

R   L

R2 R2 L2 ( 2  2 ) L

e

 

2

R R      L  e   R    L   e   2 2 L R R 2   2   L L

 

R   L  2 R 2    L

  R    R    e  L    e L 

  

_____________________________________________________________________________________