Unit IV: Correlation & Spectral Densities PDF

UNIT IV CORRELATION AND SPECTRAL DENSITIES PART – A 1. Define correlation of the process { X t }. Answer: If the process X t is stationary either in the strict sense or in the wide...

UNIT IV CORRELATION AND SPECTRAL DENSITIES PART – A 1. Define correlation of the process { X t }. Answer: If the process X t is stationary either in the strict sense or in the wide sense, then E[ X t X t   ] is a function of  , denoted by R  or RXX  . This function RXX   is called the autocorrelation function of the process X t . 2. State any two properties of an autocorrelation function. Answer: i. R  is an even function of . ii. If R  is the autocorrelation function of a stationary process X t with no periodic component, then lim R    x2 , provided the limit exists.   3. Prove that for a WSS process { X t } , RXX    RXX   . Answer: RXX    E[ X t X t   ] RXX     E[ X t X t   ]  E[ X t   X t ]  RXX   Therefore R  is an even function of . 4. Show that the autocorrelation function RXX   is maximum at   0. Answer: RXX   is maximum at   0 i.e. R   R0    Cauchy-Schwarz inequality is E[ XY ]  E X 2 E Y 2 2 Put X  X t  and Y  X t    , then EX t X t   2  EX 2 t E X 2 t    i.e. R 2  EX 2  2 [Since E  X t  and Var  X t  are constants for a stationary process] R 2  R02 Taking square root on both sides,   R   R0. [Since R0  E X 2 t  is positive] 5. Statistically independent zero mean random processes X t  and Y t  have autocorrelation functions RXX    e and RYY    cos 2 respectively.  Find the autocorrelation function of the sum Z t   X t   Y t . Answer: 104 ------------------------------------------------------------------------------------------------------------------------------------------------ Given RXX    e , RYY    cos 2  E  X t   EY t   0 If Z t   X t   Y t  , then RZZ    RXX    RYY    RXY    RYX   RXY    E[ X t Y t   ] [Since the processes are independent] RXY    0 0  0 Similarly RYX    0 RZZ    e    cos 2  0  0   e  cos 2 6. The autocorrelation function of a stationary process is RXX    16  9. Find the mean and variance of the process. 1  6 2 Answer: Given RXX    16  9 1  6 2  x2  lim R     9   lim16      1  6 2   9   16  lim    1  6 2    16  0  16 Mean   x  E X t   4  E X 2 t    RXX 0  9  16  1  60   16  9  25 Variance 2    E X t   E  X t  2  25  4   25  16  9 2 7. If the autocorrelation function of a stationary processes is RXX    36  4. Find the mean and variance of the process. 1  3 2 Answer: Given RXX    36  4 1  3 2  x2  lim R     4   lim 36      1  3 2   4   36  lim    1  3 2    36  0  36 105 ------------------------------------------------------------------------------------------------------------------------------------------------ Mean   x  E  X t   6  E X 2 t    RXX 0  4  36  1  30   36  4  40 Variance 2    E X t   E  X t  2  40  6   40  36  4. 2 8. The random process X t  has an autocorrelation function RXX    18  1  4 cos12 . Find E[ X t ] and E[ X 2 t ]. 2 6   2 Answer: Given RXX    18  1  4 cos12  2 6  2 E X 2 t     RXX 0  2  18  [1  4 cos 0] 60 2 59  18  [1  4]  6 3 Removing the periodic components of RXX   , then  x2  lim RXX      2   lim18      6  2   2   18  lim    6   2    18  0  18 Mean   x  E  X t   18 9. Define cross correlation function and state any two of its properties. Answer: If the process X t and Y t  are jointly wide sense stationary, then E  X t Y t    is a function of  , denoted by RXY  . This function RXY   is called the cross correlation function of the process X t and Y t . Properties of cross correlation function are: i. RXY     RYX  . ii. If the process X t and Y t  are orthogonal, then RXY    0. iii. If the process X t and Y t  are independent, then RXY    E X t E Y t   . 106 ------------------------------------------------------------------------------------------------------------------------------------------------ 10. Define power spectral density function of stationary random processes X t . Answer: If X t is a stationary process with autocorrelation function R  , then the Fourier transform of R  is called the power spectral density function of X t and denoted as S   or S XX  .  i.e. S     R e  i d.  11. Define cross spectral density. Answer: If X t and Y t  are two jointly stationary random processes with cross correlation function RXY   , then the Fourier transform of RXY   is called the cross spectral density of X t and Y t  and denoted as S XY  .  i.e. S XY     R  eXY  i d.  12. State and prove any one of the properties of the cross spectral density function. Answer: Cross spectral density function is not an even function of  , but it has a symmetry relationship. i.e. SYX    S XY    Proof:  S XY     R  eXY  i d   S XY      R  e d  XY i  Putting   u when   , u   d  du when   , u    S XY      RXY  u e  iu  du    S XY      R u e YX  iu du  RXY     RYX      R  e  i  YX d   SYX   i.e. SYX    SYX    107 ------------------------------------------------------------------------------------------------------------------------------------------------ 13. Find the power spectral density of a random signal with   autocorrelation function e. Answer: Given RXX    e    S     R  e XX  i d    e   e  i d    e   cos  i sin  d     e   cos d  i  e    sin  d     2 e   cos  d (Since the first integrand is even 0 and the second integral is odd)       2 e   cos   sin         2 2 0   2 0  2 1    0    2  2 S    2  2 14. The autocorrelation function of the random telegraph signal process is given by R   a 2e 2  . Determine the power density spectrum of the random telegraph signal. Answer: Given R   a 2e 2    S     R  e XX  i d     a 2e  2  e  i d    a2  e  2  cos  i sin  d     a2  e  2  cos d  ia 2  e 2  sin  d     2a 2  e  2  cos  d (Since the first integrand is even 0 and 108 ------------------------------------------------------------------------------------------------------------------------------------------------ the second integral is odd)   2    2a 2  e  2 cos    sin     2    2 2 0   2a 2 0  2 1  2  0  4   2  4 S    2 4   2 15. Find the power spectral density of a WSS process with autocorrelation function R   e . 2 Answer: Given R   e  2  S     R e  i d     e  e  i d 2    i     2    e    d   i  i   i   2 2     2            e     d  2 2    i  2  2             e  2   4 d 2  e  i  2 2         e 4 e   2  d  i  dx Put      x   d  dx  d   2   When   , x   When   , x   2   S   dx e  x2 e 4   2  4  e e  x2  dx    i.e. S XY     R  e XY  i d 2   4 e    2   4  e  109 ------------------------------------------------------------------------------------------------------------------------------------------------ 16. IF the power spectral density of a WSS process is give b  a    ,  a by S     a. 0 ,  a  Answer:  R   S  e 1 i  d 2  1   a a    S  e i d    S  e i d    S  ei d  2    a a  1  b  a   a   e d  i  2   a a  a b a   cos  i sin d 2a a  a a  b  a   cos d  i b  a   sin d 2a  a 2a  a a 2  a   cos d  i 0 b b  2a 0 2a a b a   cos d a 0  a    sin   cos a b     a     0 2 b  cos a   1  0     0  2  a       2 b  1 cos a    2  a   2  R   1  cos a . b a 2 17. The power spectral density function of a zero mean wide sense 1 ;   0 stationary process X t is given by S    . Find R . 0 ; Elsewhere Answer:  R   S  ei d 1 2    0 0     S  e d   S  e d   S  e d  1 i i i  2     0 0  1  0 i      1.e d  2   0  110 ------------------------------------------------------------------------------------------------------------------------------------------------ 0  cos  i sin d 1  2  0 0 0  cos d  i  sin d 1 1  2  0 2  0  1 0 2  cos d  i 0 [ The 1st integrand is 1  2 0 2 even and the 2nd is odd]  1  sin   0       0  1 sin 0  0  sin  0 R  .  18. The power spectral density of the random process X t is given  ; 1 by S    . Find its autocorrelation function. 0 ; Elsewhere Answer:  R   S  e 1 i  d 2  1   1 1    S  e d   S  e d   S  e d  i i i  2    1 1  1   1   .e d  i  2  1  1  cos  i sin  d 1 2 1  1 1  1  cos d  i 1  sin  d 2 1 2 1 1  2  cos d  i 0 [ The 1st integrand is 1 1 2 0 2 even and the 2nd is odd]  sin   1     0  sin   0 1  sin  R  .  111 ------------------------------------------------------------------------------------------------------------------------------------------------ Given the power spectral density S XX    1 19. , find the average 4 2 power of the process. Answer:  R   S  e 1 i  d 2   1 1  2  4   2 ei d  1 1  2   4  z 2 eiz dz [Where C is the closed contour  consisting of the real axis from  R to R and the upper half of the circle z  R ]  eiz  2i  lim  z  2i  1  2  z 2 i z  2i z  2i   ei.2i   i   2i  2i   e 2   i   4i  e 2 R   4 e0  1 Average power  R0    4 4 112 ------------------------------------------------------------------------------------------------------------------------------------------------ PART – B 1. If X t is a WSS process with autocorrelation function RXX   and if Y t   X t  a   X t  a . Show that RYY    2 RXX    RXX   2a   RXX   2a . Answer: Given Y t   X t  a   X t  a  RYY t   E Y t Y t     E  X t  a   X t  a  X t    a   X t    a   E[ X t  a  X t    a   X t  a X t    a   X t  a X t    a   X t  a X t    a   E X t  a  X t  a     E  X t  a  X t  a     E  X t  a X t  a    a  a   E X t  a  X t  a    a  a   E X t  a X t  a     E X t  a X t  a     E X t  a  X t  a    2a   E X t  a  X t  a    2a   E X t  a  X t  a     E X t  a X t  a     E  X t  a X t  a    2a   E X t  a X t  a    2a   RXX    RXX    RXX   2a   RXX   2a  RYY    2 RXX    RXX   2a   RXX   2a . 2. A stationary random process X t  has autocorrelation function given 24 2  36 by RXX   . Find the mean and variance of X t . 6 2  4 Answer: 24 2  36 Given RXX    6 2  4  x2  lim R     24 2  36   lim   6  4    2  2 36      24  2      lim    2 4    6  2        24  0  4 60 Mean   x  E  X t   2 113 ------------------------------------------------------------------------------------------------------------------------------------------------  E X 2 t    RXX 0  240   36  60   4 36  9 4 Variance    E X 2 t   E  X t  2  9  2   9  4  5. 2 3. State the properties of autocorrelation function for a WSS process. Answer: i. R  is an even function of . i.e. RXX     RXX   ii. R  is maximum at   0. i.e. R   R 0 iii. If the autocorrelation function R  of a real stationary function X t is continuous at   0 , it is continuous at every other point. iv. If R  is the autocorrelation function of a stationary process X t with no periodic component, then lim R    x2 , provided the limit exists.   4. Given a stationary random process X t   10 cos100t    where     ,  followed uniform distribution. Find the autocorrelation function of the process. Answer: Given  is uniformly distributed in   ,  , the PDF of  is f    1 1  ,    .      2 RXX    E X t X t    RXX    E10 cos100t   .10 cos100t       RXX    100 E cos100t   cos100t  100      100  cos100t   cos100t  100    f  d    100  cos100t   cos100t  100    1 d  2  100 1 2 cos100t   cos100t  100   d 2  2   100 cos100t    100t  100   cos100t    100t  100   d 4   25        cos200t  100  2 d   cos100d       114 ------------------------------------------------------------------------------------------------------------------------------------------------ 25  sin 200t  100  2          cos100  d    2     25  sin 200t  100  2   sin 200t  100  2         cos100      2    sin200t  100 cos 2  cos200t  100 sin 2      25  sin200t  100 cos 2  cos200t  100 sin 2     cos100        2         25  2 cos100  RXX    50 cos100. 5. Consider two random processes X t   3 cost    and Y t   2 cost     where     and  is uniformly distributed random variable 2 over 0,2 . Verify whether RXX    RXX 0 RYY 0 . Answer: Given  is uniformly distributed in 0,2  , the PDF of  is f    1 1  ,0    2.      2   X t   3 cost    and Y t   2 cos t      2 RXX 0   E X t  2   2   9 cos 2 t    f  d 0  1  cos 2t     1 2  9   d 0 2  2 2 2 9     d   cos2t  2 d  4  0 0  9  2  sin 2t  2    2   0     4   2 0    2  0   2 sin 2t  4   sin 2t  0  9 1  4   2  2 sin 2t cos 4  cos 2t sin 4  sin 2t  9 1  4   2  2 sin 2t  sin 2t  9 1  4  1  9 2  2 0   4 2  2 9 9  4 115 ------------------------------------------------------------------------------------------------------------------------------------------------ RYY 0   E Y 2 t   2     4 cos 2  t     f  d 0  2      1  cos 2 t      2  4   2  1 d  2  2 0     2 2 4      d   cos2t  2   d  4  0 0  1  2  sin 2t  2      2   0        2 0  1  2  0  1 sin 2t  4     sin 2t     2  1  2  0  1 sin2t       sin2t     2  1  2  sin 2t cos3  cos2t sin 3  sin 2t cos  cos2t sin  1    2  1   2   sin 2t  sin 2t  1  2  1 1  1  2  0   2  2  2   RXX 0 RYY 0  .2  9 9 2 RXX 0 RYY 0   3 RXY    E  X t Y t    2     3 cost   2 cos t       f  d 0  2 2   1  6  cost   cos t       d 0  2  2 2   2 cost   cos t       d 31    20  2 2 3        2  cost    t      2   cost    t      2 d 0 2 2 3           cos 2t     2  d   cos     d  2  0  2   20  116 ------------------------------------------------------------------------------------------------------------------------------------------------ 2       sin  2t     2    3        2   cos    0 2  2  2   2      0  1             sin 2t     4   sin 2t       cos  2  0 3  2 2   2   2   2  1   7          sin 2t      sin 2t       cos  2  3  2 2   2   2   2  1   7           sin 2t      sin 2t       cos cos  sin sin 2  3  2 2   2  2   2 2    7 7    1  sin2t    cos 2  cos2t    sin 2    3     0  sin 2  2  2   sin2t    cos  cos2t   sin      2  2   3 1     cos2t     cos2t     2 sin  2  2  3 1  3   0  2 sin    2 sin   3sin 2  2  2 RXY     3 sin   3 sin  i.e. 3  3 sin  Since the function takes the maximum and minimum values to be 1 to -1, 3  3 sin . Hence RXX 0 RYY 0   RXY  . 6. The autocorrelation function for a stationary process X t  is given by RXX    9  2e . Find the mean value of the random 2 variable Y   X t dt and variance of X t . 0 Answer: Given RXX    9  2e.     x2  lim RXX    lim 9  2e    9  20   9      x  E  X t   3 2 E Y    E X t dt 0 2   3dt  3t 0  32  0   6 2 0  Mean of Y  E Y   6 117 ------------------------------------------------------------------------------------------------------------------------------------------------   2   2   R  d 2 EY XX 2  2   9  2e d 2   2  d 2  2  2    9  2e  0   2  2 2    9  2e  d 0   2  2  14  4e   9  2e  d 0 2  2   214  4e   9  2  e   e      2 0     2 18.2  4e  2  9  2 2e  2  e  2   0  4  0  20  1 22  2     2 36  4e  18  2 3e    2  2 2   220  2e   40  4e 2 2  E Y   E Y  2 Variance of Y 2  40  4e 2  62  4e 2  4  4e 3  1. 7. If X t   5 cost    and Y t   2 cost    where  is a constant     and  is a random variable uniformly distributed in 0,2  , 2 find RXX   , RYY   , RXY   and RYX  . Verify two properties of autocorrelation function and cross correlation function. Answer: Given  is uniformly distributed in 0,2  , the PDF of  is f    1 1  ,0    2.      2   X t   3 cost    and Y t   2 cos t      2 RXX    E X t X t    RXX    E 5 sin t   5 sin  t       2   5 sin t   5 sin  t       f  d 0 2  25  sin t   sin  t       1 d 0 2 2 2 sin t   sin t     d 25 1 2 2 0  118 ------------------------------------------------------------------------------------------------------------------------------------------------ 2  cost    t       cost    t     d 25  4 0 2  cos   cos2t    2 d 25  4 0 25   2 2  cos   d   cos2t    2 d  4  0 0   sin 2t    2   2 25   cos   0   2    4   2 0  25   cos  2  0   sin 2t    4   sin 2t    1   4  2  25  1  sin 2t   cos 4  cos2t   sin 4   2 cos     4  2   sin 2t     25   2 cos   sin 2t     sin 2t    1  4  2  25  1    2 cos   0  4  2  RXX    25 2 cos   25 cos 4 2 RYY    E Y t Y t    RYY    E2 cost   2 cos t              RYY    E 2 cos t    2 cos  t          2   2  2         2 cos t    2 cos  t        f  d 0  2   2  2       1  4  cost     cos  t        d 0  2   2  2 2 4        2  cost  2    cost    2   d 0 2     sin t    sin t     d 2  [ cos    cos ]  0  2 2 2 sin t   sin t     d 21  2 0  2  cost    t       cost    t     d 1   0 2  cos   cos2t    2 d 1   0 119 ------------------------------------------------------------------------------------------------------------------------------------------------ 2 2 1   cos   d   cos2t    2 d   0 0   sin 2t    2   2 1   cos  0   2      2 0  1  2 cos   sin 2t    4   sin 2t    1    2  1   2 cos   sin 2t     sin 2t    1  2  1 1   2 cos   0   2   2 cos   1  RYY    2 cos  RXY    E X t Y t        RXY    E 5 sin t   2 cos t         2       10 E sin t   cos t         2   10 E sin t    sin t      2  10  sin t   sin t      f  d 0 2 2 sin t   sin t      d 10 1   2 0 2 2  cost    t       cost    t     d 5  2 0 2  cos  2   cos2t   d 5  2 0 2 2 5      cos  2 d  cos2t     d  2  0 0  5  sin   2   2    cos2t    0  2   2  2 0  1   2 sin   4   sin   0   2 cos2t    5  2 1   2 sin   sin    2 cos2t    5  2  5  2 cos2t    2 120 ------------------------------------------------------------------------------------------------------------------------------------------------ RXY    5 cos2t    RYX    E Y t X t         RYX    E 2 cos t    5 sin t        2         10 E cos t     sin t        2    10 E  sin t   sin t      2  10  sin t   sin t      f  d 0 2 2 sin t   sin t      d 10 1   2 0 2 2  cost    t       cost    t     d 5  2 0 2  cos  2   cos2t   d 5  2 0 5   2 2    cos  2 d  cos2t     d  2  0 0  5  sin   2   2    cos2t    0  2   2  2 0  5 1   sin   4   sin   0  2 cos2t    2  2  5  1   sin  sin    2 cos2t    2  2     2 cos2t    5 2 RYX    5 cos2t    RXX     cos      cos   RXX   25 25 2 2 RXX     RXX   (1) RXX 0   cos  0   25 25 2 2 RXX    25 cos   25 2 RXX    RXX 0 (2) 8. Define power spectral density and cross spectral density of a random process. State their properties. 121 ------------------------------------------------------------------------------------------------------------------------------------------------ Answer: Power spectral density: If X t is a stationary process with the autocorrelation function R  , then the Fourier transform of R  is called the power spectral density function of X t and denoted as S XX  .  i.e. S XX     R e  i d.  Properties of power spectral density function: i. The value of the spectral density function at zero frequency is equal to the total area under the graph of the autocorrelation function. ii. The mean square value of a wide sense stationary process is equal to the total area under the graph of the spectral density. iii. The spectral density function of a real random process is an even function. iv. The spectral density of a process X t , real or complex is a real function of  and nonnegative. v. The spectral density and the autocorrelation function of a real WSS process form a Fourier cosine transform pair. Cross spectral density: If X t and Y t are two jointly stationary random processes with cross correlation function RXY   , then the Fourier transform of RXY   is called the cross spectral density of X t and Y t and denoted as S XY  .  i.e. S XY     R  e XY  i d  Properties of cross spectral density function: i. S XY    SYX    ii. ReS XY   and ReSYX   are even functions of  iii. ImS XY   and ImSYX   are odd functions of  iv. S XY    0 and SYX    0 if X t  and y t  are orthogonal. v. If X t  and y t  are uncorrelated, then S XY    SYX    2E X E Y S  . 9. State and prove Wiener-Khinchine theorem. Answer: Statement: If X T   is the Fourier transform of the truncated  X t  , for t  T random process defined as X T     where X t is a real 0 , for t  T WSS process with power spectral density function S   , then S    lim 1 T  2T E X T  . 2   122 ------------------------------------------------------------------------------------------------------------------------------------------------ Proof:  Given X T     X t eT  it dt  T   X t e  it dt [ X t is real] T X T    X T  X T    2 T T  X t e dt1  X t2 eit 2 dt2  it1  1