Chapter 1 Random Process: 1.1 Introduction (Physical Phenomenon)
Chapter 1 Random Process: 1.1 Introduction (Physical Phenomenon)
Chapter 1 Random Process: 1.1 Introduction (Physical Phenomenon)
S X (t,s )
-T t T
x j (t )= sample function
(1.1)
(1.2)
Y (t ' )Y (t ' )
1
X (t ) E X (t )
xf X ( t ) ( x ) d x
for all t
(1.6)
(1.7)
- -
- -
x1 x2 f X ( t1 ) X ( t2 ) ( x1 , x2 )dx1dx2
x1 x2 f X (0) X ( t2 t1 ) ( x1 , x2 )dx1dx2
RX (t2 t1 )
(1.8)
(1.10)
RX ( ) E X (t ) X (t ) ,
for all t
(1.11)
RX (0) E X 2 (t ) , 0
(1.12)
2. RX ( ) R( )
(1.13)
3. RX ( ) RX (0)
(1.14)
10
Proof of property 3:
E [( X (t ) X (t )) ] 0
2
Consider
E[ X 2 (t )] 2 E[ X (t ) X (t )] E[ X 2 (t )] 0
2 E [ X (t )] 2 RX ( ) 0
2
2 RX (0) 2 RX ( ) 0
RX (0) RX ( ) RX (0)
| RX ( ) | RX (0)
11
12
(1.15)
(1.16)
A2
RX ( ) E X (t ) X (t )
cos(2f ct )
2
(1.17)
13
14
15
(t ) 0,
and
t0
(t ) dt 1
(A2.3)
(A2.4)
g (t ) (t t0 )dt g (t0 )
(A2.5)
g ( ) (t )d g (t )
(A2.6)
17
18
E X (thence,
X (tindependent.
k ) X ( ti ) kE X ( tk ) E
i ) 0
19
Figure 1.6
Sample function of random binary wave.
20
21
5. For tk ti T , tk 0 , ti tk
X(tk) and X(ti) occur in the same pulse interval
iff td T- tk -ti
i.e., td -ti T
A2 , td T - tk - ti
E X (tk ) X (ti ) td
0,
E X (tk ) X (ti )
T- tk -ti
0
elsewhere
A 2 f Td (td )dtd
A2
dtd
0
T
tk ti
2
A (1
)
T
T- tk -ti
tk ti T
22
A (1 ), T
RX ( )
, where tk -ti
T
0,
T
23
Cross-correlation Function
RXY (t,u ) E X (t )Y (u )
(1.19)
(1.20)
and RYX (t,u ) E Y (t ) X (u )
Note RXY (t , u ) and RYX (t , u ) are not general even
functions.
The correlation matrix is
RX (t , u ) RXY (t , u )
R (t , u )
R
(
t
,
u
)
R
(
t
,
u
)
Y
YX
RYX ( ) RY ( )
where t u
24
(1.22)
25
X 1 (t ) X (t ) cos( 2f c t )
X 2 is(ta) stationary
X (t ) process
sin( 2f
),
where X(t)
andc tis
uniformly
distributed over [0, 2].
R12 () E X 1 (t ) X 2 (t )
E X (t ) X (t ) E cos(2f c t ) sin( 2f c t 2f c )
1
RX ( ) E sin( 2f c t 2f c 2) E sin( 2f c )
2
1
=0
RX ( ) sin(2f c )
2
(1.23)
At 0, sin(2f c ) 0, R12 ( ) 0 ,
X 1 (t ) and X 2 (t ) are orthogonal.
26
2T T
? 1
If X(t) is stationary,
E x (T )
2T
1
2T
X
E x (t ) dt
X dt
27
(1.25)
lim x (T ) X
b. lim var x (T ) 0
T
2 T
Rx (, T ) is a random variable.
(1.26)
28
(t )impulse
h(response
1 )X (t of1the
) dsystem
where h(t) is Ythe
1
-
Y ( t ) E Y (t )
If E[X(t)] is finite
and system is stable
h ( 1 ) X ( t 1 ) d1
h ( 1 ) E x ( t 1 ) d1
If X(t) is stationary,
h ( 1 ) X ( t 1 ) d1
H(0) :System DC response.
-
(1.27)
Y X h( 1 ) d1 X H (0),
-
(1.28)
(1.29)
29
h( 1 )X (t 1 ) d1 h( 2 ) X ( 2 ) d 2
(1.30)
2
E[
X
(t )] is finite and the system is stable,
If
RY (t,)
d1 h( 1 )
If
d 2 h ( 2 ) RX (t 1, 2 )
(1.31)
RX (t 1 , 2 ) RX (t 1 2 ) (stationary)
RY ( )
h( 1 )h ( 2 ) RX ( 1 2 ) d1 d2
(1.32)
RY (0) E Y (t )
2
G( f )
g (t )
g (t ) exp( j 2ft ) dt
G ( f ) exp( j 2ft ) df
2 -1
h( 1 ) H ( f ) exp( j 2f1 ) df
h( ) R ( ) d d
H
(
f
)
exp(
j
2
f
)
df
1
2
X
2
1
1
2
E Y (t )
2
(1.34)
(1.35)
(1.36)
31
E Y (t ) df H ( f )
2
RX ( ) exp( j 2f ) d
(1.37)
S X ( f ) RX ( ) exp( 2f ) d
E Y (t ) H ( f ) S X ( f ) df
2
(1.38)
(1.39)
Recall E Y (t ) h( 1 )RX ( 2 1 ) d1 d 2
(1.33)
- -
Let H ( f ) be the magnitude response of an ideal narrowband filter
1, f f c 1 f
2
|H ( f )|
(1.40)
1
0, f f c 2 f
2
f : Filter Bandwidth
If f
f c and S X ( f ) is continuous,
E Y 2 (t ) 2f S X ( f c ) in W/Hz
32
S X ( f ) RX ( ) exp( j 2f ) d
RX ( ) S X ( ) exp( j 2f ) df
(1.42)
(1.43)
Einstein-Wiener-Khintahine relations:
S X ( f ) RX ( )
S X ( f ) is more useful than RX ( ) !
33
a. S X (0) RX ( ) d
(1.44)
b. E X (t ) S X ( f ) df
2
(1.45)
c. If X (t ) is stationary,
E Y 2 (t ) (2f ) S X ( f ) 0
SX ( f ) 0
for all f
(1.46)
d. S X ( f ) RX ( ) exp( j 2f ) d
SX ( f )
u
(1.47)
pX ( f )
SX ( f )
S X ( f ) df
(1.48)
34
X (t ) A cos(2fct ), ~ U ( , )
A2
RX ( )
cos(2fc )
2
S X ( f ) RX ( ) exp( j 2f ) d
A2
exp( j 2f c )d exp( j 2f c ) exp( j 2f ) d
4
A2
( f f c ) ( f f c )
4
Appendix 2,
exp j 2 ( f c f ) d ( f f c )
35
A,
X (t )
A,
A2 (1 )
RX ( )
T
T
T
) exp( j 2f ) d
T
A 2T sinc 2 ( f T )
SX ( f )
A2 (1
(1.50)
(1.51)
g ( f )
T
(1.52)
SX ( f )
36
(1.53)
E X (t ) X (t ) E cos(2f ct 2f c ) cos(2fct )
1
RX ( ) E cos(2f c ) cos(4f c t 2f c 2)
2
1
RX ( ) cos(2f c )
2
(1.54)
SY ( f ) RY ( ) exp( j 2f ) d
1
RX ( ) exp( j 2 ( f f c )) exp( j 2 ( f f c )) d
4
1
S X ( f f c ) S X ( f f c )
(1.55)
4
We shift the S X ( f )to the right by f c , shift it to the left by f c ,
add them and divide by 4.
37
Relation Among The PSD of The Input and Output Random Processes
X(t)
SX
(f)
h(t)
Y(t)
SY
(f)
Recall (1.32)
RY ( )
SY ( f )
SY ( f )
h( 1 )h( 2 ) RX ( 1 2 ) d 1 d 2
(1.32)
h( 1 )h ( 2 ) RX ( 1 2 ) exp( j 2f ) d 1 d 2 d
Let 1 2 0 , or 0 1 2
SX ( f )H ( f )H * ( f )
2
H ( f ) SX ( f )
(1.58) 38
x(t ) dt
(1.59)
)
x
(
t
)
dt
X
(
f
,
T
)
2T T
2T
Time-averaged autocorrelation
periodogram function
(1.61)
(1.62)
39
1
2
1
2
T x(t )x(t )dt 2T X ( f , T ) exp( j 2 fT )df
(1.63)
1
2
R
(
lim
X
(
f
,
T
)
exp( j 2f )df
NoteXthat forT any
2T x(t) periodogram does not converge as
given
(1.64)
1
2
E X ( f T ) exp( j 2f )df
T 2T
1
2
RX ( ) lim
E X ( f T ) exp( j 2f )df
T 2T
E RX ( ) RX ( ) lim
(1.66)
1
2
S X ( f ) lim
E X ( f ,T )
T 2T
(1.67) is used to estimate the PSD of x(t)
2
1 T
lim
E x (t ) exp( j 2ft )dt
T 2T
T
(1.67)
40
Cross-Spectral Densities
S XY ( f ) RXY ( ) exp( j 2f )d
(1.68)
(1.69)
RXY ( ) RYX ( )
S XY ( f ) SYX ( f ) SYX
(f)
(1.22)
(1.72)
41
Example 1.8 X(t) and Y(t) are zero mean stationary processes.
Consider Z (t ) X (t ) Y (t )
S Z ( f ) S X ( f ) SY ( f )
(1.75)
Example 1.9 X(t) and Y(t) are jointly stationary.
RVZ (t , u ) E V (t ) Z (u )
E
h1 ( 1 ) X (t 1 )d 1 h2 ( 2 )Y (u 2 )d 2
h1 ( 1 )h2 ( 2 )RXY (t 1 , u 2 )d 1d 2
Let t u
RVZ ( )
SVY ( f ) H1 ( f ) H 2 ( f ) SXY ( f )
42
( e.g g(t):
(1.79)
(e) )
( y Y ) 2
1
fY ( y )
exp
2
2
2 Y
Y
1
y2
Normalized fY ( y )
exp( ) , as N (0,1)
2
2
(1.80)
(1.81)
43
1
Yi
( X i X )
X
i 1,2,...., N
Hence, EYi 0,
Var Yi 1.
1
Define VN
N
Y
i 1
44
X(t)
Gaussian
h(t)
Y(t)
Gaussian
Y (t ) h(t )X ( )d
0
Define
Z gY (t ) h (t ) X ( ) d dt
T
0
gY (t )h(t ) dt X ( ) d
g ( ) X ( ) d
0
where g ( ) gY (t )h (t )dt
0
Y (t ) h(t )X ( )d , 0 t is Gaussian
0
45
2. If X(t) is Gaussisan
Then X(t1) , X(t2) , X(t3) , ., X(tn) are jointly Gaussian.
Let X ( ti ) E X (ti ) i 1,2 ,....,n
and the set of covariance functions be
1
T
1
exp(
(
x
( x )) (1.85)
n
1
2
( 2 ) 2 2
46
E [( X (t k ) X ( tk ) )( X (ti ) X ( ti ) )] 0
Then they are independent
Proof : 2uncorrelated
1
2
2
,
where
E
[(
X
(
t
)
E
(
X
(
t
))
] , i 1,2 ,n.
i
i
i
n2
1
n
( 2 )
2
exp(
1
x T 1 (x ))
2
f X ( x ) f X i ( xi )
i 1
xi X i
1
2 i
i
47
1.9 Noise
Shot noise
Thermal noise
1
1
E I
E V 4kT f 4kTGf
R
R
E VTN2 4kTRf
2
TN
2
TN
volts2
amps2
White noise
N0
SW ( f )
2
N 0 kTe
(1.93)
(1.94)
(1.95)
49
N 0
-B f B
SN ( f ) 2
f B
0
B N
0
RN ( )
exp( j 2 f ) df
B 2
N 0 B sinc( 2 B )
(1.96)
(1.97)
50
X
T
2
cos(2fct )
T
fc
2 T
w' ( t )
w( t ) cos( 2f c t )dt
0
T
The varance of w' ( t ) is
2
2 E
T
k
, k is integer
T
(1.98)
2 T T
0
0
T
From (1.95)
2 T T N0
2
(t1 t2 ) cos( 2 f c t1 ) cos( 2 f c t2 ) dt1 dt2
0
0
T
2
N0 T
N0
2
51
cos
(
2
f
t
)
dt
(1.99)
c
0
T
2
Two representations
a. in-phase and quadrature components (cos(2 fct) ,sin(2 fct))
b.envelope and phase
1.11 In-phase and quadrature representation
n (t ) nI (t ) cos(2 f c t ) nQ (t ) sin( 2 f c t )
(1.100)
52
Important Properties
1.nI(t) and nQ(t) have zero mean.
2.If n(t) is Gaussian then nI(t) and nQ(t) are jointly Gaussian.
3.If n(t) is stationary then nI(t) and nQ(t) are jointly stationary.
SN ( f fc ) SN ( f fc ) ,
4.S N I ( f ) S NQ ( f )
0
-B f B
otherwise
(1.101)
N0
2 .
-B f B
otherwise
(1.102)
7.If n(t) is Gaussian, its PSD is symmetric about fc, then nI(t) and
nQ(t) are statistically independent.
53
fc B N
N0
0
RN ( )
exp( j 2 f )df
exp( j 2 f )df
fc B 2
fc B 2
N 0 B sinc( 2 B ) exp( j 2 f c ) exp( j 2 f c )
fc B
2 N 0 B sinc( 2 B ) cos(2 f c )
(1.103)
54
r (t ) n (t ) n (t )
2
I
2
Q
(1.106)
Phase
nQ (t )
(t ) tan
n
(
t
)
I
(1.107)
Let NI and NQ be R.V.s obtained (at some fixed time) from nI(t)
and nQ(t). NI and NQ are independent Gaussian with zero mean and
variance 2.
55
nI2 nQ2
1
f N I , N Q (nI , nQ )
exp(
)
2
2
2
2
f N I , NQ ( nI , nQ )dnI dnQ
Let nI r cos
nI2 nQ2
1
exp(
) dnI dnQ
2
2
2
2
(1.108)
(1.109)
(1.110)
nQ r sin
(1.111)
dnI dnQ r dr d
(1.112)
56
exp( 2 ) rdrd
2
2
2
r
r2
f R , ( r, )
exp( 2 )
2
2
2
1
0 2
0 2 , f ( ) 2
elsewhere
0
r
r2
f R ( r ) 2 exp( 2 2 ) , r 0
0
elsewhere
f R ( r ) is Rayleigh distribution.
For convenience , let
fV ( )
(1.113)
(1.114)
(1.115)
r
. fV ( ) f R ( r )
2
exp( ) , 0
2
0
elsewhere
(1.118)
57
(1.119)
( nI A) nQ
1
f N I , N Q (nI , nQ )
exp
2
2
2
2
Let
r (t ) nI (t ) nQ2 (t )
2
1
2
nQ (t )
nI (t )
(t) tan -1
(1.123)
(1.124)
59
f R (r)
2
0
f R , ( r, )d
r
r 2 A2 2
Ar
exp(
) exp( 2 cos )d
2
2
0
2
2
(1.126)
2 0
Ar
r
r 2 A2
Ar
Let x 2 , f R(r) 2 exp(
) I0 ( 2 )
2
(1.127)
(1.128)
r
A
Normalized , a
fV ( v ) f R ( r )
v2 a2
v exp(
) I 0 ( av )
2
(1.131)
(1.132)
61