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The concepts of autocorrelation and crosscorrelation functions, including their measurement, examples, properties, and the difference between them. It also covers the ergodic hypothesis and the relationship between ensemble and time averages.
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Measurement of Autocorrelation Functions
Examples of Autocorrelation Functions
Crosscorrelation Functions
Properties of Crosscorrelation Functions
Reading:
G. R. Cooper & C. D. McGillem 6.4 - 6.
EE/STAT 322, #
X
Assume
t )
is an ergodic random process, so
X
(^) ( τ (^) )
can be estimated
using the time average, ˆ
X
(^) ( τ (^) ) =
1
T (^) −
τ
∫
T (^) −
τ
0 x ( t ) x ( t + τ
dt
, 0 ≤ τ T
Divide
to
intervals.
t , τ → n ∆ t ,
dt
t .
X
(^) ( n ∆
t ) =
1
N
(^) −
n
N
(^) −
n
x k x k + n , n
X
(^) ( n ∆
t )
is an unbiased estimate of
X
(^) ( τ (^) ) .
R X ( n ∆ t
1
N
(^) −
n
N
(^) −
n
E [ x k x k + n ]
1
N
(^) −
n
N
(^) −
n
X
(^) ( n ∆
t ) =
X
(^) ( n
∆
t ) .
EE/STAT 322, #
Solution:
Let
X 1 = X ( t 1 ) , X 2 = X ( t 1 + τ
2 A
. Let
t ) =
t ) ,
where
t )
is the zero-mean part.
If
< τ < t
a , R
N
(^) ( τ (^) ) =
A
2
4
Pr(
t 1
and
t 1
τ
are in the same interval
A 2
4
Pr(
t 1 + τ − t a
< t
0
≤
t 1 ) =
A
2
4 · t a − τ
t a
If
τ (^) |
> t
a ,
x ( τ (^) ) =
4
,
x ( τ (^) )
A 2
4
A
2
4
| τ (^) |
t a )
τ (^) | ≤
t a
τ (^) |
> t
a
τ
2
/
2
A
)
( τ
X
R
a
t
a
t
−
0
4
/
2
A
EE/STAT 322, #
4
X
Example:
(Ex.
6-5.2) Find out which of the following functions can
not
(a) be a valid autocorrelation function.
e − τ (^2) ; (b)
τ (^) | e −|
τ (^) | ; (c)
e −
( τ (^) +2)
; (d)
sin
(^) πτ
πτ
2 ; (e)
τ (^2)
τ (^2)
Solution:
(a)
X
(^) (0) = 1
X
(^) (0)
X
(^) ( τ (^) )
for
τ (^) |
>
. Valid.
(b)
X
(^) (0) = 0
power of process is zero. Not valid.
(c)
X
(^) (0) = 10
e − 2 , R X
X
(^) (0)
X
(^) (2)
. Not valid.
(d)
X
(^) (0) = 1
X
(^) (0)
X
(^) ( τ (^) )
for
τ (^) |
>
. Valid.
(e)
X
(^) ( τ (^) ) = 1 +
−
4
τ (^2)
τ (^) ) , but
N
(^) ( τ (^) ) =
−
4
τ (^2)
is not a valid
autocorrelation function, because
N
(^) (0) =
EE/STAT 322, #
R Time average: crosscorrelation function:
xy
τ (^) ) = lim
T (^) →∞
1
2 T
T − T x ( t ) y ( t + τ
dt
yx
τ (^) ) = lim
T (^) →∞
1
2 T
T − T y ( t ) x ( t + τ
dt
For jointly ergodic processes
t )
and
t ) ,
xy
τ (^) ) =
XY
τ (^) )
yx
τ (^) ) =
Y X
τ (^) ) .
EE/STAT 322, #
t
1 t
)
( t
X
1
ξ
2
ξ
n
ξ
2
t
t
1
t
)
( t
Y
1
ξ 2
ξ
m
ξ
2
t
)
(τ
XY
R
EE/STAT 322, #
(b)
XY
τ (^) ) =
〈 X ( t ) Y ( t + τ
= lim
T (^) →∞
1
2 T
T − T x ( t ) y ( t + τ
dt
= lim
T (^) →∞
1
2 T
∫
T −
T
2 cos(
t
θ )10 sin(5(
t
τ (^) ) +
θ ) dt
XY
τ (^) ) = 10(0 +
2 T
2 T
sin(
τ (^) )) = 10 sin(
τ (^) ) .
So
XY
τ (^) ) =
XY
τ (^) ) , ⇒ X ( t )
and
t )
are jointly ergodic.
EE/STAT 322, #
Y X
XY
XY
τ (^) ) =
Y X
τ (^) ) .
XY
τ (^) ) =
E [ X ( t ) Y ( t + τ
, and
Y X
τ (^) ) =
E [ Y ( t ) X ( t − τ
E [ X ( t − τ
t )] =
E [ X ( t ) Y ( t + τ
(due to stationarity).
XY
τ (^) ) | ≤
X
(^) (0)
Y
(0)]
1 / 2 .
Use Schwartz inequality:
XY
τ (^) ) |
X
(^) (0)
Y
(0)]
1 / 2
XY
τ (^) ) | 2
X
(^) (0)
Y
(0)
E [ X ( t ) Y ( t + τ
2
≤
2 ( t )]
E [ Y 2 ( t + τ
Corollary:
E [ X ( t ) Y ( t
2 ≤ E [ X 2 ( t
E [ Y 2 ( t ]
EE/STAT 322, #
We can also show that
˙
X
(^) ( τ (^) ) =
˙
X
˙
X
(^) ( τ (^) ) =
d 2 R
X (^) ( τ (^) )
dτ
(^2)
Proof:
˙
X
˙
X
(^) ( τ (^) ) =
t )
˙
t
τ (^) )]
e lim → 0 X ( t + e ) − X ( t )
e · X ( t + e ) − X ( t )
e
= lim
e →
0
e 1
{
R
X
(^) ( τ (^) )
−
X
(^) ( τ
−
e )
e
X
(^) ( τ + e ) − R X
τ (^) )
e
= lim
e →
0
e 1
{
dR
X
(^) ( τ
−
e )
dτ
dR
X
(^) ( τ (^) )
dτ
elim →
0
e^1
{
dR
X
( τ (^) )
dτ
dR
X
(^) ( τ
−
e )
dτ
} = − d 2 R X
τ (^) )
dτ
(^2)
EE/STAT 322, #
Example:
(Ex. 6-7.1) Prove that
XY
τ (^) ) | ≤
X
(0)
Y
(0)]
1 / 2 .
Let Proof:
X 1 = X ( t ) , Y 2 = Y ( t + τ
(^12) (^) ] =
X
(^) (0)
, and
2
2
] =
Y
(0)
X
1
R X (^) (0)
Y 2
R Y (0)
2 } ≥
E
[ X
1 Y 2 ]
R X (^) (0)
R Y (0)
XY
τ (^) )
≤
X
(^) (0)
Y (^) (0)
EE/STAT 322, #
