Transcript Lecture Random input output

Example 1 A random signal X(t) with an autocorrelation function given as
R XX (τ) =  (τ)
( White Noise )
If the signal is passed to an RC low pass filter as shown below
X(t)
Y(t)
Find the following
The output autocorrelation RYY(t) ?
(b) The average output power
(a)
The output autocorrelation RYY(t) was found to be
R YY (τ) = R XX (τ)  h(  τ)  h(τ)
This method involve doing a double convolution which some time is a cumbersome.
However we can use the power spectrum method as follows:
SYY ( )= SXX ( ) H( )
2
where
X(t)
SXX ( ) =
Y(t)
{ R XX ( )}  1
F.T
Fourier Transform
H( ) =
1
1 + j
 SYY ( )=
1
1+ 2

H( ) =
2
Therefore
1
1+ 2
R YY (τ) =
1 
e
2
(b) The average output power E[Y2(t)] is defined as
1
E[Y (t)] = R YY (0) =
2
2



SYY ( )d
Using the output autocorrelation RYY(t) we have
E[Y 2 (t)] = R YY (0) =
1 0
1
e

W
2
2
Using the output PSD SYY(w) we have
1 
E[Y (t)] =
SYY ( )d


2
2


1
=
2

1
 1+ 2 d

1
1  
1
1
=
tan   =
(  )=
2
2 2 2
2
Example 2
If the input X(t) to a linear filter is a zero mean white noise with
an average power N0 watt and the output of the filter is a
random process Y(t) with an autocorrelation function given as
R YY (τ) = 2e 
R XX (τ) = N0 (τ)
Y(t)
X(t)
τ
h(t)
R YY (τ) = 2e 
Find the following
(a) The filter impulse response h(t) ?
(b) The variance of the input X(t) ?
(c) The mean and variance of the output Y(t) ?
Example 3 Let X(t) be a differentiable (derivative exists) WSS random process, and define
d
Y(t) = X(t)
dt
If the input process X(t) has an autocorrelation function given as
R XX (τ) = (1 

T
)
for  < T
Find SYY (f) and RYY () ?
and zero elsewhere
Solution
Since the out put random process Y(t) is the first derivative
of the input random process as shown below
d
dt
X(t)
Y(t)
The output Y() is related to the input X() as
Y( ) = jX( )
 H( )  j
 H( )   2
2
Therefore the output power spectral density SYY() is related to
input power spectral density SXX() as
SYY ( )= SXX ( ) H( ) =  2SXX ( )
2
Therefore the output autocorrelation function RYY() is related To
input autocorrelation RXX() as
d2
R YY ( ) =  2 R XX ( )
d
Now if RXX() is the triangle pulse given below
R XX (τ) = (1 

T
for  < T
)
R XX ( )
and zero elsewhere
 SXX ()  sinc2 (T/2)
1
H( )   2
2
2
2
2
 SYY ( ) = SXX ( ) H()   sinc (T/2)
T

0
T
 R YY ( ) =
2
1
 ( )  [ (  T) +  (  T)]
T
T
The result can also be obtain by finding RYY() from RXX() as follows
d2
R YY ( ) =  2 R XX ( )
d
Which can be shown graphically as follows
dR XX ( )
d
R XX ( )
1
d2R XX ( )
d 2
1
1
T
1
T
T
0
d R XX ( )
d 2
2
2
T
T
T

0
1
T

0
T
T
R YY ( )  

T

1
T

1
T

T

0
T
1
T

2
T
2
1
 R YY ( ) =  ( )  [ (  T) +  (  T)]
T
T
2
2
 SYY ( ) =
 cos(T) =  2sinc 2 (T/2)
T
T