sm8_33

Prévia do material em texto

Problem 8.33. A random pulse has amplitude A and duration T but starts at an arbitrary time t0. That is, 
the random process is defined as 
 
)(rect)( 0ttAtX += 
 
where rect(t) is defined in Section 2.9. The random variable t0 is assumed to be uniformly distributed over 
[0,T] with density 
 
⎪⎩
⎪⎨
⎧ ≤≤=
otherwise0
01
)(
0
Ts
Tsft 
(a) What is the autocorrelation function of the random process X(t)? 
(b) What is the spectrum of the random process X(t)? 
 
 
Solution 
First note that the process X(t) is not stationary. This may be demonstrated by computing 
the mean of X(t) for which we use the fact that 
 
∫∞
∞−
= dssfsxfxf tXX )()|()( 0 
combined with the fact that 
 
[ ]
⎩⎨
⎧ +≤≤=
= ∫∞
∞−
otherwise0
)|(|)(
00
00
TtttA
dxtxxfttX XE
 
 
Consequently, we have 
[ ]
( )
⎪⎪⎩
⎪⎪⎨
⎧
>
≤<−
≤≤
<
=
= ∫∞
∞−
Tt
TtTTtA
TtTAt
t
dssfstXtX t
20
2/2
0/
00
)(]|)([)(
0
EE
 
 
Thus the mean of the process is dependent on t, and the process is nonstationary. 
 
 
 
Continued on next slide 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
page…8-40 
Problem 8.33 continued 
 
 
We take a similar approach to compute the autocorrelation function. First we break the 
situation into a number of cases: 
 
i) For any t < 0, s < 0, t > 2T, or s > 2T, we have that 
 [ ] 0)()( =sXtXE 
 
ii) For 0 ≤ t < s ≤ 2T, we first assume t0 is known 
 
[ ]
⎩⎨
⎧ <<−=
⎩⎨
⎧ <<+<>=
otherwise0
),min()0,max(
otherwise0
0,,
|)()(
0
2
000
2
0
TttTsA
TtTtsttA
tsXtXE
 
 
 
Evaluating the unconditional expectation, we have 
 
[ ] [ ]
{ }{ }0,),0max(),min(max
1
)(|)()()()(
2
),min(
),0max(
2
0
TsTt
T
A
dw
T
A
dwwfwsXtXsXtX
Tt
Ts
t
−−=
⎟⎠
⎞⎜⎝
⎛=
=
∫
∫
−
∞
∞−
EE
 
where the second maximum takes care of the case where the lower limit on the 
integral is greater than the upper limit. 
 
iii) For 0 ≤ s < t ≤ 2T, we use a similar argument to obtain 
 
[ ]
⎩⎨
⎧ <<−=
otherwise0
),min()0,max(
|)()( 0
2
0
TstTtA
tsXtXE 
 and 
 
[ ] { }{ }0,),0max(),min(max)()( 2 TtTs
T
AsXtX −−=E 
 
Continued on next slide 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
page…8-41 
Problem 8.33 continued 
 
Combining all of these results we have the autocorrelation is given by 
 
[ ]
{ }{ }
{ }{ }
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
≤<≤−−
≤<≤−−
=
otherwise0
200,),0max(),min(max
200,),0max(),min(max
)()(
2
2
TtsTtTs
T
A
TstTsTt
T
A
sXtXE 
This result depends upon both t and s, not just t-s, as one would expect for a non-
stationary process. 
 
 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only 
to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that 
permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 
page…8-42