This preview shows pages 1–3. Sign up to view the full content.
ECEN 661  Modulation Theory
Homework #3 Solutions
1.
(a)
Since the shifted pulses in each series are nonoverlapping, all cross terms will be zero and hence
this simplifies to:
.
Taking the data symbols to be
, this simplifies to
.
Focus attention on the interval
, and we get
.
Note, an identical result is obtained for any other interval of length
we look at. Hence,
for all .
(b) Write
, where
,
.
Note that since both
and
are zeromean and independent, then
.
ut
()
2
a
n
gt
2
nT
–
n
∑
2
b
n
2
–
T
–
n
∑
2
+
=
2
a
n
2
g
2
t
2
–
n
∑
b
n
2
g
2
t
2
–
T
–
n
∑
+
=
a
n
b
n
,
1
±
{}
∈
2
g
2
t
2
–
g
2
t
2
–
T
–
+
[]
n
∑
=
0
tT
<
≤
2
g
2
t
g
2
+
+s
i
n
2
π
t
2
T

sin
2
π
+
2
T

+
==
sin
2
π
t
2
T
cos
2
π
t
2
T
+1
T
2
1
=
t
xt
jy t
+
=
a
n
2
–
n
∑
=
yt
b
n
2
–
T
–
n
∑
–
=
φ
uu
τ
1
2

Ex
t
–
τ
+
τ
+
+
=
1
2
Ext
τ
+
1
2
Eyt
τ
+
+
=
1
2
φ
xx
τ
1
2
φ
yy
τ
+
φ
τ
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentNote in the last step above we take advantage of the fact that
is of the same nature as
(just delayed in time by
) and hence has the same autocorrelation. From the above result we find
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '11
 Miller

Click to edit the document details