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EE 230B Quiz 3,
Spring 2008 Solutions
Prof. G. Pottie
For the channel
f
(
z
)
=
1
2
+
1
2
z
−
1
:
1/
2
2
X
Σ
2
X
z
1
a) Show
X
(
e
j
ω
T
)
=
1
+
cos
T
.
Why is this channel problematic?
Solution:
Since this is a minimum phase channel (no precursors), we can easily compute
the folded spectrum using the factorization of the equivalent noisewhitened matched
filter channel:
X
(
z
)
=
f
(
z
)
f
*(
z
−
1
)
=
(1/ 2
+
z
−
1
/2
)(1/ 2
+
z
)
=
1
2
(1
+
z
+
z
−
1
+
1)
=
1
+
1
2
(
z
+
z
−
1
)
X
(
e
j
T
)
=
1
+
1
2
(
e
j
T
+
e
−
j
T
)
=
1
+
cos
T
This channel has a null at the bandedges, and so equalizers will considerably enhance the
noise.
b) Compute
J
min
for the ideal MMSE linear equalizer and DFE and compare, using the
trapezoid rule for integration with
Δ
=
π
/4
T
and N
0
=0.05.
You may use the following
data:
ω

π
/
T
3
π
/4
T

π
/2
T

π
/4
T
0
π
/
T
π
/2
T
3
π
/4
T
π
/
T
X
(
e
j
T
)
0
0.293
1
1.707
2
1.707
1
0.293
0
Solution:
For the linear equalizer,
J
=
T
2
No
X
(
e
j
T
)
+
No
d
≈
T
2
N
0
4
T
1
0.05
+
2
.298
+
2
1.05
+
2
1.757
+
1
2.05
⎛
⎝
⎜
⎞
⎠
⎟
−
/
T
/
T
∫
J
=
.05
8
1
0.05
+
2
.298
+
2
1.05
+
2
1.757
+
1
2.05
⎛
⎝
⎜
⎞
⎠
⎟
=
0.189
Similarly, for the DFE
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This note was uploaded on 03/29/2011 for the course EECS 100 taught by Professor Dun during the Spring '11 term at CSU Channel Islands.
 Spring '11
 Dun

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