ECEN 455: Assignment 8
Problems:
1. (CSE: 8.3) In a binary PAM system, the clock that specifies the sampling of the correlator output is
offset from the optimum sampling time by 10%.
(a) If the signal pulse used is rectangular, determine the loss in SNR due to mistiming.
If the transmitted signal is
r
(
t
) =
∞
summationdisplay
n
=
∞
a
n
h
(
t

nT
) +
n
(
t
)
,
then the output of the receiving filter is
y
(
t
) =
∞
summationdisplay
n
=
∞
a
n
x
(
t

nT
) +
ν
(
t
)
where
x
(
t
) =
h
(
t
)
⋆ h
(
t
)
and
ν
(
t
) =
n
(
t
)
⋆ h
(
t
)
. If the sampling time is off by 10%, then
the samples at the output of the correlator are taken at
t
= (
m
±
0
.
1)
T
. Assuming that
t
=
(
m

0
.
1)
T
without loss of generality, then the sampled sequence is
y
m
=
∞
summationdisplay
n
=
∞
a
n
x
((
m

0
.
1

n
)
T
+
ν
((
m

0
.
1)
T
)
.
If the signal pulse is rectangular with amplitude
A
and duration
T
, then
∞
summationdisplay
n
=
∞
a
n
x
((
m

0
.
1

n
)
T
is nonzero only for
n
=
m
and
n
=
m

1
. The sampled sequence is therefore given by
y
m
=
a
m
x
(

0
.
1
T
) +
a
m

1
x
(
T

0
.
1
T
) +
ν
((
m

0
.
1)
T
)
= 0
.
9
a
m
A
2
T
+ 0
.
1
a
m

1
A
2
T
+
ν
((
m

0
.
1)
T
)
.
The power spectral density of the noise at the output of the correlator is
S
ν
(
f
) =
S
n
(
f
)

H
(
f
)

2
=
N
0
2
A
2
T
2
sinc
2
(
fT
)
.
Thus, the variance of the noise is
σ
n
u
2
=
integraldisplay
∞
∞
N
0
2
A
2
T
2
sinc
2
(
fT
)
df
=
N
0
2
A
2
T
2
1
T
=
N
0
2
A
2
T
and the SNR is
SNR
= (0
.
9)
2
2(
A
2
T
)
2
N
0
A
2
T
= 0
.
81
2
A
2
T
N
0
.
As it is observed, there is a loss of
10log
10
0
.
81 =

0
.
9151
dB due to the mistiming.
1
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(b) Determine the amount of intersymbol interference introduced by the mistiming and determine
its effect on performance.
Recall from the previous part that the sampled sequence is
y
m
= 0
.
9
a
m
A
2
T
+ 0
.
1
a
m

1
A
2
T
+
ν
m
.
The term
0
.
1
a
m

1
A
2
T
expresses the ISI introduced to the system. If
a
m
= 1
is transmitted,
then the probability of error is
P
(
e

a
m
= 1) = 0
.
5
P
(
e

a
m
= 1
,a
m

1
= 1) + 0
.
5
P
(
e

a
m
= 1
,a
m

1
=

1)
=
1
2
√
πN
0
A
2
T
integraldisplay

A
2
T
∞
e

ν
2
N
0
A
2
T
dν
+
1
2
√
πN
0
A
2
T
integraldisplay

0
.
8
A
2
T
∞
e

ν
2
N
0
A
2
T
dν
= 0
.
5
Q
radicalBigg
2
A
2
T
N
0
+ 0
.
5
Q
radicalBigg
parenleftbigg
8
10
parenrightbigg
2
2
A
2
T
N
0
.
Since the symbols of the binary PAM system are equiprobable the previous derived expression
is the probability of error when a symbol by symbol detector is employed. Comparing this with
the probability of error of a system with no ISI, we observe that there is an increase of the
probability of error by
0
.
5
Q
radicalBigg
parenleftbigg
8
10
parenrightbigg
2
2
A
2
T
N
0

0
.
5
Q
radicalBigg
2
A
2
T
N
0
.
2. (CSE: 8.7) Consider a fourphase PSK signal that is represented by the equivalent lowpass signal
v
(
t
) =
summationdisplay
n
a
n
g
(
t

nT
)
,
where
a
n
takes on one of the four possible values
±
1
±
j
√
2
with equal probability.
The sequence of
information symbols
{
a
n
}
is statistically independent.
(a) Determine and sketch the powerspectral density of
v
(
t
)
when
g
(
t
) =
braceleftbigg
A,
0
≤
t
≤
T
0
,
otherwise
The autocorrelation function of the information symbols
{
a
n
}
is
R
a
(
k
) =
E
[
a
*
n
a
n
+
k
] =

a
n

2
4
δ
(
k
) =
δ
(
k
)
Thus, the power spectral density of
v
(
t
)
is
S
V
(
f
) =
1
T
S
a
(
f
)

G
(
f
)

2
=
1
T

G
(
f
)

2
where
G
(
f
) =
F
[
g
(
t
)]
. If
g
(
t
) =
A
Π
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 Spring '08
 Staff
 Signal Processing, spectral density, Autocorrelation

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