ECE 5670 : Digital Communications
ECE department, Cornell University, Spring 2011
Homework 2 Solutions
Instructor: Salman Avestimehr
Ofﬁce 325 Rhodes Hall
1. We have doubled our transmit energy constraint – so the transmit voltage range has in
creased by a factor of
√
2
. Let us consider an increase in data rate of
r
from the existing rate
of
R
bits per unit time. So, the distance between the nearest voltage levels is now
2
√
2
E
2
R
+
r
(1)
as compared to the previous distance of
2
√
E
2
R
.
(2)
The new error probability is
Q
2
√
2
E
σ
2
R
+
r
!
≈
1
2
e

4
E
2
σ
2
2
R
+
r
(3)
as compared to the previous level of
p
e
def
=
Q
2
√
2
E
σ
2
2(
R
+
r
)
!
(4)
≈
1
2
e

2
E
2
σ
2
2
2
R
.
(5)
So, the new error probability (cf. Equation (3)) can be written as
p
2
1

2
r
e
.
(6)
We then conclude that the set
(
r,d
)
is one that satisﬁes:
0
≤
r
≤
0
.
5
(7)
0
≤
d
≤
2
1

2
r
.
(8)
2. The overall unreliability level for sequential communication is given by:
P
[
E
] = 1

(1

p
e
)
n
(9)
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View Full Documentwhere
p
e
=
Q
√
SNR
2
R

1
!
(10)
Setting
P
[
E
]
equal to
10

5
and solving for SNR, we get:
10

5
= 1

1

Q
√
SNR
2
R

1
!!
n
(11)
1

Q
√
SNR
2
R

1
!
= (1

10

5
)
1
/n
(12)
√
SNR
2
R

1
=
Q

1
(
1

(1

10

5
)
1
/n
)
(13)
SNR
= [(2
R

1)
Q

1
(
1

(1

10

5
)
1
/n
)
]
2
(14)
If you tried plotting
10 log
10
SNR
, you would realize that some numbers are so large that
MATLAB cannot handle it (especially values like
2
8192
). Thus we need to simplify by hand.
So when we take the
10 log
10
SNR
, and approximate
2
R

1
by
2
R
, we get the following:
10 log
10
SNR
= 10 log
10
[2
R
Q

1
(
1

(1

10

5
)
1
/n
)
]
2
(15)
= 20 log
10
[2
R
Q

1
(
1

(1

10

5
)
1
/n
)
]
(16)
= 20 log
10
2
R
+ 20 log
10
Q

1
(
1

(1

10

5
)
1
/n
)
(17)
= 20
R
log
10
2 + 20 log
10
Q

1
(
1

(1

10

5
)
1
/n
)
.
(18)
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 Spring '11
 SCAGLIONE
 Normal Distribution, Volt, Probability theory, Natural logarithm, ML rule

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