ECE 562
Fall 2011
December 6, 2011
HOMEWORK ASSIGNMENT 7 SOLUTIONS
1.
True or False.
Determine if the following statements are True or False. You need to
provide a brief justification for your answer to get credit.
(a) If the mobile speed is 72 km/hr, the carrier frequency is 900 MHz, and symbol
rate for communication is 10 symbols a second, then the mobile experiences slow
fading.
False:
f
max
= 60 Hz, and
T
c
= 0
.
1
/f
max
0
.
1 =
T
s
.
(b) In wireless communication, if the speed of the mobile is doubled the delay spread
is also doubled.
False:
The mobile speed has little effect on the delay spread. It is the Doppler
speed that gets doubled.
(c) A wireless channel with
τ
ds
= 10

4
seconds and
f
max
= 100 Hz operating with a
bandwidth of 1
.
25 MHz is a flat fading channel.
False:
W
τ
ds
(d) For a fixed
γ
b
,
P
b
for binary DPSK signaling over a slow flat fading channel is
always larger than
P
b
without fading, irrespective of the distribution of the fading.
True:
By Jensen’s inequality since exp is convex.
2.
Useful Result.
Show that
∞
0
Q
(
√
x
)
e

x/γ
γ
dx
=
1
2
1

γ
2 +
γ
.
Hint:
Start with:
∞
x
=0
Q
(
√
x
)
e

x/γ
γ
dx
=
∞
x
=0
∞
t
=
√
x
e

t
2
/
2
√
2
π
e

x/γ
γ
dt dx
=
∞
t
=0
t
2
x
=0
e

t
2
/
2
√
2
π
e

x/γ
γ
dx dt
and use the fact that the Gaussian pdf integrates to 1 to conclude the result.
Ans:
Starting with the hint,
∞
x
=0
Q
(
√
x
)
e

x/γ
γ
dx
=
∞
0
1
√
2
π
e

1
2
t
2

1
√
2
π
e

1
2
(
γ
+2
γ
)
t
2
dt
=
1
2

∞
0
1
√
2
π
e

1
2
(
γ
+2
γ
)
t
2
dt
=
1
2

γ
2 +
γ
∞
0
1
2
π
γ
2+
γ
e

1
2
(
γ
+2
γ
)
t
2
dt
=
1
2
1

γ
2 +
γ
c V.V. Veeravalli, 2011
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 Fall '09
 Rayleigh fading, V.V. Veeravalli

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