COLUMBIA UNIVERSITY
Department of Electrical Engineering
EE E4703: Wireless Communications (Spring 2005)
Solution for Homework Assignment No. 4
COLUMBIA UNIVERSITY
Department of Electrical Engineering
EE E4703: Wireless Communications (Spring 2004)
Solution for Homework Assignment No. 4
Problem 1
Rappaport, p. 249, Problem 5.6
We must first find the first and second moment of the excess delay (and use only
linear scale – not dB – to do it). We use them to get the delay spread.
For the indoor channel:
� P (t )t = 1 � 0 + 1 � 50 + 0.1 � 75 + 0. 01 � 100 = 27. 73ns
1 + 1 + 0 .1 + 0. 01
� P (t )
� P(t )t = 1 � 0 + 1 � 50 + 0 .1 � 75 + 0 .01 � 100 = 1499 ns
=
1 + 1 + 0 .1 + 0. 01
� P (t )
t =
k
k
k
2
t
k
2
2
2
2
2
k
k
st = t 2 -t
= 27 .02 ns
2
It is given that:
st
Ts
£ 0.1
\ Ts ‡ 10s t = 270 .2 ns
Rs =
1
£ 3 .7 Msym/ sec
Ts
For the outdoor channel:
� P (t )t = 0.01 � 0 + 0.1 � 5 + 1 �10 = 9.46ms
0 .01 + 0 .1 + 1
� P (t )
� P(t )t = 0.01 � 0 + 0.1� 5 + 1 �10 = 92.34 ms
=
0.01 + 0.1 + 1
� P(t )
t =
k
k
k
2
t2
k
k
k
s t = t 2 -t 2 = 1 .69 ms
2
2
2
2
2
st
Ts
£ 0 .1
\ Ts ‡ 10s t = 16 .9 ms
Rs =
1
£ 59 .17 Ksym / sec
Ts
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Problem 2
Rappaport, p. 250, Problem 5.8
The Coherence bandwidths are given in equations 5.38 & 5.39 in the text:
BC ,90% »
1
50s t
BC ,50% »
1
5s t
Indoor channel:
s t = 27 .02 ns
BC ,90% »
1
= 740 KHz
50s t
BC ,50% »
1
= 7.4 MHz
5s t
Outdoor channel:
s t = 1 .69 ms
BC ,90% »
1
= 11 .83 KHz
50s t
BC ,50% »
1
= 118 .3 KHz
5s t
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Problem 3
Rappaport, pp. 252 -253 , Problem 5.28
(a)
� P (t )t = 1� 0 + 0.1� 1 + 1� 2 = 1ms
1 + 0 .1 + 1
� P (t )
� P(t )t = 1 � 0 + 0.1� 1 + 1 � 2 = 1.952 ms
=
1 + 0.1 + 1
� P(t )
t =
k
k
k
2
t
2
k
2
2
k
2
2
k
s t = t 2 -t 2 = 0.976 ms
(b) The maximum delay spread (20dB) definition is given in page 199 in the text. It is
defined as the time delay between the first arriving signal and the maximum delay at
which a multipath component is within 20 dB of the strongest arriving multipath
signal.
In our case, the last multipath component arrives 2µs after the first arriving signal, so
the maximum delay spread is 2µs .
(c)
T s ‡ 10s t = 9.76 ms
Rs =
1
= 102 .5 Ksym / sec
Ts
(d) We are looking for the Doppler Coherence Time.
v = 30 kmhr = 8 .33 m s
c
= 0 .333 m
f
v
\ f m = = 25Hz
l
0 .423
TC =
= 16 .92 ms
fm
l=
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Problem 4
Rappaport, p. 250, Problem 5.11
Rayleigh distribution pdf:
r2
r - 2
f r (r ) = 2 e 2 s
s
R
p (r < R) = �
0
r2
R
r2
R
r - 2
1
2
f r (r )dr = � 2 e 2 s dr =
e 2s 2rdr
2 �
2s 0
0s
We replace the integration variable with u=r2
u = r2
du = 2rdr
R2
u
u
1
1
2
2s 2
2s 2
p (r < R) =
du =
2 �e
2 2s e
2s 0
2s
0
=1- e
-
R2
2s
2
R2
The rms value (square root of the second moment ) of a Rayleigh distribution is
2s 2 = 2s and 10 dB b e low the rms value , in amplitude , is
p (r < 0 .2s ) = 1 - e
-
(
0.2s
2s 2
)2
2
s.
10
= 1 - e - 0.1 = 0.0952 = 9.52 %
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Problem 5
A = 15
B = 10
a = 3000
b = 104
w = 2p �106
v(t ) = (A + B sin 2pat + C cos 2pbt) cos wt
= Acos wt + B sin 2pat cos wt + C cos 2pbt cos wt
B
C
= Acos wt + [sin( w + 2pa )t - sin( w - 2pa )t ] + [cos(w + 2pb )t + cos(w - 2pb)t ]
2
2
(a+b) The modulated wave will have frequency components at the following
frequencies:
Amplitude
Frequency [Hz]
A = 15
w
= 1MHz
2p
B
=5
2
w
+ a = 1 .003 MHz
2p
B
=5
2
w
- a = 0 .997 MHz
2p
C
=4
2
w
+ b = 1.01MHz
2p
C
=4
2
w
- b = 0 .99 MHz
2p
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