EE E4703 Wireless Communications Spring 2005
Professor Diament
Solution to Homework Assignment #5
Problem 6.1
5
Given fm = 2π
= 0.796Hz, βf = 10 ⇒ bandwidth of the FM signal
Bt = 2(βf + 1)fm ⇒ BT = 17.5Hz.
ωc
For fc = 2π
= 796.18Hz, the upper sideband frequency is fc to fc + B2T =
796.18 to 804.93Hz.
The lower sideband frequency is fc − B2T to fc = 787.43 to 796.18Hz.
Problem 6.2
= 500Hz, we
For m(t) = sin(1000πt), ∆F = 1KHz, Am = 2V, fm = 1000π
2π
∆f
1000
have kf = Am = 2 = 500Hz/V.
Thus, for Am = 8V, fm = 2000Hz, fc = 2 × 106 Hz, Ac = 4V we have
Z
t
SF M (t) =Ac cos[2πfc t + 2πkf
m(η)d(η)]
−∞
8
cos(2π × 2000t)]
2π2000
=4 cos[4 × 106 πt − 2 cos(4000πt)].
=4 cos[2π × 2 × 106 t − 2π × 500
(1)
Problem 6.3
3 /3600
v
Maximum Doppler Spread: fm = = 80×1.6×10
= 52.1Hz.
300/440
λ
And the notch width is 2fm = 104.2Hz.
Problem 6.4
Interpreting W as fm (max audio frequency) and fd as ∆f (frequency
deviation) we get:
βf = fwd = 12
= 3.
4
Problem 6.13
For SNR=30dB=1000,
B=200KHz, the maximum possible data rate, C =
¡
¢
S
B log2 1 + N = 1.99Mbps.
The GSM data rate us 270.833kbps, which is only about 0.136C.
Problem 6
By applying the operator F(B) ,
function B(x, θ) = ejx sin(θ) we get
∂2B
1 ∂ 2B
1 ∂B
+
B
+
+
to the
∂x2
x ∂x
x2 ∂θ2
sin(θ) jx sin(θ)
1
F(B) = − sin2 (θ)ejx sin(θ) + j
e
+ ejx sin(θ) + 2 (−jx sin(θ) − x2 cos(θ))ejx sin(θ)
x
·
¸ x
sin(θ)
sin(θ)
= − sin2 (θ) + j
+1−j
− cos2 (θ) ejx sin(θ) = 0
x
x
(2)
By applying the same operator to B(x, θ) =
the same result:
n
Jn (x)ejnθ we should get
1 X ∂Jn (x) jnθ X
1 X 2
jnθ
e
+
J
(x)e
−
n Jn (x)ejnθ
n
2
2
∂x
x
∂x
x
n
n
n
n
¸
X · ∂ 2 Jn (x) 1 ∂Jn (x)
n2
=
+
+ Jn (x) − 2 Jn (x) ejnθ = 0
2
∂x
x
∂x
x
n
(3)
F(B) =
X ∂ 2 Jn (x)
P
ejxθ +
Therefore,
∂ 2 Jn (x) 1 ∂Jn (x)
n2
+
+
J
(x)
−
Jn (x) = 0
n
∂x2
x ∂x
x2
(4)