Solution 2
3.1The modulated signal is
u(t) = m(t)c(t) = Am(t)cos(2π4× 103t)
= A
[2cos(2π
200
πt) + 4sin(2π
250
πt+
π
3)
]cos(2π4× 103t)
= Acos(2π(4× 103 +200
π)t) +Acos(2π(4× 103 − 200
π)t)
+2Asin(2π(4×103+250
π)t+
π
3)−2Asin(2π(4×103− 250
π)t− π
3)
Taking the Fourier transform of the previous relation, we obtain
u(f) = A
[δ(f − 200
π) + δ(f +
200
π) +
2
jej
π3 δ(f − 250
π)− 2
jej
π3 δ(f +
250
π)
]∗ 1
2
[δ(f − 4× 103) + δ(f + 4× 103)
]=
A
2[δ(f − 4× 103 − 200
π) + δ(f − 4× 103 +
200
π)
+ 2e−j π6 δ(f − 4× 103 − 250
π) + 2ej
π6 δ(f − 4× 103 +
250
π)
δ(f + 4× 103 − 200
π) + δ(f + 4× 103 +
200
π)
+ 2e−j π6 δ(f + 4× 103 − 250
π) + 2ej
π6 δ(f + 4× 103 +
250
π)]
To find the power content of the modulated signal we write u2(t) as
u2(t) = A2cos2(2π(4× 103 +200
π)t) +A2cos2(2π(4× 103 − 200
π)t)
+ 4A2sin2(2π(4× 103 +250
π)t+
π
3) + 4A2sin2(2π(4× 103 − 250
π)t− π
3)
+ terms of cosine and sine functions in the first power
Hence,
P = limT→∞
∫ T2
−T2
u2(t)dt =A2
2+
A2
2+
4A2
2+
4A2
2= 5A2
3.71. The spectrim of u(t) is
U(f) =20
2[δ(f − fc) + δ(f + fc)]
+2
4[δ(f − fc − 1500) + δ(f − fc + 1500) + δ(f + fc − 1500) + δ(f + fc + 1500)]
+10
4[δ(f − fc − 3000) + δ(f − fc + 3000) + δ(f + fc − 3000) + δ(f + fc + 3000)]
1
2. The square of the modulated signal is
u2(t) =400cos2(2πfct) + cos2(2π(fc − 1500)t)
+ cos2(2π(fc + 1500)t) + 25cos2(2π(fc − 3000)t) + 25cos2(2π(fc + 3000)t)
+ terms that are multiples of cosines
If we integrate u2(t) from −T2 to T
2 , normalize the integral by 1T and take the
limit as T → ∞, then all the terms involving cosines tend to zero, whereasthe squares of the cosines give a value of 1
2 . Hence, the power content at thefrequency fc = 105 Hz is Pfc = 400
2 = 200, the power content at the frequencyPfc+1500 is the same as the power content at the frequency Pfc−1500 and equalto 1
2 , whereas Pfc−3000 = Pfc+3000 = 252 .
3.
u(t) = (20 + 2cos(2π1500t) + 10cos(2π3000t))cos(2πfct)
= 20(1 +1
10cos(2π1500t) +
1
2cos(2π3000t))cos(2πfct)
This is the form of a conventional AM signal with message signal
m(t) =1
10cos(2π1500t) +
1
2cos(2π3000t)
= cos2(2π1500t) +1
10cos(2π1500t)− 1
2
The minimum of g(z) = z2 + 110z − 1
2 is achieved for z = − 120 and it is
min(g(z)) = − 201400 . Since z = − 1
20 is in the range of cos(2π1500t), we con-clude that the minimum value of m(t) is − 201
400 . Hence, the modulation index isα = − 201
400 .
4.
u(t) = 20cos(2πfct) + cos(2π(fc − 1500)t) + cos(2π(fc + 1500)t)
+ 5cos(2π(fc − 3000)t) + 5cos(2π(fc + 3000)t)
The power in the sidebands is
Psidebands =1
2+
1
2+
25
2+
25
2= 26
The total power is Ptotal = Pcarrier + Psidebands = 200 + 26 = 226. The ratioof the sidebands power to the total power is
Psidebands
Ptotal=
26
226=
13
113
3.14
u(t) = 5 cos 1800πt+ 20 cos 2000πt+ 5 cos 2200πt
= 20 cos 2000πt (1 + 0.5 cos 200πt) .
2
1. m(t) = cos 200πt, c(t) = 20 cos 2000πt.
2. modulation index a is 0.5.
3. ratio is 12a
2 = 18
3.241. Spectrum illustration
2. For KL = 60, when K = 10, L = 6 orK = 6, L = 10, K + L is minimized.
3. For group 1, to modulate the signal [10, 10 + 4K]kHz to [300, 300 + 4K]kHz,we find fc1 = 290kHZ. Similarly, for group l, fcl = 290+4K(l−1)kHz, where1 ≤ l ≤ L.
4.41. The average transmitted power is 1
21002 = 5000.
2. The peak-phase deviation ∆pmax = kp max |m(t)| = max |4 sin 2000πt| = 4
3. The peak-frequency deviation is given by
∆fmax = kf max |m(t)| = max | 12π
dϕ(t)
dt| = max | 1
2π8000π cos 2000πt| = 4000Hz.
4. It can be an FM or PM signal.
For PM, kpm(t) = 4 sin 2000πt; for FM, kfm(t) = 4000 cos 2000πt.
4.61. For the narrowband FM signal, ∆fn = 0.1 × 15kHz = 1.5kHz. To achieve
∆f = 75kHz, the frequency multiplier (the upper one) factor n1 = ∆f∆fn
= 50.
Then, according to fc = (n1 + n2)f0, where f0 = 100kHz, we have n2 = 990.
2. The maximum allowable drift of the 100kHz oscillator ∆ = 250+990 = 1.923 ×
10−3Hz.
3
4.191. The instantaneous frequency is
f1(t) = fc + kfm1(t).
The maximum of f1(t) is
max{f1(t)} = 1.5MHz.
2. The phase of the PM modulated signal is ϕ(t) = kpm1(t) and the instantaneousfrequency
fPM1 (t) = fc +
1
π
d
dtϕ(t) =
kpπ
d
dtm1(t).
Then we have
max{fPM1 (t)} = 106 +
3
2π,
min{fPM1 (t)} = 106 − 3
2π.
3. The maximum value of m2(t) is 1 and it is achieved for t = 0. Hence,
max{f2(t)} = 106 + 103 = 1.001MHz.
Since, F [sinc(2 × 104t)] = 12×104Π(
f2×104 ), the bandwidth of the message is
W = 104. Thus, using Carson’s rule, we obtain
B = 2
(kf max[|m(t)|]
W+ 1
)W = 22KHz.
6.81. PT = 40kW , since the channel attenuation is 80dB, we have PR = 10−8PT =
4 × 10−4W If the noise limiting filter has bandwidth B, then the predetectionnoise power is
Pn = 2
∫ fc+B2
fc−B2
N0
2df = N0B = 2× 10−10W
In the case of DSB modulation, B = 2W = 2 × 104Hz, whereas in SSBmodulation B = W = 104Hz. Thus, the predetection signal to noise ratio inDSB and conventional AM is(
S
N
)i
=PR
Pn=
4× 10−4
2× 10−10 × 2× 104= 100 = 20dB.
and for SSB(S
N
)i
=PR
Pn=
4× 10−4
2× 10−10 × 104= 200 = 23dB.
4
2. For DSB, the demodulation gain is 2, hence(S
N
)o,DSB
= 2
(S
N
)i,DSB
= 200 = 23dB
3. For SSB, the demodulation gain is 1, hence(S
N
)o,DSB
=
(S
N
)i,SSB
= 200 = 23dB
4. For conventional AM with α = 0.8 and Pmn = 0.2, we have(S
N
)o,AM
=α2Pmn
1 + α2Pmn
(S
N
)i,AM
= 0.126× 200 = 25.23 = 14dB
6.91. For FM system, according to B = 2(1 + βf )W , we have βf = 11.5.(
S
N
)oFM
=3
2A2
c
PM
N0W
β2f
{max |m(t)|}2,(
S
N
)oAM
=A2
ca2PM
2N0W.
Then, we have(S
N
)oFM
/
(S
N
)oAM
=3β2
f
a2{max |m(t)|}2= 549.1 = 27.4dB.
2. (S
N
)oFM
=3
2
β2fPM
{max |m(t)|}2
(S
N
)b
,(S
N
)oPM
=β2pPM
{max |m(t)|}2
(S
N
)b
.
We have β2p = 3β2
f . Thus BWPM
BWFM=
2(βp+1)W2(βf+1)W completes the proof.
5