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