Telcommunication laboratory ybu engineering solution homework quizz exam (17)

Solution 2

3.1 The modulated signal is u(t) = m(t)c(t) = Am(t)cos(2π4 × 103 t) ] [ 250 π 200 t) + 4sin(2π t + ) cos(2π4 × 103 t) = A 2cos(2π π π 3 200 200 = Acos(2π(4 × 103 + )t) + Acos(2π(4 × 103 − )t) π π π π 250 250 +2Asin(2π(4×103 + )t+ )−2Asin(2π(4×103 − )t− ) π 3 π 3 Taking the Fourier transform of the previous relation, we obtain [ ] 250 250 200 200 2 jπ 2 jπ 3 3 u(f ) = A δ(f − ) + δ(f + ) + e δ(f − ) − e δ(f + ) π π j π j π ] 1[ δ(f − 4 × 103 ) + δ(f + 4 × 103 ) ∗ 2 A 200 200 = [δ(f − 4 × 103 − ) + δ(f − 4 × 103 + ) 2 π π π π 250 250 + 2e−j 6 δ(f − 4 × 103 − ) + 2ej 6 δ(f − 4 × 103 + ) π π 200 200 ) + δ(f + 4 × 103 + ) δ(f + 4 × 103 − π π π π 250 250 + 2e−j 6 δ(f + 4 × 103 − ) + 2ej 6 δ(f + 4 × 103 + )] π π To find the power content of the modulated signal we write u2 (t) as 200 200 )t) + A2 cos2 (2π(4 × 103 − )t) π π π 250 π 250 )t + ) + 4A2 sin2 (2π(4 × 103 − )t − ) + 4A2 sin2 (2π(4 × 103 + π 3 π 3 + terms of cosine and sine f unctions in the f irst power

u2 (t) = A2 cos2 (2π(4 × 103 +

Hence, ∫

T 2

P = lim

T →∞

u2 (t)dt =

− T2

A2 A2 4A2 4A2 + + + = 5A2 2 2 2 2

3.7 1. The spectrim of u(t) is U (f ) =

20 [δ(f − fc ) + δ(f + fc )] 2 2 + [δ(f − fc − 1500) + δ(f − fc + 1500) + δ(f + fc − 1500) + δ(f + fc + 1500)] 4 10 + [δ(f − fc − 3000) + δ(f − fc + 3000) + δ(f + fc − 3000) + δ(f + fc + 3000)] 4 1

2. The square of the modulated signal is u2 (t) =400cos2 (2πfc t) + 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 T2 , normalize the integral by T1 and take the limit as T → ∞, then all the terms involving cosines tend to zero, whereas the squares of the cosines give a value of 21 . Hence, the power content at the frequency fc = 105 Hz is Pfc = 400 2 = 200, the power content at the frequency Pfc +1500 is the same as the power content at the frequency Pfc −1500 and equal to 21 , whereas Pfc −3000 = Pfc +3000 = 25 2 .

3. u(t) = (20 + 2cos(2π1500t) + 10cos(2π3000t))cos(2πfc t) 1 1 = 20(1 + cos(2π1500t) + cos(2π3000t))cos(2πfc t) 10 2 This is the form of a conventional AM signal with message signal 1 1 cos(2π1500t) + cos(2π3000t) 10 2 1 1 2 = cos (2π1500t) + cos(2π1500t) − 10 2

m(t) =

1 1 The minimum of g(z) = z 2 + 10 z − 12 is achieved for z = − 20 and it is 201 1 min(g(z)) = − 400 . Since z = − 20 is in the range of cos(2π1500t), we con201 clude that the minimum value of m(t) is − 400 . Hence, the modulation index is 201 α = − 400 .

4. u(t) = 20cos(2πfc t) + 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 1 25 25 + + + = 26 2 2 2 2

The total power is Ptotal = Pcarrier + Psidebands = 200 + 26 = 226. The ratio of the sidebands power to the total power is 26 13 Psidebands = = Ptotal 226 113

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 12 a2 =

1 8

3.24 1. 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, where 1 ≤ l ≤ L.

4.4 1. The average transmitted power is 12 1002 = 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 |

1 dϕ(t) 1 | = max | 8000π cos 2000πt| = 4000Hz. 2π dt 2π

4. It can be an FM or PM signal.

For PM, kp m(t) = 4 sin 2000πt; for FM, kf m(t) = 4000 cos 2000πt.

4.6 1. For the narrowband FM signal, ∆fn = 0.1 × 15kHz = 1.5kHz. To achieve ∆f ∆f = 75kHz, the frequency multiplier (the upper one) factor n1 = ∆f = 50. n Then, according to fc = (n1 + n2 )f0 , where f0 = 100kHz, we have n2 = 990.

2. The maximum allowable drift of the 100kHz oscillator ∆ = 10−3 Hz.

3

2 50+990

= 1.923 ×

4.19 1. The instantaneous frequency is f1 (t) = fc + kf m1 (t).

The maximum of f1 (t) is max{f1 (t)} = 1.5M Hz. 2. The phase of the PM modulated signal is ϕ(t) = kp m1 (t) and the instantaneous frequency kp d 1 d ϕ(t) = m1 (t). f1P M (t) = fc + π dt π dt Then we have max{f1P M (t)} min{f1P M (t)}

3 , 2π 3 = 106 − .

2π = 106 +

3. The maximum value of m2 (t) is 1 and it is achieved for t = 0. Hence, max{f2 (t)} = 106 + 103 = 1.001M Hz. f 1 Since, F[sinc(2 × 104 t)] = 2×10 4 Π( 2×104 ), the bandwidth of the message is W = 104 . Thus, using Carson’s rule, we obtain ( ) kf max[|m(t)|] B=2 + 1 W = 22KHz. W

6.8 1. PT = 40kW , since the channel attenuation is 80dB, we have PR = 10−8 PT = 4 × 10−4 W If the noise limiting filter has bandwidth B, then the predetection noise power is ∫ Pn = 2

fc + B 2 fc − B 2

N0 df = N0 B = 2 × 10−10 W 2

In the case of DSB modulation, B = 2W = 2 × 104 Hz, whereas in SSB modulation B = W = 104 Hz. Thus, the predetection signal to noise ratio in DSB and conventional AM is ( ) S PR 4 × 10−4 = = = 100 = 20dB. N i Pn 2 × 10−10 × 2 × 104 and for SSB (

S N

) = i

PR 4 × 10−4 = = 200 = 23dB. Pn 2 × 10−10 × 104

4

2. For DSB, the demodulation gain is 2, hence ( ) ( ) S S =2 = 200 = 23dB N o,DSB N i,DSB 3. For SSB, the demodulation gain is 1, hence ( ) ( ) S S = = 200 = 23dB N o,DSB N i,SSB 4. For conventional AM with α = 0.8 and Pmn = 0.2, we have ( ) ( ) α 2 P mn S S = = 0.126 × 200 = 25.23 = 14dB 2 N o,AM 1 + α Pmn N i,AM

6.9 1. For FM system, according to B = 2(1 + βf )W , we have βf = 11.5. ( (

S N S N

) =

βf2 3 2 PM Ac , 2 N0 W {max |m(t)|}2

=

A2c a2 PM .

2N0 W

)oF M oAM

Then, we have ( ) ( ) 3βf2 S S / = 2 = 549.1 = 27.4dB. N oF M N oAM a {max |m(t)|}2 2. ( (

S N S N

) = )

oF M

= oP M

We have βp2 = 3βf2 . Thus

BWP M BWF M

( ) βf2 PM 3 S , 2 {max |m(t)|}2 N b ( ) βp2 PM S .

2 {max |m(t)|} N b =

2(βp +1)W 2(βf +1)W

5

completes the proof.