THEORY EXAMINATION (SEM–VI) 2016-17 COMMUNICATION ENGINEERING

COMMUNICATION ENGINEERING (EEC608)

Time: 3 Hours  Max Marks: 100

SECTION – A (Short Answer Questions)

(10 × 2 = 20 Marks)

(a) RMS noise voltage at amplifier input

Thermal noise voltage is given by:

Vn=4kTRBV_n = \sqrt{4kTRB}Vn​=4kTRB​

Where
k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \, J/Kk=1.38×10−23J/K,
T=300KT = 300 KT=300K (27°C),
R=10kΩR = 10 k\OmegaR=10kΩ,
B=20−18=2 MHzB = 20-18 = 2 \, MHzB=20−18=2MHz

Vn≈1.82 μVV_n \approx 1.82 \, \mu VVn​≈1.82μV 

(b) Effect on maximum deviation in FM

Maximum frequency deviation remains unchanged, because deviation depends on amplitude of modulating signal, not on modulating frequency.

(c) Isotropic source and isotropic medium

Isotropic source: Radiates power equally in all directions.

Isotropic medium: Has identical properties in all directions.

(d) Interference of radio waves

Interference occurs when two or more waves of same frequency overlap.

Conditions:                Same frequency                  Constant phase difference

Same polarization

(e) Advantages of PCM                                        High noise immunity

Error detection and correction                               Suitable for digital transmission

Better signal quality

(f) Reflection and refraction

Reflection: Bouncing back of waves from a surface

Refraction: Bending of waves when passing from one medium to another

(g) Classification of optical fibers

Step-index fiber

Graded-index fiber

Single-mode fiber

Multimode fiber

(h) Bandwidth of FM signal

Given:
fm=2 kHzf_m = 2 \, kHzfm​=2kHz, Δf=10 kHz\Delta f = 10 \, kHzΔf=10kHz

Using Carson’s rule:

BW=2(Δf+fm)=2(10+2)=24 kHzBW = 2(\Delta f + f_m) = 2(10 + 2) = 24 \, kHzBW=2(Δf+fm​)=2(10+2)=24kHz 

(i) Sensitivity and selectivity

Sensitivity: Ability of RF amplifier to detect weak signals

Selectivity: Ability to reject unwanted adjacent frequencies

(j) Critical angle calculation

sin⁡θc=n2n1=1.461.5\sin \theta_c = \frac{n_2}{n_1} = \frac{1.46}{1.5}sinθc​=n1​n2​​=1.51.46​ θc=sin⁡−1(0.973)≈76.7∘\theta_c = \sin^{-1}(0.973) \approx 76.7^\circθc​=sin−1(0.973)≈76.7∘ 

SECTION – B (Long Answer Questions)

(Attempt any FIVE – 5 × 10 = 50 Marks)

2(a) Superheterodyne receiver & AM power numerical

Working:
A superheterodyne receiver converts incoming RF signals to a fixed Intermediate Frequency (IF) using a mixer and local oscillator, improving selectivity and sensitivity.

Power calculation:

Given:
Unmodulated power Pc=9kWP_c = 9 kWPc​=9kW
Modulated power Pt=10.125kWP_t = 10.125 kWPt​=10.125kW

Pt=Pc(1+m22)P_t = P_c \left(1 + \frac{m^2}{2}\right)Pt​=Pc​(1+2m2​) 10.125=9(1+m22)⇒m=0.510.125 = 9 \left(1 + \frac{m^2}{2}\right) \Rightarrow m = 0.510.125=9(1+2m2​)⇒m=0.5

Percentage modulation = 50%

With additional 40% modulation:

mtotal2=0.52+0.42=0.41m_{total}^2 = 0.5^2 + 0.4^2 = 0.41mtotal2​=0.52+0.42=0.41 Ptotal=9(1+0.41/2)=10.845kWP_{total} = 9(1 + 0.41/2) = 10.845 kWPtotal​=9(1+0.41/2)=10.845kW 

2(b) Phase-shift method of SSB-SC & numerical

Given:

em=10sin⁡(2π1000t),ec=20sin⁡(2π104t)e_m = 10\sin(2\pi1000t), \quad e_c = 20\sin(2\pi10^4 t)em​=10sin(2π1000t),ec​=20sin(2π104t)

(i) Modulation index:

m=1020=0.5m = \frac{10}{20} = 0.5m=2010​=0.5

(ii) Sideband frequencies:

fc±fm=10kHz±1kHz=9kHz,11kHzf_c \pm f_m = 10kHz \pm 1kHz = 9kHz, 11kHzfc​±fm​=10kHz±1kHz=9kHz,11kHz

(iii) Sideband amplitude:

=mEc2=5V= \frac{mE_c}{2} = 5V=2mEc​​=5V

(iv) Bandwidth = 1 kHz

(v) Transmission efficiency:

η=m22+m2=11.1%\eta = \frac{m^2}{2+m^2} = 11.1\%η=2+m2m2​=11.1% 

2(c) Angle modulation & PM equation

Angle modulation includes FM and PM.

Phase modulated wave:

s(t)=Accos⁡(ωct+kpm(t))s(t) = A_c \cos(\omega_ct + k_p m(t))s(t)=Ac​cos(ωc​t+kp​m(t))

Pre-emphasis boosts high-frequency signals before transmission.
De-emphasis attenuates them at receiver, improving SNR.

2(d) Noise and SNR

Noise is unwanted disturbance affecting signal quality.

External noise:

Atmospheric noise

Cosmic noise

Industrial noise

SNR=Signal PowerNoise PowerSNR = \frac{Signal \ Power}{Noise \ Power}SNR=Noise PowerSignal Power​

Noise figure: Measure of degradation of SNR by a system.

2(e) Definitions

Quantization error: Difference between actual and quantized value

DPCM: Encodes difference between samples

ADM: Variable step-size delta modulation

PPM: Information conveyed by pulse position

2(f) FSK generation & detection, TDM

FSK represents binary data using different frequencies.

Coherent detection uses synchronized carrier.

TDM: Multiple signals share same channel using different time slots.

2(g) Uplink frequency higher than downlink

Higher uplink frequency reduces antenna size and atmospheric losses.

Earth station design considerations:

High power amplifier

Low noise amplifier

Antenna gain

Frequency stability

SECTION – C (Very Long Answer Questions)

(Attempt any TWO – 2 × 15 = 30 Marks)

3. Ground wave & sky wave propagation

Ground wave: Travels along earth surface, suitable for low frequencies.

Angle of tilt: Tilt of wavefront due to earth conductivity, reduces field strength.

Sky wave terms:

Virtual height: Apparent reflection height

Critical frequency: Highest frequency returned vertically

Skip distance: Minimum distance for sky wave return

4. Monochrome TV receiver & raster scanning

TV receiver blocks include:

Tuner

IF amplifier

Video detector

Sync separator

Deflection circuits

Picture tube

Raster scanning: Electron beam scans screen line by line horizontally and vertically to form image.

5. Optical vs electrical communication & numerical

Advantages of optical communication:

High bandwidth

Low attenuation

No electromagnetic interference

Secure transmission

Critical angle (n₂ = air):

sin⁡θc=11.5⇒θc=41.8∘\sin \theta_c = \frac{1}{1.5} \Rightarrow \theta_c = 41.8^\circsinθc​=1.51​⇒θc​=41.8∘