About this previous year paper
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
sinθc=n2n1=1.461.5\sin \theta_c = \frac{n_2}{n_1} = \frac{1.46}{1.5}sinθc=n1n2=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)=Accos(ωct+kpm(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):
sinθc=11.5⇒θc=41.8∘\sin \theta_c = \frac{1}{1.5} \Rightarrow \theta_c = 41.8^\circsinθc=1.51⇒θc=41.8∘