THEORY EXAMINATION (SEM–VI) 2016-17 ANTENNA AND WAVE PROPAGATION
ANTENNA AND WAVE PROPAGATION (NEN023)
Time: 3 Hours Max Marks: 100
SECTION – A (Short Answer Questions)
(10 × 2 = 20 Marks)
(a) Directional antenna efficiency & directivity of isotropic antenna
A directional antenna concentrates radiated power in a particular direction, hence it has higher efficiency than an isotropic antenna which radiates equally in all directions.
Directivity of isotropic antenna:
D=1(0 dB)D = 1 \quad (0\,\text{dB})D=1(0dB)
(b) Parameters for receiving antenna design
Important parameters are: Gain
Directivity Bandwidth
Polarization Input impedance
Effective aperture
(c) Disadvantages of binomial array
Low directivity Broad beamwidth
Reduced gain compared to uniform arrays Complex amplitude distribution
(d) Relation for electric field strength
Electric field strength at distance d: E∝1dE \propto \frac{1}{d}E∝d1
Thus, as distance increases, electric field strength decreases.
(e) Why sky wave propagation is not possible above 30 MHz
At frequencies above 30 MHz, ionospheric layers cannot reflect radio waves back to earth due to insufficient electron density.
(f) Broadside vs End-fire array
| Broadside Array | End-fire Array |
|---|---|
| Maximum radiation perpendicular to array axis | Maximum radiation along array axis |
| Elements in phase | Progressive phase shift |
(g) Radiation pattern for U=UmcosθU = U_m \cos \thetaU=Umcosθ
Unidirectional: Single main lobe Bidirectional: Two lobes opposite to each other
(h) Directivity over half-sphere
Directivity: D=2D = 2D=2
(i) Power gain of paraboloidal reflector
Given aperture = 10λ10\lambda10λ:
G=(πDλ)2=(10π)2G = \left(\frac{\pi D}{\lambda}\right)^2 = (10\pi)^2G=(λπD)2=(10π)2 GdB=20log10(10π)≈29.9 dBG_{dB} = 20\log_{10}(10\pi) \approx 29.9\,\text{dB}GdB=20log10(10π)≈29.9dB
(j) HPBW and BWFN
HPBW: Half Power Beam Width BWFN: Beam Width Between First Nulls
Relation:
D∝1BeamwidthD \propto \frac{1}{\text{Beamwidth}}D∝Beamwidth1
SECTION – B (Long Answer Questions)
(Attempt any FIVE – 5 × 10 = 50 Marks)
2(a) Aperture & Effective Aperture
Aperture is the physical area that intercepts energy.
Effective aperture: Ae=Gλ24πA_e = \frac{G\lambda^2}{4\pi}Ae=4πGλ2
If gain is reduced to ¼, effective aperture also becomes ¼.
2(b) Maximum received power (Friis equation)
Given:
Distance = 0.5 km
Frequency = 1 GHz Pt=150 WP_t = 150\,WPt=150W
Gt=25 dBG_t = 25\,dBGt=25dB Gr=20 dBG_r = 20\,dBGr=20dB
Using Friis formula:
Pr=PtGtGr(λ4πR)2P_r = P_t G_t G_r \left(\frac{\lambda}{4\pi R}\right)^2Pr=PtGtGr(4πRλ)2
Substituting values gives the maximum received power.
2(c) Antenna array & normalized field
An antenna array is a group of radiating elements arranged to obtain desired radiation characteristics.
For uniform linear array:
Enorm=sin(Nϕ/2)Nsin(ϕ/2)E_{norm} = \frac{\sin(N\phi/2)}{N\sin(\phi/2)}Enorm=Nsin(ϕ/2)sin(Nϕ/2)
2(d) MUF & ionospheric refraction
MUF (Maximum Usable Frequency):
MUF=fcsecθ\text{MUF} = f_c \sec \thetaMUF=fcsecθ
Radio waves bend due to gradual change in refractive index of ionosphere.
Ionosphere consists of D, E, F₁, F₂ layers.
2(e) Corner vs Parabolic reflector & reciprocity theorem
Comparison:
| Corner Reflector | Parabolic Reflector |
|---|---|
| Moderate gain | Very high gain |
| Simple structure | Complex structure |
Reciprocity theorem:
Transmission and reception characteristics of an antenna are identical.
Mutual impedance:
Zm=V2I1Z_m = \frac{V_2}{I_1}Zm=I1V2
2(f) Binomial array & end-fire numerical
Far-field pattern derived using pattern multiplication.
Given directivity = 30 (end-fire array):
Array length and beamwidth calculated using standard formulas.
For broadside array, length is smaller and beamwidth is wider.
2(g) Small loop antenna & Yagi antenna
Small loop antenna far-field:
Eθ∝IAsinθrE_\theta \propto \frac{I A \sin\theta}{r}Eθ∝rIAsinθ
Yagi antenna:
Highly directional antenna with one reflector, one driven element, and multiple directors.
Design for 61–68 MHz includes spacing ≈ 0.2λ.
2(h) Rhombic antenna
Rhombic antenna works on traveling wave principle.
Maximum field intensity is obtained by proper choice of side length and termination resistance.
SECTION – C (Very Long Answer Questions)
(Attempt any TWO – 2 × 15 = 30 Marks)
3(a) Directivity of end-fire array
For two isotropic sources:
D=2[1+sin(2πdλ)]1+(2πdλ)D = \frac{2\left[1+\sin\left(\frac{2\pi d}{\lambda}\right)\right]}{1+\left(\frac{2\pi d}{\lambda}\right)}D=1+(λ2πd)2[1+sin(λ2πd)]
3(b) Beamwidth and directivity
Directivity:
D=41257θEθHD = \frac{41257}{\theta_E \theta_H}D=θEθH41257
4(a) Radiation resistance
Radiation resistance of short dipole:
Rr=80π2(Lλ)2R_r = 80\pi^2\left(\frac{L}{\lambda}\right)^2Rr=80π2(λL)2
4(b) Critical frequency and skip distance
Critical frequency: Highest frequency returned vertically
Virtual height: Apparent reflection height
Skip distance: Minimum distance for sky wave return
Silence zone: Region without reception
5(a)–(c) Short Notes
VLF & LF antennas: Very large size, used for submarine communication
Ground wave propagation: Earth conductivity affects attenuation
Conical spiral antenna: Frequency independent, wide bandwidth
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