Chapter 2 · 5 hours
Antenna Parameters and Arrays
IOE past exam questions
Past questions and answers
33 questions set from this chapter, 9 of them more than once. Most asked first.
- Asked 8 times
- 2081 Chaitra · 3 marks
- 2080 Chaitra · 3 marks
- 2079 Chaitra · 3 marks
- 2077 Chaitra · 2 marks
- 2075 Bhadra · 3 marks
- 2074 Bhadra · 2 marks
- 2070 Magh · 2 marks
- 2069 Bhadra · 2 marks
Explain the polarization of an antenna.
Answer
Polarization of an antenna is the polarization of the wave it radiates in a given direction (usually the direction of maximum gain): the figure traced by the tip of the electric field vector with time at a fixed point in the far field.
Types:
- Linear: E moves along a straight line. Occurs when there is one component, or two components in phase (or 180° apart). Vertical (e.g. vertical monopole) or horizontal (e.g. horizontal dipole).
- Circular: E rotates with constant magnitude. Needs two orthogonal components of equal amplitude and 90° phase difference. Right-hand (RHCP) or left-hand (LHCP), e.g. axial-mode helix.
- Elliptical: general case; two orthogonal components with unequal amplitude and/or a phase difference other than 0° or 180°.
Linear Circular Elliptical
| .-. .--.
| ( ) ( )
| '-' '--'
Axial ratio AR = major axis / minor axis: 1 for circular, ∞ for linear.
Polarization loss factor: PLF = cos²ψ, where ψ is the angle between the wave and antenna polarizations. A vertical antenna receiving a horizontal wave gets PLF = 0 (no signal).
- Asked 4 times
- 2077 Chaitra · 2 marks
- 2076 Baisakh · 2 marks
- 2074 Bhadra · 2 marks
- 2070 Magh · 2 marks
Explain the radiation pattern (radiation pattern lobes) of an antenna.
Answer
A radiation pattern is a graph (2-D or 3-D) of the radiated field strength or power of an antenna versus direction (θ, φ) in the far field. It is usually normalized to the maximum and drawn in the E-plane and H-plane.
Radiation pattern lobes:
- Main (major) lobe: lobe containing the direction of maximum radiation.
- Minor lobes: all other lobes; side lobes are next to the main lobe and the back lobe is opposite it.
- Nulls: directions of zero radiation between lobes.
- HPBW: angle between −3 dB points of the main lobe; FNBW: angle between first nulls.
- Side lobe level: side lobe peak relative to main lobe (dB).
main lobe
.---.
/ \ HPBW
side | | side
lobe .| |. lobe
( \ o / )
' '-+-' '
back lobe
- Asked 4 times
- 2080 Asoj · 3 marks
- 2077 Chaitra · 4 marks
- 2073 Magh · 4 marks
- 2072 Magh · 3 marks
Write a short note on effective isotropic radiated power (EIRP).
Answer
Effective Isotropic Radiated Power (EIRP) is the power that an ideal isotropic antenna would have to radiate to give the same power density, in the direction of maximum radiation, as the actual antenna.
EIRP = Pt · Gt / L
EIRP (dBW) = Pt (dBW) + Gt (dBi) − L (dB)
where Pt = transmitter power, Gt = transmit antenna gain relative to isotropic, L = feeder/cable losses.
The power flux density at distance d is then
PFD = EIRP / (4π d²) W/m²
Example: Pt = 100 W (20 dBW), Gt = 30 dBi, feeder loss = 2 dB EIRP = 20 + 30 − 2 = 48 dBW ≈ 63.1 kW.
Importance:
- Used in link budget calculations for satellite and terrestrial links.
- Lets transmitters with different power and antennas be compared with one number.
- Regulatory bodies set limits on EIRP (e.g. Wi-Fi, satellite footprints are drawn as EIRP contours).
- A high-gain antenna can give a large EIRP with a small transmitter power.
ERP (effective radiated power) is similar but relative to a half-wave dipole: EIRP = ERP + 2.15 dB.
- Asked 3 times
- 2077 Chaitra · 2 marks
- 2076 Baisakh · 2 marks
- 2070 Bhadra · 2 marks
Explain the beamwidth (half power beamwidth) of an antenna.
Answer
Beamwidth is the angular width of the main lobe of the radiation pattern. The Half Power Beamwidth (HPBW) is the angle between the two directions in the main lobe where the radiated power falls to half (−3 dB) of its maximum, i.e. where the field is 1/√2 = 0.707 of maximum.
max
.---. -3 dB
0.707/ | \0.707
/<-HPBW->\
/ | \
null | null
<---FNBW---->
- FNBW (first null beamwidth): angle between the first nulls; FNBW ≈ 2 × HPBW.
- Narrow beamwidth means high directivity: D ≈ 41253/(θE° × θH°).
- Examples: short dipole 90°, λ/2 dipole 78°.
- Asked 3 times
- 2075 Bhadra · 3 marks
- 2072 Asoj · 4 marks
- 2069 Bhadra · 2 marks
Explain antenna efficiency.
Answer
Antenna (radiation) efficiency is the ratio of the power radiated by the antenna to the power accepted at its input terminals. It accounts for conductor and dielectric losses.
η = P_rad / P_in = P_rad / (P_rad + P_loss)
= Rr / (Rr + RL)
where Rr = radiation resistance and RL = loss resistance.
The total efficiency also includes mismatch loss:
e0 = e_r · e_c · e_d, e_r = 1 − |Γ|²
(e_r reflection, e_c conductor, e_d dielectric efficiency).
Example: Rr = 73 Ω, RL = 2 Ω → η = 73/75 = 0.973 (97.3 %). With |Γ| = 0.2, e_r = 0.96 and total efficiency ≈ 0.934 (93.4 %).
Efficiency links gain and directivity: G = η D. Electrically short antennas have very small Rr, so their efficiency is low.
- Asked 3 times
- 2079 Chaitra · 3 marks
- 2072 Asoj · 4 marks
- 2070 Bhadra · 2 marks
Explain the directivity of an antenna.
Answer
Directivity (D) is the ratio of the radiation intensity of an antenna in a given direction to the radiation intensity averaged over all directions. Usually the maximum value is quoted:
D = U_max / U_avg = 4π U_max / P_rad
where U = radiation intensity (W/sr) and U_avg = P_rad/4π (isotropic source).
- It depends only on the shape of the pattern, not on losses.
- Isotropic antenna D = 1 (0 dBi); short dipole 1.5 (1.76 dBi); λ/2 dipole 1.64 (2.15 dBi).
- For a pencil beam with half-power beamwidths θ1, θ2:
D ≈ 4π / (θ1 θ2) (radians)
≈ 41253 / (θ1° θ2°)
Example: θ1 = θ2 = 30° → D ≈ 41253/900 ≈ 45.8 (16.6 dBi).
Relation with gain: G = η D, where η is the radiation efficiency.
- Asked 2 times
- 2079 Chaitra · 8 marks
- 2075 Baisakh · 6 marks
Derive an expression for the total field in case of two isotropic point sources with equal amplitude and opposite phase.
Answer
Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P.
to far point P
/
/ φ
(1)-----------O-----------(2)
|<--- d/2 --->|<--- d/2 --->|
path from (2) shorter by (d/2)cosφ
path from (1) longer by (d/2)cosφ
Path difference between the two waves = d cosφ, so the phase difference due to path is
ψ = β d cosφ, β = 2π/λ
With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2.
Field for equal amplitude, opposite phase
Currents equal in magnitude but opposite in phase (180°). By superposition:
E = −E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2jE0 sin(ψ/2)
|E| = 2E0 sin[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ].
- Maxima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire).
- Minima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°.
- Half power: sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe.
φ=90° (null)
|
.--. | .--.
180°( )---(1)O(2)---( ) 0°
'--' | '--'
|
End-fire figure-of-eight
General conditions (any d):
- Maxima: (βd/2)cosφ = ±(2n + 1)π/2 → cosφ = ±(2n + 1)λ/(2d)
- Minima: (βd/2)cosφ = ±nπ → cosφ = ±nλ/d
So the opposite-phase pair is a simple end-fire array.
- Asked 2 times
- 2080 Chaitra · 6+2 marks
- 2076 Baisakh · 6+3 marks
Derive an expression for the total field in case of two isotropic point sources with same amplitude and out of phase (currents equal in magnitude but opposite in phase). Also show the condition of maxima and minima direction of intensity and draw the radiation pattern.
Answer
Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P.
to far point P
/
/ φ
(1)-----------O-----------(2)
|<--- d/2 --->|<--- d/2 --->|
path from (2) shorter by (d/2)cosφ
path from (1) longer by (d/2)cosφ
Path difference between the two waves = d cosφ, so the phase difference due to path is
ψ = β d cosφ, β = 2π/λ
With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2.
Total field
Currents equal in magnitude but opposite in phase (180°). By superposition:
E = −E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2jE0 sin(ψ/2)
|E| = 2E0 sin[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ].
- Maxima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire).
- Minima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°.
- Half power: sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe.
φ=90° (null)
|
.--. | .--.
180°( )---(1)O(2)---( ) 0°
'--' | '--'
|
End-fire figure-of-eight
General conditions for maxima and minima
- Maxima when sin[(βd/2)cosφ] = ±1:
(βd/2) cosφ = ±(2n + 1) π/2
cosφ_max = ±(2n + 1) λ / (2d), n = 0, 1, 2 ...
- Minima when sin[(βd/2)cosφ] = 0:
(βd/2) cosφ = ±n π
cosφ_min = ±n λ / d, n = 0, 1, 2 ...
n = 0 always gives a null at φ = 90° (broadside), since the two equal and opposite fields cancel on the perpendicular bisector.
For d = λ/2 this gives maxima at 0° and 180° and nulls at 90° and 270°, as shown in the pattern.
- Asked 2 times
- 2077 Chaitra · 2+6 marks
- 2076 Baisakh · 2+6 marks
What is the importance of antenna array? Describe the principle of End fire and Broadside array, and differentiate between them.
Answer
An antenna array is a group of similar antennas arranged in a regular geometry and fed with controlled amplitudes and phases so that their fields add in desired directions.
Importance of antenna array
- Gives high directivity and gain that one small element cannot give.
- Narrow, steerable beam: the beam direction can be changed electronically by changing phase (phased arrays, radar).
- Pattern can be shaped: low side lobes, nulls towards interference.
- Higher total power handling, since power is shared by many elements.
- Better reliability: failure of one element does not stop operation.
For a uniform linear array of N elements, spacing d, progressive phase δ:
ψ = βd cosθ + δ
AF = sin(Nψ/2) / [N sin(ψ/2)]
Maximum when ψ = 0
Broadside array
- All elements fed in phase (δ = 0).
- ψ = 0 at θ = 90°, so maximum radiation is perpendicular to the array axis.
- Pattern is a disk-like beam around the axis (narrow in the plane containing the axis).
End-fire array
- Progressive phase δ = −βd, i.e. each element lags the previous one by the phase delay over spacing d.
- ψ = βd(cosθ − 1) = 0 at θ = 0°, so maximum radiation is along the array axis.
Broadside End-fire
^ ^ ^ ^
| | | | o o o o o ====>
o--o--o--o--o max along axis
max perpendicular
Difference
| Point | Broadside | End-fire |
|---|---|---|
| Phase shift δ | 0 (in phase) | −βd (progressive) |
| Max direction | ⊥ to array axis (θ = 90°) | Along axis (θ = 0°) |
| Pattern | Fan/disk shaped | Cigar/pencil along axis |
| BWFN | ≈ 2λ/(Nd) | ≈ 2√(2λ/(Nd)) |
| Directivity | ≈ 2Nd/λ | ≈ 4Nd/λ |
| Example N = 10, d = λ/2 | BWFN ≈ 23°, D ≈ 10 | BWFN ≈ 72°, D ≈ 20 |
| Typical use | Broadcast, base stations | Yagi-Uda, log-periodic, point-to-point |
- 2074 Bhadra · 2 marks
Explain antenna gain.
Answer
Antenna gain (G) is the ratio of the radiation intensity in a given direction to the intensity that would be produced if the power accepted by the antenna were radiated isotropically:
G = 4π U(θ,φ) / P_in = η · D
It includes the antenna's losses through the efficiency η. Gain is given in dBi (vs isotropic) or dBd (vs λ/2 dipole; dBi = dBd + 2.15). Example: λ/2 dipole 2.15 dBi; parabolic dish G = η_a(πD/λ)², often 30–50 dBi.
- 2076 Baisakh · 2 marks
Explain antenna gain and directivity.
Answer
Directivity (D) is the ratio of maximum radiation intensity to the average radiation intensity: D = 4πU_max/P_rad. It depends only on the pattern shape.
Gain (G) is the same ratio but referred to the input power, so it includes losses: G = 4πU_max/P_in = η·D, where η is the radiation efficiency. Hence G ≤ D. Example: λ/2 dipole D = 1.64; with η = 0.97, G ≈ 1.59 (2.0 dBi).
- 2079 Chaitra · 3 marks
Explain the input impedance of an antenna.
Answer
Input impedance is the impedance seen at the antenna's feed terminals, i.e. the ratio of voltage to current there (or of E to H at a reference point).
Z_in = R_in + j X_in
R_in = Rr + RL
- Rr (radiation resistance) represents power radiated; RL represents ohmic and dielectric loss.
- X_in represents energy stored in the near field.
- The antenna is resonant when X_in = 0.
Z_in depends on frequency, antenna length, feed point, and nearby objects/ground. Examples: λ/2 dipole ≈ 73 + j42.5 Ω (slightly shortened → 73 Ω), λ/4 monopole ≈ 36.5 Ω, folded dipole ≈ 292 Ω.
For maximum power transfer, Z_in must match the line Z0 (Γ = (Z_in − Z0)/(Z_in + Z0) → 0). Mismatch gives VSWR > 1 and reflected power; matching networks or baluns are used.
- 2075 Bhadra · 3 marks
Explain the bandwidth of an antenna.
Answer
Bandwidth of an antenna is the range of frequencies over which its performance (input impedance/VSWR, gain, pattern, beamwidth, polarization) stays within an acceptable limit of its value at the centre frequency.
Narrowband: BW% = (fH − fL)/fc × 100
Broadband: ratio = fH : fL (e.g. 10:1)
Resonant: BW ≈ fc / Q
Common criterion: VSWR ≤ 2 (return loss ≥ 10 dB).
Example: an antenna centred at 900 MHz with VSWR ≤ 2 from 860 to 940 MHz has BW = 80 MHz = 8.89 %.
Types: impedance bandwidth and pattern bandwidth. Thick conductors, travelling-wave and frequency-independent designs (log-periodic, spiral, biconical) give wide bandwidth; high-Q resonant antennas (thin dipole, small loop) are narrowband.
- 2081 Chaitra · 3 marks
Explain the antenna parameter: receiving antenna gain (Gr).
Answer
Receiving antenna gain (Gr) expresses how well an antenna captures power from an incoming wave in its direction of maximum response compared with an isotropic antenna. By reciprocity it equals the transmitting gain, and it is related to the effective aperture Ae:
Gr = 4π Ae / λ²
Ae = η_a × A_physical (aperture antennas)
Pr = PFD × Ae = Pt Gt Gr (λ / 4πd)²
Example: 1.2 m dish at 12 GHz (λ = 0.025 m), η_a = 0.55: Ae = 0.55 × π(0.6)² ≈ 0.622 m²; Gr = 4π × 0.622/0.025² ≈ 12 507 ≈ 41.0 dBi.
Gr is higher for larger aperture and higher frequency; it appears in the receiver term G/T of satellite links.
- 2072 Magh · 3 marks
Write a short note on linear polarization.
Answer
An antenna or wave is linearly polarized when the tip of its electric field vector, at a fixed point, always moves along a straight line as time passes.
Conditions (with components Ex, Ey):
- Only one component exists, or
- Two orthogonal components are in phase or 180° out of phase (Δφ = nπ), with any amplitudes.
Vertical Horizontal Slant (45°)
| /
| ------- /
| /
Types:
- Vertical polarization: E perpendicular to the earth, e.g. vertical monopole, AM/mobile.
- Horizontal polarization: E parallel to the earth, e.g. horizontal dipole, TV broadcast.
- Slant (±45°): used for polarization diversity in cellular base stations.
Axial ratio = ∞. A linearly polarized receiving antenna gets PLF = cos²ψ, where ψ is the angle between wave and antenna polarizations; cross-polarized (ψ = 90°) reception gives zero signal. A linear antenna receives half of the power (3 dB loss) of a circularly polarized wave.
- 2080 Chaitra · 3 marks
Write a short note on power flux density (PFD).
Answer
Power flux density (PFD) is the power flowing per unit area normal to the direction of propagation, in W/m² (or dBW/m²). It is the magnitude of the average Poynting vector.
For an isotropic radiator, power Pt spreads uniformly over a sphere of area 4πd²:
PFD = Pt / (4π d²)
With antenna gain: PFD = Pt Gt / (4π d²) = EIRP / (4π d²)
In terms of field: PFD = E²rms / η0 = E²rms / 120π
Example: EIRP = 1 kW, d = 10 km: PFD = 1000/(4π × 10⁸) = 7.96 × 10⁻⁷ W/m² (−61.0 dBW/m²), giving E_rms = √(PFD × 120π) = 17.3 mV/m.
Uses: received power Pr = PFD × Ae; ITU sets PFD limits for satellites to protect terrestrial services. PFD falls as 1/d² (inverse square law).
- 2076 Bhadra · 6 marks
Describe the antenna gain, antenna efficiency and antenna polarization with mathematical relation if necessary.
Answer
Antenna gain
Gain is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the input power were radiated isotropically:
G(θ,φ) = 4π U(θ,φ) / P_in
P_rad = η P_in → G = η D
G(dBi) = 10 log10 G
It includes losses, so G ≤ D. Example: λ/2 dipole, D = 1.64; if η = 0.97, G = 1.59 (2.0 dBi). For aperture antennas G = 4πAe/λ².
Antenna efficiency
Radiation efficiency is the ratio of power radiated to power accepted at the terminals:
η = P_rad / P_in = Rr / (Rr + RL)
Rr = radiation resistance, RL = ohmic/dielectric loss resistance. Total efficiency also includes mismatch: e0 = (1 − |Γ|²) · η.
Example: Rr = 73 Ω, RL = 2 Ω → η = 73/75 = 97.3 %.
Antenna polarization
Polarization is the orientation of the E-field vector traced with time at a fixed far-field point in the direction of maximum radiation. Write the field as two orthogonal components:
Ex = Ex0 cos(ωt − kz)
Ey = Ey0 cos(ωt − kz + δ)
- Linear: δ = 0 or π (or one component zero).
- Circular: Ex0 = Ey0 and δ = ±90° (RHCP / LHCP).
- Elliptical: all other cases.
Axial ratio AR = major/minor axis (1 for circular, ∞ for linear). Polarization loss factor PLF = |ρ̂w · ρ̂a|² = cos²ψ.
- 2075 Baisakh · 6 marks
Describe the antenna gain, antenna efficiency and directivity of antenna with mathematical derivation if necessary.
Answer
Antenna gain
Gain is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the input power were radiated isotropically:
G(θ,φ) = 4π U(θ,φ) / P_in
P_rad = η P_in → G = η D
G(dBi) = 10 log10 G
It includes losses, so G ≤ D. Example: λ/2 dipole, D = 1.64; if η = 0.97, G = 1.59 (2.0 dBi). For aperture antennas G = 4πAe/λ².
Antenna efficiency
Radiation efficiency is the ratio of power radiated to power accepted at the terminals:
η = P_rad / P_in = Rr / (Rr + RL)
Rr = radiation resistance, RL = ohmic/dielectric loss resistance. Total efficiency also includes mismatch: e0 = (1 − |Γ|²) · η.
Example: Rr = 73 Ω, RL = 2 Ω → η = 73/75 = 97.3 %.
Directivity
Directivity is the ratio of maximum radiation intensity to average radiation intensity:
U_avg = P_rad / 4π
D = U_max / U_avg = 4π U_max / P_rad
P_rad = ∮ U dΩ = ∫₀²π ∫₀π U(θ,φ) sinθ dθ dφ
Derivation for a short dipole (U = U0 sin²θ):
P_rad = U0 ∫₀²π ∫₀π sin³θ dθ dφ = U0 · 2π · 4/3 = 8πU0/3
D = 4π U0 / (8πU0/3) = 1.5 (1.76 dBi)
Approximation for narrow beams: D ≈ 41253/(θ1° θ2°).
- 2072 Magh · 6 marks
Describe antenna gain, antenna efficiency and beam width of antenna with mathematical derivations if necessary.
Answer
Antenna gain
Gain is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the input power were radiated isotropically:
G(θ,φ) = 4π U(θ,φ) / P_in
P_rad = η P_in → G = η D
G(dBi) = 10 log10 G
It includes losses, so G ≤ D. Example: λ/2 dipole, D = 1.64; if η = 0.97, G = 1.59 (2.0 dBi). For aperture antennas G = 4πAe/λ².
Antenna efficiency
Radiation efficiency is the ratio of power radiated to power accepted at the terminals:
η = P_rad / P_in = Rr / (Rr + RL)
Rr = radiation resistance, RL = ohmic/dielectric loss resistance. Total efficiency also includes mismatch: e0 = (1 − |Γ|²) · η.
Example: Rr = 73 Ω, RL = 2 Ω → η = 73/75 = 97.3 %.
Beamwidth
HPBW is the angle between the two directions in the main lobe where power falls to half (field to 0.707) of maximum. FNBW is the angle between the first nulls.
Derivation for a short dipole (E ∝ sinθ, power ∝ sin²θ):
sin²θ = 1/2 → sinθ = 1/√2
θ1 = 45°, θ2 = 135°
HPBW = 135° − 45° = 90°
For a λ/2 dipole, E = cos[(π/2)cosθ]/sinθ = 0.707 gives θ ≈ 51° and 129° → HPBW ≈ 78°.
max
.---.
0.707/ | \0.707
/<HPBW>\
/ <-FNBW->\
A narrower beam means higher directivity: D ≈ 41253/(θE° θH°).
- 2080 Asoj · 3+3+3 marks
Describe the radiation pattern, antenna gain and power-flux density with mathematical derivation if necessary.
Answer
Radiation pattern
The radiation pattern is a graph of the far-field radiation (field or power) of an antenna as a function of direction (θ, φ). The field pattern plots |E(θ,φ)|; the power pattern plots U(θ,φ) ∝ |E|². Normally it is normalized and drawn in the E-plane and H-plane.
For a short dipole, from the far field Eθ = jηkI0 l sinθ e^(−jkr)/(4πr):
E_n(θ) = sinθ, P_n(θ) = sin²θ
main lobe
.---.
/ \ HPBW
side | | side
lobe . \ o / . lobe
'-+-'
back lobe
Parts: main lobe, side lobes, back lobe, nulls, HPBW, FNBW, side lobe level.
Antenna gain
Gain is the ratio of the radiation intensity in a given direction to the radiation intensity that would be obtained if the input power were radiated isotropically:
G(θ,φ) = 4π U(θ,φ) / P_in
P_rad = η P_in → G = η D
G(dBi) = 10 log10 G
It includes losses, so G ≤ D. Example: λ/2 dipole, D = 1.64; if η = 0.97, G = 1.59 (2.0 dBi). For aperture antennas G = 4πAe/λ².
Derivation of D for the short dipole: U = U0 sin²θ, P_rad = U0 ∫∫ sin³θ dθ dφ = 8πU0/3, D = 4πU0/P_rad = 1.5; with η = 1, G = 1.5 (1.76 dBi).
Power flux density (PFD)
PFD is the power crossing unit area normal to the direction of propagation (W/m²), the magnitude of the average Poynting vector.
Power Pt radiated isotropically spreads over a sphere of area 4πd²:
PFD_iso = Pt / (4π d²)
With gain Gt: PFD = Pt Gt / (4π d²) = EIRP / (4πd²)
Also: PFD = E²rms / 120π
Example: EIRP = 1 kW, d = 10 km → PFD = 1000/(4π × 10⁸) = 7.96 × 10⁻⁷ W/m², E_rms = 17.3 mV/m.
Received power: Pr = PFD × Ae.
- 2073 Magh · 2+6 marks
List the parameters of antenna and explain any three of them.
Answer
Antenna parameters are the quantities used to describe the performance of an antenna.
List of antenna parameters
- Radiation pattern
- Directivity
- Gain
- Radiation efficiency
- Beamwidth (HPBW, FNBW)
- Bandwidth
- Input impedance
- Radiation resistance
- Polarization
- Effective aperture (effective area)
- Front-to-back ratio, side lobe level
- EIRP
Explanation of three parameters
1. Directivity
D = U_max / U_avg = 4π U_max / P_rad
It is the ratio of the maximum radiation intensity to the average intensity. For a short dipole U = U0 sin²θ, P_rad = 8πU0/3, so D = 1.5 (1.76 dBi). For narrow beams D ≈ 41253/(θ1° θ2°). Isotropic D = 1; λ/2 dipole 1.64.
2. Gain
G = 4π U_max / P_in = η D
Gain includes losses, so G ≤ D. Expressed in dBi; dBi = dBd + 2.15. Also G = 4πAe/λ² for aperture antennas. Example: a 1.2 m dish at 12 GHz with 55 % efficiency has G ≈ 41.0 dBi.
3. Half power beamwidth (HPBW) The angle between the two directions of the main lobe where power is half (−3 dB) of maximum. For a short dipole, sin²θ = 0.5 gives θ = 45° and 135°, so HPBW = 90°; λ/2 dipole ≈ 78°.
max
.---.
0.707/ | \0.707
/<HPBW>\
/ \
A narrow beamwidth means high directivity and better rejection of interference from other directions.
- 2071 Bhadra · 5 marks
Explain any five parameters of antenna.
Answer
Antenna parameters describe how well an antenna radiates or receives. Five important ones:
-
Radiation pattern: a graph of the radiated field or power versus direction (θ, φ) in the far field. It shows the main lobe, side lobes, back lobe and nulls. E.g. short dipole: E ∝ sinθ (doughnut).
-
Directivity (D): ratio of maximum radiation intensity to average radiation intensity:
D = 4π U_max / P_rad
Short dipole 1.5; λ/2 dipole 1.64 (2.15 dBi).
-
Gain (G): directivity including losses: G = ηD = 4πU_max/P_in. Expressed in dBi.
-
Radiation efficiency (η):
η = P_rad / P_in = Rr / (Rr + RL)
e.g. Rr = 73 Ω, RL = 2 Ω → 97.3 %.
- Half power beamwidth (HPBW): angle between the −3 dB points of the main lobe; short dipole 90°, λ/2 dipole 78°. Narrow beam → high directivity.
Other parameters: bandwidth, input impedance, polarization, effective aperture.
- 2081 Chaitra · 4+4 marks
Derive the expressions for the total field in case of two isotropic point sources with (i) an equal amplitude and opposite phase (ii) an equal amplitude and same phase.
Answer
Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P.
to far point P
/
/ φ
(1)-----------O-----------(2)
|<--- d/2 --->|<--- d/2 --->|
path from (2) shorter by (d/2)cosφ
path from (1) longer by (d/2)cosφ
Path difference between the two waves = d cosφ, so the phase difference due to path is
ψ = β d cosφ, β = 2π/λ
With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2.
(i) Equal amplitude, opposite phase
Currents equal in magnitude but opposite in phase (180°). By superposition:
E = −E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2jE0 sin(ψ/2)
|E| = 2E0 sin[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ].
- Maxima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire).
- Minima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°.
- Half power: sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe.
φ=90° (null)
|
.--. | .--.
180°( )---(1)O(2)---( ) 0°
'--' | '--'
|
End-fire figure-of-eight
(ii) Equal amplitude, same phase
Currents equal in magnitude and in phase. By superposition:
E = E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2E0 cos(ψ/2)
E = 2E0 cos[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = cos[(π/2) cosφ].
- Maxima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270° (broadside).
- Minima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180°.
- Half power: cos[(π/2)cosφ] = 1/√2 → cosφ = ±1/2 → φ = 60°, 120° → HPBW = 60°.
φ=90°
.-.
( )
(1)---O---(2) φ=0° (null on axis)
( )
'-'
φ=270°
Broadside figure-of-eight
| Case | Field | Max (d = λ/2) | Type |
|---|---|---|---|
| Same phase | 2E0 cos[(βd/2)cosφ] | 90°, 270° | Broadside |
| Opposite phase | 2E0 sin[(βd/2)cosφ] | 0°, 180° | End-fire |
- 2072 Magh · 2+6 marks
Why do we need antenna array rather than changing the length of single antenna? Derive the expression and draw the pattern for an array of two isotropic radiators with: i) Equal amplitude and phase ii) Equal amplitude and opposite phase
Answer
Why an array instead of a longer single antenna
- Making a single wire longer than about 1.25λ does not keep increasing gain: current phase reversals appear and the beam splits into many lobes (e.g. 3λ/2 dipole has lower directivity than a λ dipole).
- One element gives a fixed, broad pattern; its beam cannot be steered without moving it.
- An array of elements fed with controlled amplitude and phase gives high directivity, narrow beam, low side lobes, nulls towards interference and electronic beam steering, while each element stays a convenient size (e.g. λ/2).
- Power is shared among elements, and failure of one element only slightly degrades performance.
Two isotropic radiators
Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P.
to far point P
/
/ φ
(1)-----------O-----------(2)
|<--- d/2 --->|<--- d/2 --->|
path from (2) shorter by (d/2)cosφ
path from (1) longer by (d/2)cosφ
Path difference between the two waves = d cosφ, so the phase difference due to path is
ψ = β d cosφ, β = 2π/λ
With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2.
i) Equal amplitude and phase
Currents equal in magnitude and in phase. By superposition:
E = E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2E0 cos(ψ/2)
E = 2E0 cos[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = cos[(π/2) cosφ].
- Maxima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270° (broadside).
- Minima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180°.
- Half power: cos[(π/2)cosφ] = 1/√2 → cosφ = ±1/2 → φ = 60°, 120° → HPBW = 60°.
φ=90°
.-.
( )
(1)---O---(2) φ=0° (null on axis)
( )
'-'
φ=270°
Broadside figure-of-eight
ii) Equal amplitude and opposite phase
Currents equal in magnitude but opposite in phase (180°). By superposition:
E = −E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2jE0 sin(ψ/2)
|E| = 2E0 sin[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ].
- Maxima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire).
- Minima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°.
- Half power: sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe.
φ=90° (null)
|
.--. | .--.
180°( )---(1)O(2)---( ) 0°
'--' | '--'
|
End-fire figure-of-eight
- 2071 Magh · 3+3 marks
Why are antenna radiation pattern lobes important during antenna design? Derive the relation for the field intensity of linear array of 2 isotropic radiators with equal potential and 180° out of phase.
Answer
Importance of radiation pattern lobes in design
- The main lobe sets the direction and width of coverage (HPBW) and hence directivity and gain.
- Side lobes and back lobe waste power in unwanted directions, lower gain, and pick up interference and noise (e.g. ground noise in satellite receivers).
- Low side lobes are needed in radar to avoid false targets and in communication to meet interference limits; a good front-to-back ratio is needed in point-to-point links.
- Nulls can be placed towards interfering sources.
- Designers therefore control element spacing, amplitude taper (e.g. binomial, Dolph–Chebyshev) and phase to shape the lobes.
Field of 2 isotropic radiators, equal amplitude, 180° out of phase
Consider two isotropic point sources 1 and 2 spaced a distance d apart on the x-axis, with the reference (origin) at the midpoint. A far point P is in direction φ measured from the array axis. Each source alone gives field E0 at P.
to far point P
/
/ φ
(1)-----------O-----------(2)
|<--- d/2 --->|<--- d/2 --->|
path from (2) shorter by (d/2)cosφ
path from (1) longer by (d/2)cosφ
Path difference between the two waves = d cosφ, so the phase difference due to path is
ψ = β d cosφ, β = 2π/λ
With respect to the midpoint, wave 2 leads by ψ/2 and wave 1 lags by ψ/2.
Currents equal in magnitude but opposite in phase (180°). By superposition:
E = −E0 e^(−jψ/2) + E0 e^(+jψ/2)
= 2jE0 sin(ψ/2)
|E| = 2E0 sin[(βd/2) cosφ]
For d = λ/2 (βd/2 = π/2), normalized: E_n = sin[(π/2) cosφ].
- Maxima: (π/2)cosφ = ±π/2 → cosφ = ±1 → φ = 0°, 180° (along the axis, end-fire).
- Minima: (π/2)cosφ = 0 → cosφ = 0 → φ = 90°, 270°.
- Half power: sin[(π/2)cosφ] = ±1/√2 → cosφ = ±1/2 → φ = 60°, 120°, 240°, 300° → HPBW = 120° per lobe.
φ=90° (null)
|
.--. | .--.
180°( )---(1)O(2)---( ) 0°
'--' | '--'
|
End-fire figure-of-eight
- 2071 Bhadra · 5+6 marks
Derive a relation for the field intensity for the array of two elements isotropic radiators in various conditions. Show the condition for broad side and end fire array with necessary diagrams.
Answer
An antenna array is a group of similar antennas whose fields add at a distant point. For two isotropic point sources, the total field is the element field multiplied by an array factor that depends on spacing d and current phase difference α.
Derivation of field intensity
Assumptions: two isotropic sources 1 and 2 on the x-axis, spacing d, equal amplitude E₀, current in source 2 leads source 1 by α. A far-field point P is at angle θ from the array axis. Take the midpoint O as phase reference.
to far point P
/
/ θ
1 ●--------O--------● 2 -> array axis
<-- d/2 -><- d/2 -->
path difference between 1 and 2 = d cosθ
Phase difference of the two waves at P:
ψ = βd cosθ + α β = 2π/λ
Field of source 1 lags by ψ/2 and field of source 2 leads by ψ/2 relative to O:
E = E₀ e^(-jψ/2) + E₀ e^(+jψ/2)
= 2E₀ cos(ψ/2)
E = 2E₀ cos[(βd cosθ + α)/2]
Normalised array factor: AF(θ) = |cos[(βd cosθ + α)/2]|.
Case 1: equal amplitude and phase (α = 0) — broadside
ψ = βd cosθ, so E = 2E₀ cos[(πd/λ) cosθ].
- Maximum when cosθ = 0, i.e. θ = 90° and 270°: perpendicular to the array axis.
- For d = λ/2: E = 2E₀ cos(π/2 cosθ); nulls at θ = 0° and 180°.
d = λ/2, α = 0 (figure-8, broadside)
90°
.-"""-.
\ /
0° ----1--O--2---- 180° (nulls on axis)
/ \
'-...-'
270°
Case 2: equal amplitude, opposite phase (α = π)
E = 2E₀ cos[(βd cosθ + π)/2] = 2E₀ |sin(πd/λ · cosθ)|. For d = λ/2 the maximum is along the axis (θ = 0°, 180°) and nulls at θ = 90°: a bidirectional end-fire pattern.
Case 3: equal amplitude, phase α = −βd — end fire
ψ = βd(cosθ − 1). Maximum (ψ = 0) at θ = 0°, so the beam is along the array axis in one direction. For d = λ/4, α = −π/2 the pattern is a cardioid:
d = λ/4, α = -90° (cardioid, end fire)
.--.
____/ \___
| 1 2 > max at θ = 0°
‾‾‾‾\ /‾‾‾
'--' null at θ = 180°
Conditions summarised
| Array | Phase condition | Direction of max |
|---|---|---|
| Broadside | α = 0 (ψ = 0 at θ = 90°) | θ = 90°, normal to axis |
| End fire | α = −βd (ψ = 0 at θ = 0°) | θ = 0°, along axis |
| General | βd cosθₘ + α = 0 | cosθₘ = −α/(βd) |
In general the beam points where the path difference is exactly cancelled by the current phase, so changing α steers the beam (the basis of phased arrays).
- 2069 Bhadra · 6 marks
Define antenna arrays and also derive a mathematical expression for the array of two element isotropic radiators.
Answer
An antenna array is an arrangement of two or more similar antennas (elements), placed with definite spacing and fed with definite current amplitude and phase, so that their fields add in wanted directions and cancel in others. Arrays give high directivity, narrow beams and beam steering that a single element cannot give.
Array of two isotropic point sources
Let two isotropic sources lie on the x-axis with spacing d, equal amplitude E₀, and the current of source 2 leading source 1 by phase α. Point P is in the far field at angle θ from the array axis. The midpoint O is the phase reference.
P (far field)
/
/ θ
1 ●-----------O-----------● 2 ---> axis
<--- d/2 ---><--- d/2 --->
extra path from 1 compared to 2 = d cosθ
Total phase difference between the two waves at P:
ψ = βd cosθ + α , β = 2π/λ
With O as reference, source 1 contributes a phase −ψ/2 and source 2 a phase +ψ/2:
E = E₀ e^(-jψ/2) + E₀ e^(jψ/2)
= E₀ [2 cos(ψ/2)]
= 2E₀ cos(ψ/2)
E = 2E₀ cos[(βd cosθ + α)/2], normalised E_n = cos(ψ/2).
Special cases
| Case | Field | Maximum |
|---|---|---|
| α = 0, d = λ/2 | cos(π/2 · cosθ) | θ = 90°, 270° (broadside) |
| α = π, d = λ/2 | sin(π/2 · cosθ) | θ = 0°, 180° (along axis) |
| α = −π/2, d = λ/4 | cos[π/4 (cosθ − 1)] | θ = 0° only (cardioid, end fire) |
Main points:
- The pattern depends only on d/λ and α, not on E₀.
- Maximum occurs where ψ = 0, i.e. cosθₘ = −α/(βd).
- If the sources are not isotropic, the total pattern = element pattern × this array factor (pattern multiplication).
- 2070 Magh · 4+4 marks
Derive a mathematical relation to calculate the field intensity of an array for two element isotropic radiators. Draw the resultant pattern for an End Fire Array with dipoles perpendicular to the axis of array with d = λ/2 and α = π, where λ is the wavelength of the operating frequency.
Answer
For two isotropic sources of equal amplitude, spacing d and current phase difference α, the far field is E = 2E₀ cos(ψ/2) with ψ = βd cosθ + α.
Derivation
Sources 1 and 2 lie on the array axis, d apart; θ is measured from the axis; the midpoint is the phase reference.
P (far)
/
/ θ
1 ●-------O-------● 2 -> axis
path difference = d cosθ
phase difference ψ = βd cosθ + α, β = 2π/λ
E = E₀ e^(-jψ/2) + E₀ e^(jψ/2)
= 2E₀ cos(ψ/2)
= 2E₀ cos[(βd cosθ + α)/2]
Maximum where ψ = 0; broadside for α = 0, end fire for α = ±βd.
Pattern for d = λ/2, α = π, dipoles ⊥ to array axis
Here βd = π, so the array factor is
AF = cos[(π cosθ + π)/2] = |sin(π/2 · cosθ)|
θ = 0° : |sin(π/2)| = 1 (max)
θ = 60° : |sin(π/4)| = 0.707
θ = 90° : sin 0 = 0 (null)
θ = 180° : 1 (max)
By pattern multiplication, total pattern = element pattern × AF.
Plane perpendicular to the dipoles (contains the array axis): a short dipole is omnidirectional here, so the total pattern = AF, a figure-8 along the axis.
Plane containing the dipoles and the array axis: the dipole field is proportional to the sine of the angle from the dipole, which is cosθ here. Total:
E(θ) = |cosθ| · |sin(π/2 · cosθ)|
θ = 0°: 1, 30°: 0.85, 45°: 0.63, 60°: 0.35, 90°: 0
The two lobes along the axis become narrower than the AF alone.
Total pattern, plane of dipoles and axis
90° (null)
|
.--. | .--.
180°( )====1-+-2====( ) 0°
max '--' | '--' max
|
270° (null)
two narrow lobes along the array axis
The result is a bidirectional end-fire pattern: maxima at θ = 0° and 180° along the array axis, nulls broadside (θ = 90°), half-power points of the AF alone at θ = 60° and 120° (|sin(π/4)| = 0.707); the total pattern is narrower after multiplying by |cosθ|.
- 2070 Bhadra · 2+3+3 marks
State the principle of pattern multiplication. Use the principle to obtain a wave pattern for array of two short dipoles for following cases where (d) = dipole separation and (α) = current phase difference. a) Dipoles aligned perpendicular to the array axis with d = λ/2, α = 0 b) Dipoles aligned perpendicular to the array axis with d = λ/2, α = π
Answer
Principle of pattern multiplication
The total field pattern of an array of identical, similarly oriented elements equals the element (unit) pattern multiplied by the array factor (the pattern of the same array made of isotropic point sources). The total phase pattern is the sum of the element phase pattern and the array phase pattern.
E_total(θ,φ) = E_element(θ,φ) × AF(θ,φ)
It lets us sketch complex patterns quickly without full integration.
Setup for both cases: two short dipoles, spacing d = λ/2 (βd = π), both perpendicular to the array axis. θ is measured from the array axis. Two planes are considered:
- Plane ⊥ to dipoles: short dipole pattern is a circle (omnidirectional).
- Plane containing dipoles and axis: short dipole pattern = |sin(angle from dipole)| = |cosθ|.
Array factor of two isotropic sources: AF = |cos[(βd cosθ + α)/2]|.
a) d = λ/2, α = 0
AF = |cos(π/2 · cosθ)| max at θ = 90°, nulls at 0°,180°
Plane ⊥ dipoles : E = AF -> broadside figure-8
Plane of dipoles: E = |cosθ| · |cos(π/2 · cosθ)|
θ=0°: 0 θ=30°: 0.18 θ=45°: 0.31
θ=60°: 0.35 θ=75°: 0.24 θ=90°: 0
In the plane of the dipoles, the AF maximum (θ = 90°) falls on the dipole null, so the result is a four-lobed pattern with lobes near θ ≈ 57°, 123°, 237°, 303°.
element AF (broadside) total (dipole plane)
( ) ___ \ /
--(+)-- x --1-2-- = ->1-2<-
( ) ‾‾‾ / \
max on axis max ⊥ axis 4 lobes
In the plane perpendicular to the dipoles the pattern is the broadside figure-8 (max at θ = 90°).
b) d = λ/2, α = π
AF = |sin(π/2 · cosθ)| max at θ = 0°,180°, null at 90°
Plane ⊥ dipoles : E = AF -> end-fire figure-8 on axis
Plane of dipoles: E = |cosθ| · |sin(π/2 · cosθ)|
θ=0°: 1 θ=30°: 0.85 θ=45°: 0.63
θ=60°: 0.35 θ=75°: 0.10 θ=90°: 0
Here both factors are maximum along the axis, so the total is a two-lobed end-fire pattern, narrower than the AF alone.
element AF (end fire) total
( )
--(+)-- x <=1-2=> = <(1-2)> narrow lobes
( ) along the axis
| Case | Plane ⊥ dipoles | Plane of dipoles |
|---|---|---|
| α = 0 | broadside figure-8 | four lobes |
| α = π | end-fire figure-8 | sharp end-fire two lobes |
- 2080 Asoj · 2+6 marks
Define pattern multiplication. Derive the expression for far field pattern of an array of two isotropic point sources having equal amplitude and quadrature phase with d = λ/4.
Answer
Pattern multiplication
Pattern multiplication states that the field pattern of an array of identical, similarly oriented non-isotropic elements is the product of the pattern of one element and the pattern of the array made of isotropic point sources (array factor). Total phase = element phase + array phase.
E(θ) = E_element(θ) × E_array(θ)
Two isotropic sources, equal amplitude, quadrature phase, d = λ/4
Let source 1 and source 2 lie on the x-axis, spacing d, with the current of source 2 lagging source 1 by 90° (α = −π/2). θ is measured from the axis, midpoint O as reference.
P (far field)
/
/ θ
1 ●-----O-----● 2 -> axis (θ = 0°)
<- d/2 -><- d/2->
path difference = d cosθ
Total phase difference at P:
ψ = βd cosθ + α
βd = (2π/λ)(λ/4) = π/2 , α = -π/2
ψ = (π/2) cosθ - π/2 = (π/2)(cosθ - 1)
Adding the two fields:
E = E₀ e^(-jψ/2) + E₀ e^(jψ/2) = 2E₀ cos(ψ/2)
E = 2E₀ cos[(π/4)(cosθ - 1)]
Normalised: E_n = cos[(π/4)(cosθ - 1)]
Pattern values
| θ | cosθ − 1 | E_n |
|---|---|---|
| 0° | 0 | 1.000 |
| 60° | −0.5 | 0.924 |
| 90° | −1 | 0.707 |
| 120° | −1.5 | 0.383 |
| 180° | −2 | 0 |
Maximum at θ = 0° (towards the lagging source, along the axis); null at θ = 180°. The pattern is a cardioid (unidirectional end fire).
90° (0.707)
.--''--.
.' '.
180° 0 1 2 > 0° (max = 1)
'. .'
'--..--'
270°
Physical reason: towards θ = 0°, the wave from source 1 travels λ/4 (90°) to reach source 2, which exactly cancels source 2's 90° lag, so the fields add. Towards θ = 180°, the travel delay and current lag add to 180°, so the fields cancel. If source 2 leads instead (α = +π/2), the cardioid points to θ = 180°.
- 2073 Bhadra · 1+4 marks
What is a linear array? Differentiate between a broadside array and end fire array.
Answer
A linear array is an array whose elements are placed along a straight line (the array axis), usually equally spaced, equal in amplitude and with a uniform progressive phase shift α between neighbours. Its array factor is
AF = sin(nψ/2) / [n sin(ψ/2)], ψ = βd cosθ + α
Broadside vs end fire array
| Point | Broadside array | End fire array |
|---|---|---|
| Direction of max | Perpendicular to array axis (θ = 90°) | Along array axis (θ = 0° or 180°) |
| Phase condition | α = 0 (all in phase) | α = −βd (progressive phase) |
| Pattern | Bidirectional fan/disc around axis | Usually unidirectional pencil along axis |
| Null-to-null beamwidth | ≈ 2λ/(nd) | ≈ 2√(2λ/(nd)) |
| Directivity (long array) | ≈ 2L/λ | ≈ 4L/λ |
| Typical spacing | λ/2 | λ/4 |
| Example | Collinear/curtain arrays for HF broadcast | Yagi-Uda, end-fire HF arrays |
Broadside: End fire:
^ max
|
o o o o o axis o o o o o ===> max
| along axis
v max
Broadside arrays give a beam at right angles to the line of elements, so they suit point-to-point links where the array can be mounted across the direction of transmission. End fire arrays give higher directivity for the same length and are used where the line of elements can point at the target.
- 2075 Baisakh · 4 marks
Write a short note comparing broadside array and endfire array.
Answer
A broadside array and an end fire array are both uniform linear arrays (equal amplitude, equal spacing d). They differ only in the progressive current phase α, which decides where the beam points. The array factor is AF = sin(nψ/2)/[n sin(ψ/2)], ψ = βd cosθ + α, with maximum where ψ = 0.
Broadside (α = 0) End fire (α = -βd)
↑ beam
● ● ● ● ● ← axis ● ● ● ● ● ⇒ beam
↓ beam along the axis
| Feature | Broadside | End fire |
|---|---|---|
| Phase shift α | 0 (all in phase) | −βd |
| Beam direction | θ = 90°, normal to axis | θ = 0°, along axis |
| Pattern | Bidirectional; disc around axis | Mostly unidirectional |
| Usual spacing | λ/2 | λ/4 |
| FNBW | 2λ/(nd) | 2√(2λ/(nd)) |
| Directivity | ≈ 2L/λ | ≈ 4L/λ (Hansen-Woodyard ≈ 7.2L/λ) |
| Examples | Collinear, curtain arrays | Yagi-Uda, helical (axial mode) |
Key points:
- For the same length L = nd, the end fire beam is broader in angle terms but its directivity is about twice that of broadside.
- Broadside arrays are used for HF broadcasting and fixed links; end fire arrays for TV reception and point-to-point VHF/UHF links.
- 2078 Chaitra · 4 marks
Write the features of end-fire array.
Answer
An end fire array is a linear array whose maximum radiation is along the array axis. It is obtained by feeding equal-amplitude elements with a progressive phase lag equal to their spacing in radians (α = −βd), so the fields add only in the direction of the array line.
● → ● → ● → ● → ● ====> main beam
phase: 0 -βd -2βd -3βd -4βd
Features
- Direction of maximum: θ = 0° (or 180°) along the axis; ψ = βd(cosθ − 1) = 0 at θ = 0°.
- Phase condition: α = −βd; for d = λ/4, α = −90° between neighbours.
- Pattern: usually unidirectional (e.g. two elements at λ/4 give a cardioid); a pencil-shaped beam for many elements.
- Beamwidth: null-to-null beamwidth ≈ 2√(2λ/(nd)) = 2√(2λ/L), wider in angle than broadside for the same length.
- Directivity: D ≈ 4L/λ (= 4nd/λ), about twice that of a broadside array of the same length.
- Hansen–Woodyard condition: extra phase δ = π/n (α = −(βd + π/n)) narrows the beam and raises directivity to about 7.2L/λ (1.8 times ordinary end fire).
- Spacing: kept small (≈ λ/4, < λ/2) to avoid a back lobe and grating lobes.
- Applications: Yagi-Uda TV antennas, axial-mode helix, VHF/UHF point-to-point links.
Questions from Old Question Collection (EX 653) (IOE EX 653 exam papers from 2069 to 2081 (20 papers)). Answers are written for this site; check them against your class notes.
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