Chapter 1 · 5 hours
Radiation and Antenna Fundamentals
IOE past exam questions
Past questions and answers
19 questions set from this chapter, 7 of them more than once. Most asked first.
- Asked 7 times
- 2079 Chaitra · 3 marks
- 2078 Chaitra · 2 marks
- 2077 Chaitra · 2 marks
- 2075 Bhadra · 2 marks
- 2074 Bhadra · 2 marks
- 2073 Magh · 2 marks
- 2072 Magh · 2 marks
State the reciprocity theorem for an antenna system.
Answer
Reciprocity theorem: If an emf V applied at the terminals of antenna 1 produces a current I at the (short-circuited) terminals of antenna 2, then the same emf V applied at the terminals of antenna 2 produces the same current I at the terminals of antenna 1.
Case 1: V --> [Ant 1] ~~~~~~> [Ant 2] --> I21
Case 2: I12 <-- [Ant 1] <~~~~~~ [Ant 2] <-- V
Reciprocity: I12 = I21, i.e. Z12 = Z21
In impedance form: V1/I2 = V2/I1, so the mutual impedances are equal, Z12 = Z21.
Conditions: the medium between the antennas must be linear, passive and isotropic (it fails in a magnetized plasma such as the ionosphere with the earth's magnetic field, or in ferrites).
Consequences for an antenna:
- The radiation pattern is the same for transmitting and receiving.
- Gain, directivity and effective aperture are the same in both modes.
- Input impedance of the antenna is the same as its internal impedance when receiving.
- Antenna patterns can be measured with the antenna under test used as a receiver.
- Asked 6 times
- 2081 Chaitra · 3 marks
- 2080 Chaitra · 3 marks
- 2078 Chaitra · 2 marks
- 2075 Bhadra · 2 marks
- 2074 Bhadra · 2 marks
- 2073 Magh · 2 marks
State the compensation theorem for an antenna system.
Answer
Compensation theorem: If the impedance Z of a branch carrying current I is changed by ΔZ, the changes in currents everywhere in the network are the same as those produced by a voltage source of value −I·ΔZ (opposing the original current) placed in series with the changed branch, with all other sources replaced by their internal impedances.
Original Changed Compensation network
--[ Z ]-- I --[Z+ΔZ]-- I' --[Z+ΔZ]--(−I·ΔZ)--
gives ΔI = I' − I
So the new current is I' = I + ΔI, where ΔI is found from the compensation source alone.
Use in antenna systems:
- To find the change in antenna current and input impedance when a nearby object, parasitic element or ground changes the impedance of a branch.
- To find the effect of changing a load or tuning impedance at the antenna terminals without solving the whole system again.
- In arrays, to study how changing one element's impedance alters currents in the other (mutually coupled) elements.
The theorem holds for linear, bilateral networks, so it applies to antennas in a linear medium.
- Asked 4 times
- 2081 Chaitra · 2+4 marks
- 2079 Chaitra · 2+4 marks
- 2075 Bhadra · 2+4 marks
- 2075 Baisakh · 2+4 marks
Define an antenna. Explain the radiation mechanism and conditions of a single wire antenna that can radiate electromagnetic waves or not.
Answer
An antenna is a metallic structure (wire, rod, horn, dish) that converts guided electromagnetic energy from a transmission line into free-space waves, and vice versa. IEEE defines it as "a means for radiating or receiving radio waves".
Radiation mechanism in a single wire
Radiation is caused by time-varying current, i.e. by accelerated or decelerated charges. For a wire of length l carrying charge density q (C/m) moving with velocity v:
I = q · v
dI/dt = q · dv/dt
l · dI/dt = l · q · dv/dt (basic radiation equation)
So a time-changing current (left side) is the same as accelerated charge (right side). Both mean radiation.
Conditions for radiation from a single wire
- No charge motion: no current, no radiation.
- Charge moving with uniform velocity:
- on a straight, infinitely long, smooth wire – no radiation;
- on a wire that is curved, bent, discontinuous, terminated or truncated – radiation occurs, because the charge changes direction or speed there.
- Charge oscillating in time (time-harmonic current), even on a straight wire – radiation occurs.
uniform v, straight -------> no radiation
bent wire ---->\ radiation at bend
\
end of wire ------>| radiation at end
oscillating current <-> <-> radiation all along
| Condition | Radiates? |
|---|---|
| Static charge | No |
| Uniform velocity, straight infinite wire | No |
| Uniform velocity, bent/ended wire | Yes |
| Time-varying (AC) current | Yes |
So a practical antenna is fed with an alternating current; the charges accelerate continuously, and the wire radiates EM waves. Radiation is strongest for wire lengths comparable to the wavelength (e.g. λ/2 dipole).
- Asked 4 times
- 2080 Chaitra · 3 marks
- 2080 Asoj · 3 marks
- 2077 Chaitra · 2 marks
- 2076 Bhadra · 3 marks
State and explain the maximum power transfer theorem used in antenna, with necessary diagrams.
Answer
Maximum power transfer theorem: A source delivers maximum power to a load when the load impedance is the complex conjugate of the source impedance: Z_L = Z_S*, i.e. R_L = R_S and X_L = −X_S.
Receiving antenna
By Thevenin's theorem the receiving antenna is a voltage source V_oc in series with its impedance Z_A = (R_r + R_L) + jX_A.
Z_A = R_A + jX_A
+----/\/\/----+
| |
(~) V_oc Z_L = R_L' + jX_L' (receiver)
| |
+-------------+
Maximum power to the receiver when Z_L = Z_A*:
P_max = |V_oc|² / (8 R_A) (V_oc peak value)
Under this condition an equal amount of power is dissipated or re-radiated in R_A, so at most half of the captured power reaches the receiver.
Transmitting antenna
The antenna is the load of the transmitter (through a transmission line). For maximum power radiation, the antenna input impedance must be matched to the line/transmitter: Z_A = Z_S*. If they are not matched, part of the power reflects, giving VSWR > 1 and mismatch loss:
Mismatch efficiency = 1 − |Γ|², Γ = (Z_A − Z0)/(Z_A + Z0)
In practice matching networks, baluns and stubs are used to satisfy this theorem.
- Asked 3 times
- 2073 Bhadra · 5 marks
- 2070 Magh · 2+4 marks
- 2069 Bhadra · 4 marks
How does a single wire act as an antenna? During the transmission mode, explain the mechanism involved by which the electric lines of force are detached from the dipole antenna to form free space waves.
Answer
A single wire acts as an antenna when an alternating current flows on it: the charges are continuously accelerated and decelerated, and accelerated charges radiate (l·dI/dt = l·q·dv/dt). A straight wire with steady DC or uniform charge motion does not radiate; a time-varying current, a bend or an open end causes radiation.
Detachment of electric lines of force from a dipole
Consider a small centre-fed dipole driven by a sinusoidal source of period T.
First quarter cycle (0 to T/4):
- Charges flow outward; positive charge builds up on one arm and negative on the other.
- Electric field lines form between them and spread outward, reaching a distance of about λ/4 by t = T/4.
Second quarter cycle (T/4 to T/2):
- The current reverses; the charges move back and neutralize.
- The old field lines, which have already travelled out, lose their ends on the conductor.
- Since there are no charges left to end on, the lines join and close on themselves, forming closed loops.
- At the same time, new lines of opposite direction start forming near the dipole.
Next half cycle: the same process repeats with reversed polarity. The new lines push the closed loops away.
t = T/4 t = T/2 later
+ ___ ___ _ _
| / \ / \ ( ) ( )
|| | --> | ( ) | --> loops travel
| \___/ \_____/ outward at c
- (lines close)
The detached closed loops of E field, with an associated H field, travel outward at the speed of light as free-space waves.
The transmission line guides energy as a TEM wave; at the open end (the dipole arms) the field lines spread out, detach and propagate. The detached fields are the radiation field; fields that stay near the antenna and return energy each cycle are the reactive (induction) field.
- Asked 3 times
- 2080 Asoj · 3 marks
- 2076 Bhadra · 3 marks
- 2072 Magh · 2 marks
State and explain the superposition theorem used in antenna, with necessary diagrams.
Answer
Superposition theorem: In a linear system with several sources, the total response (current, voltage or field) at any point is the vector (phasor) sum of the responses due to each source acting alone, with the other sources replaced by their internal impedances.
In antennas
Because free space is linear, the total field at a far point P from several radiating elements is
E_total = E1 + E2 + ... + En (phasor sum)
each term having its own amplitude and phase (due to path difference and current phase).
P (far point)
/|\
/ | \
E1 / |E2\ E3
/ | \
[1] [2] [3] elements of an array
|<-d->|<-d->|
Example – two isotropic sources spaced d with equal currents in phase:
E = E0 e^(jψ/2) + E0 e^(−jψ/2) = 2E0 cos(ψ/2)
ψ = k d cosθ
Uses:
- Array theory: array pattern = element pattern × array factor (pattern multiplication follows from superposition).
- Finding the field of a long wire or dipole by adding the fields of many infinitesimal current elements (integration).
- Finding the effect of ground using an image antenna.
- Circuit analysis of antenna feed networks with several sources.
It is valid only for linear media and linear circuit elements.
- Asked 2 times
- 2080 Chaitra · 2+6 marks
- 2070 Bhadra · 2+8 marks
Define an antenna. Explain the operation of infinitesimal dipole with the help of mathematical relations and the field pattern.
Answer
An antenna is a transducer that converts guided electromagnetic energy on a transmission line into free-space radiated waves (and vice versa).
Infinitesimal (Hertzian) dipole
An infinitesimal dipole is a very short straight wire of length l ≪ λ (l ≤ λ/50) carrying a uniform current I(z) = I0 along its length, placed at the origin along the z-axis. Real antennas can be built as a sum of such elements.
z
| P(r, θ, φ)
| /
┌┼┐/ r
l │I0│ θ
└┼┘
|
+----------- y
/
x
Vector potential
A = âz μ I0 l e^(−jkr) / (4π r)
Ar = Az cosθ, Aθ = −Az sinθ
Field components (H = ∇×A/μ, E from Maxwell)
Hφ = j k I0 l sinθ e^(−jkr)/(4πr) · [1 + 1/(jkr)]
Er = η I0 l cosθ e^(−jkr)/(2πr²) · [1 + 1/(jkr)]
Eθ = jηkI0 l sinθ e^(−jkr)/(4πr)
· [1 + 1/(jkr) − 1/(kr)²]
Hr = Hθ = Eφ = 0
Field regions:
- kr ≪ 1 (near/reactive field): 1/r² and 1/r³ terms dominate; energy is stored, not radiated.
- kr ≫ 1 (far field): only 1/r terms remain:
Eθ ≈ jη k I0 l sinθ e^(−jkr) / (4πr)
Hφ ≈ j k I0 l sinθ e^(−jkr) / (4πr)
Er ≈ 0, Eθ / Hφ = η = 120π Ω
The far fields are transverse, in phase, and form a TEM wave.
Power, radiation resistance and directivity
P_rad = η (π/3) |I0 l / λ|²
Rr = 2 P_rad / |I0|² = 80π² (l/λ)² Ω
U(θ) ∝ sin²θ → D = 1.5 (1.76 dBi), HPBW = 90°
Example: l = λ/50 → Rr = 80π²/2500 ≈ 0.316 Ω (very small, so efficiency is low).
Field pattern
The far-field pattern is |Eθ| ∝ sinθ:
- E-plane (vertical plane): figure-of-eight, maximum at θ = 90°, nulls along the axis (θ = 0°, 180°).
- H-plane (horizontal plane): circle – omnidirectional.
- 3-D: doughnut shape.
E-plane (sinθ) H-plane
z y
| .---.
.-. | .-. / \
( ) | ( ) | o | x
'-' | '-' \ /
| '---'
null on axis circle
- 2078 Chaitra · 2+3+3 marks
Define antenna. Explain the radiation mechanism in single wire antenna. Explain travelling wave long wire antenna with required diagrams.
Answer
An antenna is a device that converts guided EM energy from a transmission line into radiated free-space waves, and vice versa.
Radiation mechanism in a single wire
Radiation comes from accelerated charges, i.e. time-varying current:
l · dI/dt = l · q · dv/dt
- No charge motion, or uniform motion along a straight infinite wire → no radiation.
- Uniform motion along a bent, curved, discontinuous or terminated wire → radiation at those points.
- Oscillating (AC) current → radiation along the whole wire.
So an antenna is fed with RF alternating current; the oscillating charges radiate. The field lines formed between the charges detach every half cycle and travel outward as free-space waves.
Travelling wave long wire antenna
A long wire (L of several wavelengths) a height h above ground, fed at one end and terminated at the far end in a resistor equal to its characteristic impedance (about 400–600 Ω). Because the end is matched, there is no reflection and only a forward travelling wave exists: I(z) = I0 e^(−jkz).
feed R = Z0 (matched)
| ------> travelling wave |
(~)===========================[R]
| |
////////////// ground //////////////
|<-------- L -------->|
Pattern (θ from the wire axis):
E(θ) ∝ sinθ · sin[(kL/2)(1 − cosθ)] / (1 − cosθ)
Main lobe angle θm ≈ cos⁻¹(1 − 0.371 λ/L)
Example: L = 5λ → θm ≈ cos⁻¹(1 − 0.0742) ≈ 22.2° from the wire.
main lobes (cone about wire)
\ /
\ / ---> towards load end
feed o------\---/-------[R]
/ \
/ \
Characteristics:
- Unidirectional: radiation is in a cone towards the terminated end; no backward lobe.
- Non-resonant and broadband, since no standing wave; input impedance ≈ Z0 (resistive).
- Gain increases and the main lobe moves closer to the wire as L increases.
- Efficiency is reduced because part of the power is lost in the terminating resistor (about 30–50 %).
- Example: Beverage antenna for LF/MF reception; two such wires form V and rhombic antennas.
Compared with a resonant (standing-wave) long wire, which has lobes in both directions, the travelling-wave antenna has a single-direction pattern and wider bandwidth.
- 2072 Asoj · 4 marks
Explain how single wire can radiate electromagnetic waves with necessary radiation patterns and polarization. List the characteristics of omnidirectional antenna.
Answer
A single wire radiates when its charges are accelerated, i.e. when it carries a time-varying (AC) current: l·dI/dt = l·q·dv/dt. A static charge or a charge moving at uniform speed on a straight infinite wire does not radiate; bends, ends and oscillating current produce radiation. Field lines formed between the oscillating charges detach every half cycle and travel out as EM waves.
Pattern and polarization of a vertical wire (dipole)
E-plane (vertical) H-plane (horizontal)
z
| .---.
.-. | .-. / \
( ) | ( ) | (o) |
'-' | '-' \ /
| '---'
figure-of-eight circle (omni)
- E ∝ sinθ: maximum broadside (θ = 90°), null along the wire.
- E-field lies along the wire, so a vertical wire gives vertical (linear) polarization; a horizontal wire gives horizontal polarization.
Characteristics of an omnidirectional antenna
- Radiates uniformly in all directions of one plane (usually the horizontal/azimuth plane).
- Directional in the orthogonal (elevation) plane; 3-D pattern is a doughnut.
- Has nulls along the axis of the antenna.
- Moderate gain, e.g. λ/2 dipole 2.15 dBi, λ/4 monopole 5.15 dBi.
- Usually vertically polarized.
- Examples: vertical dipole, monopole/whip, discone, normal-mode helix.
- Used in broadcasting, mobile phones, Wi-Fi access points, base stations where users are in all directions.
- 2076 Bhadra · 2+4 marks
What is the retarded potential? Explain how electromagnetic waves are generated by a conductor.
Answer
Retarded potential: The scalar and vector potentials at a point at distance R from a time-varying source, calculated using the value of the source at an earlier time t − R/v, because the effect of the source travels at a finite speed v (= c in free space). The delay R/v is called the retardation.
V(t) = (1/4πε) ∫ ρ(t − R/v) / R dv
A(t) = (μ/4π) ∫ J(t − R/v) / R dv
E and H are then found from B = ∇×A and E = −∇V − ∂A/∂t.
Generation of EM waves by a conductor
- An alternating source drives current along a conductor (e.g. a dipole). Charges move back and forth, so they are continuously accelerated – the basic condition for radiation: l·dI/dt = l·q·dv/dt.
- In the first quarter cycle, charges collect at the ends and electric field lines form between the two arms; the moving charges (current) produce magnetic field circles around the wire.
- In the next quarter cycle the current reverses and charges neutralize. The field lines already sent out lose their charges, so they close on themselves into loops and detach.
- New lines of opposite direction form and push the loops outward. The changing E creates H and the changing H creates E (Maxwell's equations), so the disturbance travels at speed c as an electromagnetic wave.
source near field far field
(~)--||-- --> ( ( ) ) --> ( ) ( ) -->
dipole lines form loops detach
Close to the conductor the reactive (induction) field stores energy; far away the 1/r radiation field carries power away (E ⊥ H ⊥ direction of travel, E/H = 120π Ω).
- 2080 Asoj · 2+5 marks
What is the retarded time? Explain the retarded vector and scalar potential with mathematical expression.
Answer
Retarded time is the earlier time at which a signal must leave the source to reach the observation point at the present time t:
t' = t − R/v
where R = distance from source to point and v = speed of propagation (c in free space). The field at P now depends on what the source was doing at t'.
Retarded potentials
For static sources:
V = (1/4πε) ∫ ρ / R dv, A = (μ/4π) ∫ J / R dv
For time-varying sources, ρ and J are taken at the retarded time:
V(r, t) = (1/4πε) ∫ [ρ(r', t − R/v)] / R dv'
A(r, t) = (μ/4π) ∫ [J(r', t − R/v)] / R dv'
These are the retarded scalar potential and retarded vector potential. They satisfy the wave equations:
∇²V − με ∂²V/∂t² = −ρ/ε
∇²A − με ∂²A/∂t² = −μJ
with the Lorenz gauge ∇·A = −με ∂V/∂t.
Phasor (time-harmonic) form, with e^(jωt) and k = ω/v = 2π/λ:
V = (1/4πε) ∫ ρ e^(−jkR) / R dv'
A = (μ/4π) ∫ J e^(−jkR) / R dv'
The factor e^(−jkR) is the phase delay of retardation.
For a line current I (wire antenna): A = (μ/4π) ∫ I e^(−jkR)/R dl.
Fields are then found from
B = ∇ × A
E = −∇V − ∂A/∂t (= −∇V − jωA)
Importance: retarded potentials explain why time-varying currents produce fields that travel outward with finite speed and a 1/r part – the radiation field. All antenna fields (dipole, loop, arrays) are derived from them.
- 2072 Asoj · 2+4 marks
What is retarded potential? Write the expression of retarded scalar potential and vector potential of infinitesimal dipole antenna in Fraunhofer far field region.
Answer
Retarded potential is the potential (scalar V or vector A) at a point computed from the source value at the earlier time t − R/v, since the effect of a time-varying source takes R/v seconds to arrive. In phasor form the delay appears as a factor e^(−jkR).
V = (1/4πε) ∫ ρ e^(−jkR)/R dv
A = (μ/4π) ∫ J e^(−jkR)/R dv
Infinitesimal dipole
Length l ≪ λ along the z-axis, uniform current I0, end charges ±q with I0 = jωq.
Retarded vector potential (exact for small l):
Az = μ I0 l e^(−jkr) / (4π r)
Ar = Az cosθ = μ I0 l cosθ e^(−jkr)/(4πr)
Aθ = −Az sinθ = −μ I0 l sinθ e^(−jkr)/(4πr)
Retarded scalar potential (from the two end charges, or from the Lorenz gauge):
V = (q l cosθ / 4πε r²) (1 + jkr) e^(−jkr)
= (I0 l cosθ / 4πε) [1/(c r) + 1/(jω r²)] e^(−jkr)
In the Fraunhofer (far-field) region
For r ≫ λ (kr ≫ 1, r > 2D²/λ) only the 1/r terms remain:
Az ≈ μ I0 l e^(−jkr) / (4π r)
Aθ ≈ −μ I0 l sinθ e^(−jkr) / (4π r)
V ≈ I0 l cosθ e^(−jkr) / (4π ε c r)
= η I0 l cosθ e^(−jkr) / (4π r)
(using 1/(εc) = η = 120π Ω).
From these the far fields are
Eθ ≈ −jω Aθ = jη k I0 l sinθ e^(−jkr)/(4πr)
Hφ ≈ Eθ / η
which give the sinθ (doughnut) pattern of the dipole.
- 2074 Bhadra · 2+6 marks
Explain retarded potential and their importance. Describe infinitesimal dipole with the help of suitable diagram, mathematical relations and the field pattern.
Answer
Retarded potential is the scalar or vector potential at a point P computed using the source value at the earlier (retarded) time t − R/v, because changes in a source take R/v seconds to reach P.
V = (1/4πε) ∫ ρ(t − R/v)/R dv
A = (μ/4π) ∫ J(t − R/v)/R dv
Phasor form: ρ, J times e^(−jkR)
Importance
- Gives a simple way to find the fields of any current distribution: find A (and V), then H = ∇×A/μ, E = −jωA − ∇V.
- Explains finite propagation speed and the phase delay e^(−jkr) of radiated fields.
- Produces the 1/r radiation field term; static potentials give only 1/r² fields, which carry no net power.
- Basis for analysing dipoles, loops, arrays and all wire antennas.
Infinitesimal dipole
A very short wire, l ≪ λ (≤ λ/50), on the z-axis with uniform current I0.
z
| P(r,θ,φ)
┌┼┐ /
l │I0│/ θ
└┼┘----- y
|
Vector potential: Az = μ I0 l e^(−jkr)/(4πr).
Fields:
Hφ = j k I0 l sinθ e^(−jkr)/(4πr) [1 + 1/(jkr)]
Er = η I0 l cosθ e^(−jkr)/(2πr²) [1 + 1/(jkr)]
Eθ = jη k I0 l sinθ e^(−jkr)/(4πr) [1 + 1/(jkr) − 1/(kr)²]
Far field (kr ≫ 1):
Eθ ≈ jη k I0 l sinθ e^(−jkr)/(4πr), Hφ = Eθ/η
Radiation parameters:
P_rad = η(π/3)|I0 l/λ|²
Rr = 80π²(l/λ)² Ω
D = 1.5 (1.76 dBi), HPBW = 90°
Field pattern
|Eθ| ∝ sinθ.
E-plane (vertical) H-plane (horizontal)
z .---.
.-. | .-. / \
( ) | ( ) | o |
'-' | '-' \ /
| '---'
- Maximum at θ = 90°, nulls along the dipole axis.
- Omnidirectional in the H-plane; 3-D pattern is a doughnut.
- 2071 Magh · 4+4 marks
Describe the operation of infinitesimal dipole with necessary mathematical relation for electric field and magnetic field. What are basic antenna parameters? Explain briefly on any four parameters.
Answer
Infinitesimal dipole
An infinitesimal (Hertzian) dipole is a very short conductor, l ≪ λ (≤ λ/50), placed along the z-axis with a uniform current I0 along its length.
Retarded vector potential:
Az = μ I0 l e^(−jkr) / (4π r)
Magnetic field (H = ∇×A/μ):
Hφ = j k I0 l sinθ e^(−jkr)/(4πr) · [1 + 1/(jkr)]
Hr = Hθ = 0
Electric field (from ∇×H = jωεE):
Er = η I0 l cosθ e^(−jkr)/(2πr²) · [1 + 1/(jkr)]
Eθ = jηkI0 l sinθ e^(−jkr)/(4πr)
· [1 + 1/(jkr) − 1/(kr)²]
Eφ = 0
Far field (kr ≫ 1):
Eθ ≈ jη k I0 l sinθ e^(−jkr)/(4πr)
Hφ ≈ Eθ / η, η = 120π Ω, Er ≈ 0
The pattern is ∝ sinθ (doughnut), D = 1.5, Rr = 80π²(l/λ)² Ω.
Basic antenna parameters
Radiation pattern, directivity, gain, efficiency, beamwidth, bandwidth, input impedance, radiation resistance, polarization, effective aperture.
- Radiation pattern: graph of the radiated field or power versus direction (θ, φ) in the far field. It shows main lobe, side lobes, back lobe and nulls.
- Directivity (D): ratio of the maximum radiation intensity to the average radiation intensity:
D = U_max / U_avg = 4π U_max / P_rad
Isotropic: D = 1; λ/2 dipole: D = 1.64. 3. Gain (G): directivity including losses: G = η_e · D, where η_e = radiation efficiency. Expressed in dBi. 4. Radiation efficiency:
η_e = P_rad / P_in = Rr / (Rr + RL)
RL = ohmic loss resistance.
Other parameters: HPBW – angle between half-power points of the main lobe; bandwidth – frequency range where parameters stay within limits; input impedance Z_in = R_in + jX_in at the terminals; polarization – orientation of the E-field vector with time.
- 2076 Baisakh · 4 marks
Sketch the current distribution and radiation pattern of center-fed vertical dipole for following lengths: i) λ/2 ii) 3λ/2
Answer
For a centre-fed thin dipole of length L, the current is a standing wave that is zero at the ends:
I(z) = I0 sin[k(L/2 − |z|)], k = 2π/λ
E(θ) ∝ [cos((kL/2)cosθ) − cos(kL/2)] / sinθ
i) L = λ/2
Current E-plane pattern
. z
|\ |
| \ max at .-. | .-.
~ | > centre ( )|( )
| / '-' | '-'
|/ |
' max at θ = 90°
- Current: half sine loop, maximum at the feed, zero at both ends; all in phase.
- Pattern: E(θ) = cos[(π/2)cosθ]/sinθ – figure-of-eight, slightly narrower than sinθ.
- HPBW ≈ 78°, D = 1.64 (2.15 dBi), R_in ≈ 73 Ω.
ii) L = 3λ/2
Current (3 half loops) E-plane pattern
| + | z
| - | centre loop \ | /
| + | opposite phase ( )|( )
ends: zero ( )-+-( )
( )|( )
/ | \
- Current: three half-wave loops; the centre loop is in opposite phase to the two outer loops; maximum (in magnitude) at the feed.
- Pattern: 6 lobes in the vertical plane. Major lobes at θ ≈ 42.6° and 137.4° from the axis, a smaller lobe broadside at 90° (≈ 2.9 dB below), nulls at θ ≈ 70.5° and 109.5°.
- D ≈ 2.23 (3.48 dBi), R_in ≈ 105 Ω.
- 2072 Magh · 6 marks
Sketch the current distribution and radiation pattern of center-fed vertical dipole for following lengths: i) λ ii) 3λ/2 iii) 2λ
Answer
For a centre-fed thin dipole of length L, the current is a standing wave that is zero at the open ends:
I(z) = I0 sin[k(L/2 − |z|)]
E(θ) ∝ [cos((kL/2)cosθ) − cos(kL/2)] / sinθ
(θ measured from the dipole axis; patterns are omnidirectional in the horizontal plane.)
i) L = λ
Current E-plane pattern
/\ z
/ \ upper half-loop |
\ / feed: I ≈ 0 .--. | .--.
\/ lower half-loop( )|( )
(both in phase) '--' | '--'
- Two half-wave loops, in phase, with current minimum (zero) at the feed point, so input impedance is very high.
- E(θ) = [cos(π cosθ) + 1]/sinθ; single broadside lobe each side, narrower than λ/2.
- HPBW ≈ 47.8°, D ≈ 2.41 (3.82 dBi).
ii) L = 3λ/2
Current: + | − | + (3 half-loops,
centre loop opposite phase)
Pattern: \ | /
( ) | ( ) major lobes
( )---+---( ) at 42.6°, 137.4°
( ) | ( ) minor lobe at 90°
/ | \
- Three half-loops; centre loop opposite to the outer ones; current maximum at the feed (R_in ≈ 105 Ω).
- 6 lobes: major lobes at θ ≈ 42.6° and 137.4°, smaller broadside lobe at 90° (≈ −2.9 dB), nulls at ≈ 70.5° and 109.5°.
- D ≈ 2.23 (3.48 dBi).
iii) L = 2λ
Current: + − | − + → 2 loops each arm;
adjacent loops opposite phase,
zero current at feed
Pattern: \ | /
( ) | ( ) 4 lobes at
---------+--------- 57.4°, 122.6°
( ) | ( ) null at 90°
/ | \
- Four half-loops; the two loops on each arm are in opposite phase; current zero at the feed.
- E(θ) = [cos(2π cosθ) − 1]/sinθ → null broadside (θ = 90°) and along the axis.
- 4 lobes (cloverleaf), maxima at θ ≈ 57.4° and 122.6°; D ≈ 2.53 (4.03 dBi).
As L increases beyond λ, phase reversals in the current split the pattern into more lobes.
- 2070 Magh · 6 marks
Differentiate between λ/2, λ and 3λ/2 length dipoles in terms of their individual radiation pattern, self impedance and directivity; where λ is the wavelength of operation frequency.
Answer
For a centre-fed dipole of length L with sinusoidal current I(z) = I0 sin[k(L/2 − |z|)], the far-field pattern is
E(θ) ∝ [cos((kL/2)cosθ) − cos(kL/2)] / sinθ
The values below are computed from this formula (R_r referred to current maximum).
| Property | λ/2 dipole | λ dipole | 3λ/2 dipole |
|---|---|---|---|
| Current | One half-loop, max at feed | Two in-phase loops, zero at feed | Three loops, centre loop reversed |
| Pattern formula | cos(π/2·cosθ)/sinθ | [cos(π cosθ)+1]/sinθ | cos(3π/2·cosθ)/sinθ |
| Lobes (E-plane) | 2 (broadside) | 2 (broadside, narrower) | 6 (4 major at 42.6°, 2 minor at 90°) |
| Max direction | θ = 90° | θ = 90° | θ ≈ 42.6°, 137.4° |
| HPBW | ≈ 78° | ≈ 47.8° | ≈ 29° (major lobe) |
| Radiation resistance | ≈ 73 Ω | ≈ 199 Ω | ≈ 105 Ω |
| Input impedance at centre | ≈ 73 + j42.5 Ω (resonant when slightly shortened) | Very high (kΩ), current minimum | ≈ 105 Ω |
| Directivity | 1.64 (2.15 dBi) | 2.41 (3.82 dBi) | 2.23 (3.48 dBi) |
λ/2 λ 3λ/2
.-. | .-. .--.|.--. \ | /
( )|( ) ( | ) ( ) | ( )
'-' | '-' '--'|'--' ( )--+--( )
( ) | ( )
/ | \
Key points:
- The λ/2 dipole is resonant, easy to match to 75 Ω coax, and is the reference antenna.
- The λ dipole has higher directivity and narrower beam, but its very high input impedance makes it hard to feed at the centre (often end-fed instead).
- The 3λ/2 dipole has phase reversal in its current; the main beam splits into tilted cones, so broadside gain falls and directivity is lower than for λ.
- 2069 Bhadra · 6 marks
Explain the characteristics of λ/2, λ and 1.28λ length dipoles where λ is the wavelength of operating frequency.
Answer
For a centre-fed dipole of length L (current I(z) = I0 sin[k(L/2 − |z|)]), the pattern is
E(θ) ∝ [cos((kL/2)cosθ) − cos(kL/2)] / sinθ
λ/2 dipole
- Current: single half-sine loop, maximum at the feed, zero at the ends.
- Pattern: cos[(π/2)cosθ]/sinθ, figure-of-eight, maximum broadside; HPBW ≈ 78°.
- Directivity 1.64 (2.15 dBi); R_r ≈ 73 Ω, Z_in ≈ 73 + j42.5 Ω (resonant when about 5 % shorter).
- Easy to match; used as the reference antenna and as driven element in Yagi arrays.
λ (full-wave) dipole
- Current: two half-loops in phase; current zero at the feed.
- Pattern: [cos(π cosθ) + 1]/sinθ, single broadside lobe, narrower: HPBW ≈ 47.8°.
- Directivity ≈ 2.41 (3.82 dBi); R_r ≈ 199 Ω (at current maximum).
- Very high input impedance at the centre, so it is difficult to feed there.
1.28λ dipole (near 1.25λ, "extended double Zepp")
- Current: each arm is longer than λ/2, so a small part near each end carries current of opposite phase.
- The opposite-phase parts are short, so they only slightly reduce the broadside field; the long in-phase parts make the main beam very narrow.
- Calculated with the formula: main lobe at θ = 90°, HPBW ≈ 31°, minor lobes at about 33° from the axis, ≈ 8.5 dB down.
- Directivity ≈ 3.29 (5.17 dBi) – about the maximum broadside directivity a single centre-fed dipole can have (1.25λ gives 3.28).
- Beyond about 1.28λ the opposite-phase current grows, the minor lobes become major lobes and broadside gain drops (3λ/2 gives only 2.23).
λ/2 λ 1.28λ
.-. | .-. .--.|.--. \ .---|---. /
( )|( ) ( | ) ( ( | ) )
'-' | '-' '--'|'--' / '---|---' \
HPBW 78° 47.8° 31° + small side lobes
| Length | D (dBi) | HPBW | Side lobes |
|---|---|---|---|
| λ/2 | 2.15 | 78° | None |
| λ | 3.82 | 47.8° | None |
| 1.28λ | 5.17 | ≈ 31° | Small, ≈ −8.5 dB |
- 2079 Chaitra · 3 marks
State the Thevenin theorem for an antenna.
Answer
Thevenin theorem: Any linear two-terminal network containing sources and impedances can be replaced by a single voltage source V_th (the open-circuit voltage at the terminals) in series with an impedance Z_th (the impedance seen into the terminals with all sources replaced by their internal impedances).
Applied to a receiving antenna
An incident wave induces an emf on the antenna. Seen from its terminals, the antenna is a voltage source V_oc (open-circuit induced voltage) in series with the antenna impedance
Z_A = (R_r + R_L) + j X_A
where R_r = radiation resistance, R_L = loss resistance, X_A = antenna reactance.
Z_A = R_r + R_L + jX_A
+-----/\/\/\-----+
| |
(~) V_oc [Z_L] receiver
| |
+----------------+
Current in the receiver load:
I = V_oc / (Z_A + Z_L)
With conjugate matching (Z_L = Z_A*), maximum power is delivered: P = |V_oc|²/(8 R_A) (peak V_oc).
Use: finding received power, effective aperture and mismatch loss of a receiving antenna system.
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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