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Comprehension

An antenna is a key component of a wireless link which efficiently couples electromagnetic energy from the transmitter to free space and from free space to the receiver. An antenna is generally a bidirectional device, i.e, the power through the antenna can flow in both the directions, hence it works as a transmitting as well as a receiving antenna. An antenna acts as an interface between the radiated electromagnetic waves and the guided waves. It can be thought of as a mode transformer which transforms a guided wave field distribution into a radiated-wave field distribution.


Question 1
The correct answer is

Radiation pattern

The standard term is the radiation pattern. It is defined as a graphical representation of the antenna's radiation properties as a function of direction — that is, how the radiated power is distributed over the angular coordinates θ and φ at a fixed large distance. That is exactly what the question describes, so the answer is option 3.

Why option 2 is the near miss. "Power pattern" is a genuine term, but it is a subdivision of the radiation pattern rather than the general name. A radiation pattern may be plotted in several forms:

FormQuantity plotted
Field pattern\(|E(\theta,\phi)|\)
Power pattern\(|E(\theta,\phi)|^{2}\), usually in dB
Phase patternthe phase of the field

The question asks for what the distribution is "generally known as", and the general, umbrella term is radiation pattern. Note also that a 3 dB drop in the power pattern corresponds to a 0.707 drop in the field pattern — the same angular width, read on different scales.

Why the other two fail. "Angular pattern" is not standard terminology at all. An antenna array is a physical arrangement of several radiating elements, not a description of the radiated power — though it is one way of shaping a pattern.

The features a radiation pattern displays.

The main lobe contains the direction of maximum radiation; side lobes are the smaller maxima either side of it; the back lobe points opposite the main beam; and nulls are the directions of zero radiation.

The numbers read off it are the half-power beamwidth (HPBW), the beamwidth between first nulls (BWFN), the side-lobe level and the front-to-back ratio.

Where the pattern must be measured. Only in the far field, beyond

\(R \gt \dfrac{2D^{2}}{\lambda}\)

because closer in the relative phases of the contributions from different parts of the aperture are still changing with distance and the pattern has not settled into its final shape.

Reciprocity, which the passage itself notes. Because an antenna is bidirectional, its receiving pattern is identical to its transmitting pattern — so a single measurement serves both roles.

Hence, the angular distribution of transmitted power is the radiation pattern.

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Question 2
The correct answer is

300 MHz - 3 GHz

UHF runs from 300 MHz to 3 GHz — option 3.

The ITU band structure, which makes every answer derivable rather than memorised. Each band spans exactly one decade, from \(3\times10^{n}\) to \(3\times10^{n+1}\) Hz:

BandNameFrequencyWavelength
VLFVery Low3–30 kHz100–10 km
LFLow30–300 kHz10–1 km
MFMedium300 kHz–3 MHz1000–100 m
HFHigh3–30 MHz100–10 m
VHFVery High30–300 MHz10–1 m
UHFUltra High300 MHz–3 GHz1 m–10 cm
SHFSuper High3–30 GHz10–1 cm
EHFExtremely High30–300 GHz10–1 mm

Counting up from HF — High, Very High, Ultra High — each step multiplies by ten, which places UHF two decades above HF's 3–30 MHz. Option 1 is LF, option 2 is HF and option 4 is EHF, so the distractors are simply neighbouring bands.

Why the wavelength matters here. UHF wavelengths run from 1 m down to 10 cm, so a half-wave dipole is between 50 cm and 5 cm long — small enough to fit inside a handset. That is precisely why mobile telephony, Wi-Fi, Bluetooth, GPS and television all sit in this band, and it connects directly to the passage's point that antenna size is set by wavelength.

How UHF propagates. It is essentially line-of-sight. The ionosphere no longer reflects at these frequencies — the sky-wave propagation that HF relies on stops working — so coverage is limited to the radio horizon,

\(d\approx4.12\left(\sqrt{h_t}+\sqrt{h_r}\right)\ \text{km}\)

with the heights in metres. This is why UHF services need many cell sites or a satellite rather than one distant transmitter.

The trade-off that makes UHF popular. Higher frequency buys wide bandwidth and small antennas at the cost of range and of penetration through buildings and foliage — a balance that suits short-range, high-capacity services exactly.

Hence, UHF is 300 MHz to 3 GHz.

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Question 3
The correct answer is

Hertzian dipole

The Hertzian dipole is the elementary building block of antenna theory — option 1.

What it is. An infinitesimally short current element of length \(dl\ll\lambda\) carrying a uniform current I along its whole length. It is an idealisation, not a buildable antenna: no real element can carry uniform current right to its open ends, where the current must fall to zero.

Why "most basic" belongs to it. Because the current is uniform, its fields can be integrated in closed form:

\(E_\theta=\dfrac{j\eta_0 I\,dl\,\sin\theta}{2\lambda r}e^{-jkr}\)

and every other antenna is then obtained by summing Hertzian dipoles along its length with the appropriate current distribution. The half-wave dipole, for instance, is the integral of elementary dipoles carrying \(I(z)=I_0\cos kz\). So the Hertzian dipole plays the same role in antenna theory that the impulse plays in signals: the elementary response from which all others are built by superposition.

Why the other three are not "basic".

AntennaCurrent distributionDirectivityRrad
Hertzian dipoleuniform, \(dl\ll\lambda\)1.5\(80\pi^{2}(dl/\lambda)^{2}\)
Short dipoletriangular, tapering to zero1.5\(20\pi^{2}(l/\lambda)^{2}\)
Half-wave dipolesinusoidal1.6473 Ω
Monopolehalf a dipole plus its ground image3.2836.5 Ω

The short dipole is more realistic but its triangular current distribution is already an integration of Hertzian elements — note that its radiation resistance is exactly a quarter of the Hertzian value for the same length, because the average current is half as large. The half-wave dipole is the most basic practical antenna and the usual gain reference, and the monopole is derived from it by image theory.

The distinction the question is testing is between "most basic" in the theoretical sense — the elementary radiator — and "most basic" in the practical sense of what one would actually build. The passage's framing of an antenna as a mode transformer is a theoretical one, and the theoretical primitive is the Hertzian dipole.

Its pattern. The \(\sin\theta\) dependence gives a doughnut: maximum broadside, nulls along the axis, and a half-power beamwidth of 90°. That is the pattern every simple wire antenna approaches as it is made shorter.

Hence, the most basic antenna element is the Hertzian dipole.

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Question 4
The correct answer is

99%

Radiation efficiency is the fraction of the input power that actually leaves the antenna as radiation rather than being lost as heat:

\(\eta=\dfrac{P_{rad}}{P_{in}}=\dfrac{P_{in}-P_{loss}}{P_{in}}\)

Step 1 — find the radiated power. Of the 100 W supplied, 1 W is dissipated, so

\(P_{rad}=100-1=99\ \text{W}\)

Step 2 — form the ratio.

\(\eta=\dfrac{99}{100}=0.99=\mathbf{99\%}\)

which is option 3.

Where the distractors come from. Option 2, 0.99 %, is the answer obtained by taking the loss fraction 1/100 and mislabelling it — it is in fact the loss, not the efficiency. Option 1 slips a decimal place. Option 4 would require zero dissipation, which no real conductor can achieve.

The resistance form, which is how efficiency is actually designed for. The input resistance of an antenna splits into a radiating part and a lossy part:

\(\eta=\dfrac{R_{rad}}{R_{rad}+R_{loss}}\)

Here \(R_{loss}/R_{rad}=1/99\). Radiation resistance is not a physical resistor — it is the equivalent resistance that would absorb the power the antenna radiates — whereas Rloss is genuine ohmic resistance in the conductors, plus dielectric and ground losses.

Why efficiency collapses for electrically small antennas. Radiation resistance falls as the square of length in wavelengths:

\(R_{rad}=80\pi^{2}\left(\dfrac{dl}{\lambda}\right)^{2}\)

so a very short antenna may have \(R_{rad}\) of a fraction of an ohm while its conductor and ground losses stay at several ohms — efficiencies of a few per cent are common on portable low-frequency antennas. A half-wave dipole, with 73 Ω of radiation resistance against a fraction of an ohm of loss, easily achieves the 99 % of this question.

How efficiency links gain to directivity.

\(G=\eta D\)

Directivity depends only on the shape of the pattern; gain is what the antenna actually delivers. At 99 % efficiency the two differ by less than 0.05 dB, which is why for a good antenna the terms are often used interchangeably — but for a short or lossy one the distinction matters a great deal.

Hence, the radiation efficiency is 99 %.

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Question 5
The correct answer is

Helical

The helical antenna in axial mode radiates a circularly polarised wave — option 4.

Why the geometry produces circular polarisation. Current travelling round a helical turn has both an axial component and a circumferential one. When the circumference of a turn is about one wavelength,

\(C\approx\lambda \quad\Longrightarrow\quad \text{one full turn} = \text{one period}\)

the current takes exactly one RF cycle to travel round a turn. The two orthogonal field components therefore emerge equal in magnitude and 90° apart in phase — which is precisely the definition of circular polarisation:

\(\vec{E}=E_0\left(\hat{a}_x\pm j\hat{a}_y\right)e^{-jkz}\)

The sense follows the winding: a right-hand-wound helix radiates right-hand circular, and it must be received by a right-hand antenna, since the opposite sense is cross-polarised and rejected.

Why the other three are linearly polarised.

AntennaPolarisationReason
Small circular loopLinearDespite its circular shape, the current is essentially uniform round the loop and the far field has a single E component — the loop is the magnetic dual of the short dipole
Parabolic reflectorTakes the feed's polarisationThe dish only focuses; with a dipole feed the output is linear
Yagi-UdaLinearAll its elements are parallel dipoles, so the field lies along one axis

The small loop is the deliberate trap: circular geometry does not imply circular polarisation. Polarisation describes the behaviour of the electric field vector in time, not the outline of the conductor.

The general condition for circular polarisation. Two orthogonal field components of equal amplitude with a 90° phase difference. Unequal amplitudes give elliptical polarisation; zero phase difference gives linear. This is why the other common way of generating it is to feed crossed dipoles through a quarter-wavelength delay.

Why it is worth the trouble. A circularly polarised wave has no fixed plane, so it is immune to Faraday rotation in the ionosphere and to the unknown orientation of a moving or tumbling terminal. That is why satellite links, GPS and telemetry use it — and why the helix, with its wide bandwidth of about 1.7:1 and near-resistive 140 Ω input, became the standard tracking feed.

Hence, the circularly polarised antenna is the helical.

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Similar Questions

  1. If the effective area of an antenna becomes \(\frac{2A}{3}\) from its initial value of 'A', while keeping its operating frequency same. Then, the antenna gain becomes \(\left(\frac{2x+4}{15}\right)\) times of its initial value. The value of x will be:

  2. For a half wave dipole antenna

    A. The average value of current is 0.64 (unit).

    B. The ratio of electric field intensity just near to the antenna surface and the potential developed on the antenna surface after reception of signal, is equal to the effective height

    C. Effective aperture will remain same if antenna gain will be improved by some technique.

    D. The effective height will be 0.64 l, where l is the physical length of antenna.

    E. If length of antenna becomes l = 0.1λ, then its current distribution become triangular.

    Choose the correct answer from the options given below :

  3. Match the following lists in terms of radiation resistances of various antennas :

    List – IList – II  
    a. Short vertical monopolei. \(31200\left(\dfrac{\text{Area of Loop}}{\lambda^{2}}\right)^{2}\)
    b. Small loop antennaii. \(80\pi^{2}\left(\dfrac{L}{\lambda}\right)^{2}\)
    c. Dipole antennaiii. 73 ohms
    d. Radiation resistance of half wave dipoleiv. \(400\left(\dfrac{\text{Physical height}}{\lambda}\right)^{2}\)

    Choose the correct answer from the codes given below:

  4. Match List I with List II

    LIST I (Type of Aperture Antenna) LIST II (Beam widtd half power points)
    A. Uniformly illuminated linear ArrayI. \(\frac{58}{D_\lambda}\)
    B. Uniformly illuminated circular apertureII. \(\frac{56}{a_{E\lambda}}\)
    C. Optimum E-plane rectangular hornIII. \(\frac{67}{a_{E\lambda}}\)
    D. Optimum H-plane rectangular hornIV. \(\frac{51}{L_\lambda}\)

     

    Choose the correct answer from the options given below:

  5. Following statements are given :

    (a) Beam width between first nulls for a broadside long array is given by \(\dfrac{2\lambda}{nd}\).

    (b) Beam width between first nulls for an end fire long array is given by \(2\sqrt{\dfrac{2\lambda}{nd}}\).

    (c) Beam width between first nulls for a broadside long array is given by \(\dfrac{\lambda}{nd}\).

    (d) Beam width between first nulls for an end fire long array is given by \(\dfrac{\lambda}{nd}\).

    Which of the above statements are correct ?

  6. The expression given below is :

    \(P_{r}=P_{t}\dfrac{A_{et}\cdot A_{er}}{r^{2}\lambda^{2}}\ \left(\text{W}\right)\)


Important Questions from Antennas

  1. Which of the following antennas is the standard reference antenna for the directiveness?

  2. Consider the following statements:

    (a) Fiber optic cable is much lighter than copper cable

    (b) Fiber optic cable is not affected by power surges or electromagnetic interference

    (c) Optical transmission is inherently bidirectional.

    Which of the statements is (are) correct?
  3. Broadside arrays have

    A. Number of dipoles of unequal size

    B. Number of dipoles equally spaced

    C. Collinear dipoles

    D. Dipoles in phase

    E. Dipoles are 90 out of phase

    Choose the correct answer from the options given below:

  4. To match the impedance of a 'ground penetrating radar antenna' to the ground, impedance of ground is given by the expression, (if ϵ r= 14, μ r= 1, σ = 10 −2 ℧/m, operating frequency = 200 MHz)

  5. For an isotropic antenna P n(θ, φ) = 1, D = 1, for all θ and φ. The beam area for the isotropic antenna is given by:

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