Match the following List I and List II :List IList II(a) Directivity (D)(1) \(\frac{\lambda^{2}G}{4\pi}\)(b) Antenna Efficiency (n)(2) \(\geq 2D^{2}/\lambda\)(c) Fraunhofer Region (R)(3) Radiated Power / Total Input Power(d) The effective area of an antenna (Ae)(4) Max Radiation Intensity of Single Antenna / Radiation Intensity of an Isotropic AntennaCodes : (a) (b) (c) (d)

The correct code is (4) (3) (2) (1).(a) Directivity is the maximum radiation intensity of the antenna divided by the radiation intensity of an isotropic antenna.(b) Antenna efficiency is the radiated power divided by the total input power.(c) The Fraunhofer (far-field) region begins at a distance of at least 2D2/λ, where D is the largest dimension of the antenna.(d) The effective area of an antenna is λ2G/4π.Options A and D both give directivity the far-field distance, and option B gives it the effective-area formula.

Antenna Fundamentals and Parameters Questions and Answers

Q: Why does a wet radome degrade a satellite link disproportionately?

A continuous film of water absorbs strongly and reflects part of the wave, adding several decibels of loss precisely when rain attenuation on the path is already high. Hydrophobic coatings and heaters are used to break the film up. The focal length is a geometric property of the reflector, no short circuit occurs, and physical temperature is not the mechanism. The two losses coinciding is what makes the outage statistics worse than either alone would suggest.

Q: What is the main reason antenna specifications are quoted in decibels?

A link may span twelve orders of magnitude, so working logarithmically turns unwieldy products into sums and keeps every figure within a couple of significant digits. Accuracy is unaffected by the notation, regulations use decibels because the industry does rather than the reverse, and a decibel figure is meaningless without its reference, which is why dBi, dBd, dBm and dBW all exist as distinct units. Stating the reference is more important in decibels, not less.

Q: What does dBm express?

A dBm figure is power in decibels above one milliwatt, so zero dBm is one milliwatt and thirty dBm is one watt. Referring to a watt gives dBW, which is thirty decibels lower numerically, gain against a dipole is dBd, and field strength against a microvolt per metre is dBuV/m. Mixing these references is the most frequent arithmetic error in radio engineering, and writing the reference explicitly in every line of a budget is the cure.

Q: The value of directivity of isotropic antenna is :

The directivity of an isotropic antenna is 1.Directivity is the ratio of an antenna's maximum radiation intensity to the radiation intensity of an isotropic antenna radiating the same total power. An isotropic antenna radiates equally in all directions, so its maximum equals its average, and the ratio is 1 (0 dB).It cannot be 0, which would mean no radiation, or 0.5, since the maximum intensity can never be below the average. Infinite directivity would need all power in a single direction, which no real antenna achieves.

Q: High frequency resistance of an antenna of length 'l', cross-section 'A', conductivity '\(\sigma\)', perimeter 'P' (\(\mu_{0}\) is free space permeability, \(\omega\) is operating frequency) is given by :

The resistance is \(l/P \sqrt{\omega \mu_{0}/2\sigma}\) ohms.At high frequency the current flows only in a thin skin of depth \(\delta\), so the effective cross-section is the perimeter times the skin depth:\[ \begin{aligned} \delta &= \sqrt{\frac{2}{\omega \mu_{0} \sigma}} \\ R &= \frac{l}{\sigma P \delta} = \frac{l}{P}\sqrt{\frac{\omega \mu_{0}}{2\sigma}} \end{aligned} \]The cross-sectional area A no longer matters, which is why it does not appear. The other options put σ in the numerator, which would make a better conductor more resistive, or misplace the factor 2 and the perimeter.

Q: If the length of a linear dipole antenna (\(l \ll \lambda\)), Ao is peak value of power density, r is radius, \(\hat{a}\) is radial unit vector, \(\theta\) is usual spherical co-ordinate, then the radiated power density is given by :

The radiated power density is \(\hat{a}\, A_{o} \sin^{2}\theta / r^{2}\).For a short (Hertzian) dipole the radiated fields vary as \(\sin\theta / r\). Power density is proportional to the square of the field, so it varies as \(\sin^{2}\theta / r^{2}\), directed radially outward. With Ao defined as the peak value of the power density, it appears to the first power.Squaring Ao would square a quantity that is already a power. A dependence on \(\sin\theta\) is that of the field, not the power, and a 1/r fall-off would also describe the field rather than the power density, which falls as the inverse square of distance.

Q: If 1 is transmit antenna and 2 is receive antenna, the Z21 (mutual impedance between antenna 1 and 2) is given by :

The mutual impedance is \(Z_{21} = V_{2oc}/I_{1}\) with I2 = 0.Treating the two antennas as a two-port network, Z21 is the open-circuit transfer impedance: drive antenna 1 with current I1, leave antenna 2 open-circuited so that I2 = 0, and measure the voltage V2oc induced at antenna 2. The ratio is the mutual impedance.The other expressions multiply or divide two voltages, which gives the wrong units for an impedance, or take the condition I1 = 0, which would mean antenna 1 is not transmitting at all.

Q: The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is :

The radiation resistance is 4.8 ohms.Radiation resistance is the resistance that would dissipate the radiated power when carrying the antenna current:\[ \begin{aligned} P &= I^{2}R_{r} \\ R_{r} &= \frac{P}{I^{2}} = \frac{120}{5^{2}} = \frac{120}{25} \\ &= 4.8\ \Omega \end{aligned} \]9.6 is twice this, what you get with \(P = I^{2}R/2\), the form used when I is a peak rather than an rms value. 3.6 and 1.8 do not follow from the given data.

101.

Why does a wet radome degrade a satellite link disproportionately?

A.A water film is lossy and reflective at microwave frequencies
B.Water changes the focal length of the reflector
C.Water shorts out the feed to the reflector surface
D.Water raises the physical temperature of the antenna

Answer: A

A continuous film of water absorbs strongly and reflects part of the wave, adding several decibels of loss precisely when rain attenuation on the path is already high. Hydrophobic coatings and heaters are used to break the film up. The focal length is a geometric property of the reflector, no short circuit occurs, and physical temperature is not the mechanism. The two losses coinciding is what makes the outage statistics worse than either alone would suggest.

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102.

What is the main reason antenna specifications are quoted in decibels?

A.Decibels remove the need to state a reference
B.Large ratios become manageable numbers that add rather than multiply
C.Decibels are more accurate than ratios
D.Regulators require decibel notation by law

Answer: B

A link may span twelve orders of magnitude, so working logarithmically turns unwieldy products into sums and keeps every figure within a couple of significant digits. Accuracy is unaffected by the notation, regulations use decibels because the industry does rather than the reverse, and a decibel figure is meaningless without its reference, which is why dBi, dBd, dBm and dBW all exist as distinct units. Stating the reference is more important in decibels, not less.

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103.

What does dBm express?

A.Gain referred to a matched dipole
B.Field strength referred to one microvolt per metre
C.Power referred to one milliwatt
D.Power referred to one watt

Answer: C

A dBm figure is power in decibels above one milliwatt, so zero dBm is one milliwatt and thirty dBm is one watt. Referring to a watt gives dBW, which is thirty decibels lower numerically, gain against a dipole is dBd, and field strength against a microvolt per metre is dBuV/m. Mixing these references is the most frequent arithmetic error in radio engineering, and writing the reference explicitly in every line of a budget is the cure.

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104.

A transmitter delivers 20 W. What is this in dBm?

A.13 dBm
B.20 dBm
C.33 dBm
D.43 dBm

Answer: D

One watt is 30 dBm, and twenty is thirteen decibels above one:

\[ 30 + 10\log_{10} 20 \approx 30 + 13 = 43 \]

so 43 dBm. Building the answer from the familiar anchor of 30 dBm for a watt, and adding ten decibels per factor of ten and three per factor of two, avoids reaching for a calculator. Twenty watts is a common transmitter rating and 43 dBm is worth recognising on sight.

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105.

The value of directivity of isotropic antenna is :

A.1
B.0
C.∞
D.0.5

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The directivity of an isotropic antenna is 1.

Directivity is the ratio of an antenna's maximum radiation intensity to the radiation intensity of an isotropic antenna radiating the same total power. An isotropic antenna radiates equally in all directions, so its maximum equals its average, and the ratio is 1 (0 dB).

It cannot be 0, which would mean no radiation, or 0.5, since the maximum intensity can never be below the average. Infinite directivity would need all power in a single direction, which no real antenna achieves.

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106.

Match the following List I and List II :

List I	List II
(a) Directivity (D)	(1) \(\frac{\lambda^{2}G}{4\pi}\)
(b) Antenna Efficiency (n)	(2) \(\geq 2D^{2}/\lambda\)
(c) Fraunhofer Region (R)	(3) Radiated Power / Total Input Power
(d) The effective area of an antenna (Ae)	(4) Max Radiation Intensity of Single Antenna / Radiation Intensity of an Isotropic Antenna

Codes : (a) (b) (c) (d)

A.(2) (3) (4) (1)
B.(1) (2) (4) (3)
C.(4) (3) (2) (1)
D.(2) (3) (1) (4)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The correct code is (4) (3) (2) (1).

(a) Directivity is the maximum radiation intensity of the antenna divided by the radiation intensity of an isotropic antenna.
(b) Antenna efficiency is the radiated power divided by the total input power.
(c) The Fraunhofer (far-field) region begins at a distance of at least 2D²/λ, where D is the largest dimension of the antenna.
(d) The effective area of an antenna is λ²G/4π.

Options A and D both give directivity the far-field distance, and option B gives it the effective-area formula.

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107.

High frequency resistance of an antenna of length 'l', cross-section 'A', conductivity '\(\sigma\)', perimeter 'P' (\(\mu_{0}\) is free space permeability, \(\omega\) is operating frequency) is given by :

A.\(l/P \sqrt{\frac{\omega \mu_{0}}{2 \sigma}}\) ohms
B.\(l/2P \sigma \sqrt{\omega \mu_{0}}\) ohms
C.\(l/P \sigma \sqrt{\omega \mu_{0}}\)
D.\(\frac{l}{2 \sigma} \sqrt{\frac{\omega \mu_{0}}{P}}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

The resistance is \(l/P \sqrt{\omega \mu_{0}/2\sigma}\) ohms.

At high frequency the current flows only in a thin skin of depth \(\delta\), so the effective cross-section is the perimeter times the skin depth:

\[ \begin{aligned} \delta &= \sqrt{\frac{2}{\omega \mu_{0} \sigma}} \\ R &= \frac{l}{\sigma P \delta} = \frac{l}{P}\sqrt{\frac{\omega \mu_{0}}{2\sigma}} \end{aligned} \]

The cross-sectional area A no longer matters, which is why it does not appear. The other options put σ in the numerator, which would make a better conductor more resistive, or misplace the factor 2 and the perimeter.

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108.

If the length of a linear dipole antenna (\(l \ll \lambda\)), A_o is peak value of power density, r is radius, \(\hat{a}\) is radial unit vector, \(\theta\) is usual spherical co-ordinate, then the radiated power density is given by :

A.\(\hat{a}\, A_{o}^{2} \sin^{2}\theta / r^{2}\)
B.\(\hat{a}\, A_{o} \sin^{2}\theta / r^{2}\)
C.\(\hat{a}\, A_{o} \sin\theta / r^{2}\)
D.\(\hat{a}\, A_{o}^{2} \sin\theta / r\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The radiated power density is \(\hat{a}\, A_{o} \sin^{2}\theta / r^{2}\).

For a short (Hertzian) dipole the radiated fields vary as \(\sin\theta / r\). Power density is proportional to the square of the field, so it varies as \(\sin^{2}\theta / r^{2}\), directed radially outward. With A_o defined as the peak value of the power density, it appears to the first power.

Squaring A_o would square a quantity that is already a power. A dependence on \(\sin\theta\) is that of the field, not the power, and a 1/r fall-off would also describe the field rather than the power density, which falls as the inverse square of distance.

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109.

If 1 is transmit antenna and 2 is receive antenna, the Z₂₁ (mutual impedance between antenna 1 and 2) is given by :

A.\(\left. \frac{V_{1oc} . V_{2oc}}{I_{2}} \right|_{I_{1}=0}\)
B.\(\left. \frac{V_{2oc}}{I_{1}} \right|_{I_{2}=0}\)
C.\(\left. \frac{V_{1oc}}{V_{2oc} . I_{1}} \right|_{I_{2}=0}\)
D.\(\left. \frac{V_{2oc}}{V_{1oc}} \times \frac{1}{I_{2}} \right|_{I_{1}=0}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The mutual impedance is \(Z_{21} = V_{2oc}/I_{1}\) with I₂ = 0.

Treating the two antennas as a two-port network, Z₂₁ is the open-circuit transfer impedance: drive antenna 1 with current I₁, leave antenna 2 open-circuited so that I₂ = 0, and measure the voltage V_2oc induced at antenna 2. The ratio is the mutual impedance.

The other expressions multiply or divide two voltages, which gives the wrong units for an impedance, or take the condition I₁ = 0, which would mean antenna 1 is not transmitting at all.

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110.

The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is :

A.4.8
B.9.6
C.3.6
D.1.8

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The radiation resistance is 4.8 ohms.

Radiation resistance is the resistance that would dissipate the radiated power when carrying the antenna current:

\[ \begin{aligned} P &= I^{2}R_{r} \\ R_{r} &= \frac{P}{I^{2}} = \frac{120}{5^{2}} = \frac{120}{25} \\ &= 4.8\ \Omega \end{aligned} \]

9.6 is twice this, what you get with \(P = I^{2}R/2\), the form used when I is a peak rather than an rms value. 3.6 and 1.8 do not follow from the given data.

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