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The correct choice is (a) 12.33Ω

The explanation is: Radiation resistance of a HERTZIAN dipole of LENGTH l is

R=80π^2\((\frac{l}{\LAMBDA})^2=80\pi^2 (\frac{\lambda/8}{\lambda})^2=12.33\Omega\)

Right choice is (c) \(\frac{\lambda^2}{4\pi \eta}\)

To explain: For ISOTROPIC radiator, DIRECTIVITY is 1. So the effective aperture is GIVEN by \(A_{EM}=D \frac{\lambda^2}{4\pi} = \frac{\lambda^2}{4\pi}\)

Then physical aperture =\(\frac{Effective\, aperture}{Aperture\, EFFICIENCY}=\frac{\lambda^2}{4\pi \eta}\)

The correct OPTION is (a) True

Explanation: Radiation RESISTANCE doesn’t DEPEND on the direction in which the power is radiated. It DEPENDS on FREQUENCY through which we will find the wavelength and thereby the radiation resistance.

Right CHOICE is (a) E-plane pattern

For explanation I would say: Varying θat a constant φ REPRESENTS a vertical PLOT against the FIELD strength. So this pattern is called E-plane pattern. H-plane or horizontal pattern are plotted with magnitude of field strength Vs φ at a constant θ. POWER pattern is plotted with square of the magnitude of field strength.

The CORRECT option is (C) 0.01m^2

To EXPLAIN I would say: Effective APERTURE Aem=Aeη=0.02×0.5=0.01 m^2

Right OPTION is (b) 0.25m

The BEST EXPLANATION: Operating frequency f0=BW×Q=20MHz×30=600MHz

Now, \(λ=\frac{c}{f}=\frac{3×10^8 m/s}{600MHz}=0.5m\)

∴ LENGTH of half-wave dipole is \(l=\frac{λ}{2}=\frac{0.5}{2}=0.25m\)

The correct choice is (a) True

To explain I would say: The graphical representation of the RADIATION properties of the antenna as a FUNCTION of space coordinates is CALLED Radiation pattern. It is also known as antenna pattern. A radiation PROPERTY includes power flux density, directivity, and radiation INTENSITY.

The CORRECT answer is (d) 8Ω

For explanation I would say: Input power Pin=Prad+Ploss

⇨ Prad=Pin-Ploss=1000-200=800W

Now, Power RADIATED Prad=Irms^2Rrad

⇨ Rrad=\(\FRAC{P_{RAD}}{I_{rms}^2} = \frac{800}{10×10}\)=8Ω

Correct OPTION is (c) 73.12Ω

Easy EXPLANATION: Since RADIATION resistance of quarter-wave monopole (l=λ/4) is 36.56Ω, then for a half-wave DIPOLE (l=λ/2) it is given by 36.56×2 = 73.12Ω. Hertzian dipole is an ideal dipole of INFINITESIMAL dipole.

Correct OPTION is (a) 1.07m^2

To ELABORATE: Effective aperture for a Hertzian dipole is given by \(A_e=1.5\FRAC{\lambda^2}{4\pi} \)

Gain of Hertzian dipole is 1.5

\(\lambda=\frac{C}{F}=\frac{3×10^8}{100×10^6}=3m\)

\(A_e=1.5 \frac{\lambda^2}{4\pi} =1.5\frac{3^2}{4\pi} =1.07m^2\)

The CORRECT option is (a) 4π

To elaborate: The DIRECTIVITY in terms of the Beam area ΩA is GIVEN by D=\(\FRAC{4\pi}{\Omega_A}\)

⇨ \(\frac{4\pi}{D}\)=4π steradians.

The correct choice is (a) 3sinθ ar W/Steradian

Explanation: Radiation INTENSITY is defined as the power radiated PER unit SOLID angle.

U= Wr.r^2= 3sinθ ar W/Steradian.

The correct answer is (a) NEAR FIELD

The best explanation: Near Field is called Fresnel Field. Far field is called Fraunhofer zone. Near field VARIES INVERSELY with r^2. ELECTROSTATIC field varies inversely with r^3.

Right option is (a) Watts/unit Solid angle

Explanation: Radiation intensity is defined as the power RADIATED form an ANTENNA PER unit solid angle. So its UNITS are Watts/Steradian. Steradian is the unit for solid angle.

The correct option is (a) True

The best explanation: The axis of back LOBE is opposite to the main lobe. So it makes 180° with BEAM of antenna. It is ALSO a SIDE lobe which is at 180 to main lobe.

The correct answer is (a) True

For explanation I would say: The radiation resistance is defined as the equivalent resistance that DISSIPATES EQUAL AMOUNT of power that is radiated by the ANTENNA. The power radiated by the radiation resistance is given byPrad=Irms^2Rrad. Radiation resistance doesn’t depend on DIRECTION of power radiated but it depends on the frequency.

The correct OPTION is (a) True

To elaborate: For a lossless ANTENNA, OHMIC losses will be zero. So, radiation EFFICIENCY will be 100%. Hence, maximum power gain will be equal to the maximum directive gain of antenna.

The CORRECT option is (a) 4π

To ELABORATE: Given Ud(θ,∅) = 5W/steradian , Pr=5W

Directive gain \(G_d=\frac{P_{d(\theta,\emptyset)^{r^2}}}{P_r/4\pi} = \frac{U_{d(\theta,\emptyset)}}{P_r/4\pi}=4\pi\)

The correct option is (a) 1.28m^2

The explanation: The effective AREA \(A_e=\FRAC{\lambda^2}{4π}D \)

\(\lambda = \frac{C}{F}=3×\frac{10^8}{100×10^6}=3M\)

\(A_e=\frac{3^2}{4π}×1.8=1.28m^2\)

The CORRECT ANSWER is (a) 900W

For EXPLANATION: POWER RADIATED Pr=I^2 Rr=100×3^2=900Watts.

Correct choice is (a) True

For explanation: Front-to-Back ratio (FBR) =\(\FRAC{POWER\, RADIATED\, DESIRED\, direction}{Power\, radiated\, in\, backward\, direction}.\) If more power is diverted backside, then the gain of the antenna decreases.

Correct OPTION is (a) Series R, L, C

The explanation is: Antenna is represented by a series R, L, C equivalent circuit. Antenna is USED for IMPEDANCE MATCHING and acts like a transducer.

The CORRECT choice is (a) True

For explanation I would say: The bandwidth of antenna is inversely proportional to the QUALITY factor of the antenna.

\(Q=\frac{f_0}{BW}\)

The CORRECT ANSWER is (a) 65

For explanation I would say: Quality factor \(Q=\FRAC{f_0}{BW}=\frac{650MHz}{10MHz}=65\)

The correct CHOICE is (c) 10Ω

For explanation: Power RADIATED Prad=Irms^2 Rrad

⇨ Rrad=\(\frac{P_{rad}}{I_{rms}^2} = \frac{1000}{10×10}=10\Omega\)

Correct option is (a) 20π^2 \((\frac{L}{\LAMBDA})^2\)

For explanation I would say: RADIATION resistance of a Hertzian dipole of LENGTH l is given by R=80π^2 \((\frac{l}{\lambda})^2\)

Then SHORT dipole is of length l/2, so the radiation resistance is given by R=80π^2\((\frac{L/2}{\lambda})^2=20\pi^2 (\frac{L}{\lambda})^2.\)

Right choice is (a) Short monopole

For explanation: Short MONOPOLES have length LESS than λ/8 and the CURRENT distribution is TRIANGULAR. Short dipole has length less than λ/2. Half-wave dipoles have length EQUAL to λ/2. Quarter-wave monopoles have length equal to λ/4.

The correct answer is (C) 4π^2/3

The explanation: Total power radiated \(P_{rad}= ∯ w_r.a_n\overline{d}L \)

\(=\iint_{\THETA = 0}^{\pi} w_r.r^2 sin\theta d\theta d\EMPTYSET\)

\(=\iint_{\theta = 0}^π\frac{4sin\theta}{3r^2}r^2sin\theta d\theta d \emptyset\)

\(=\iint_{\theta = 0}^π \frac{4}{3}sin^2\theta d\theta d\emptyset=\iint_{\theta=0}^π\frac{4}{3}(\frac{1-cos2\theta}{2})d\theta d\emptyset\)

\(=\frac{4}{3}(\frac{1}{2})(π)(2π)=\frac{4}{3}\) π^2.

The CORRECT choice is (B) \((\frac{\lambda}{4\pi R})^2\)

Easy explanation: The free space loss factor is given by \((\frac{\lambda}{4\pi R})^2\). It is used to know the amount of losses occurred DUE to the spreading of energy by an ANTENNA.

The correct answer is (b) decrease

Easy EXPLANATION: Since the radiation resistance of a HERTZIAN dipole is R=80π^2 \((\frac{L}{\lambda})^2\), the radiation resistance will decrease as the length of the dipole decreases. It is directly PROPORTIONAL to the square of the length of the dipole.

The CORRECT ANSWER is (b) Directly PROPORTIONAL

To elaborate: The directivity D is directly proportional to the EFFECTIVE aperture of ANTENNA. \(A_e=D\frac{\lambda^2}{4\pi}\)

Correct answer is (b) 3°

To EXPLAIN I WOULD say: Resolution is half of the first null beam width. \(R=\FRAC{FNBW}{2}=\frac{6°}{2}=3°.\)

Right option is (a) Major LOBE

For explanation I would say: Major lobe contains the maximum radiated power. SOMETIMES DEPENDING on our REQUIREMENT there can be more than 2 major lobes.

Right choice is (a) Ω≈∅A.∅E

The EXPLANATION is: HALF-power beam width is the ANGULAR RANGE of antenna pattern in which half power is RADIATED. The relation between Ω, ∅A, ∅E is given by Ω≈∅ A.∅E.

Right answer is (c) 236.5μW

To elaborate: GIVEN d=20m, Pt=50W and F=900MHZ

Gain of half-wave dipoles is 1.64

\(λ = \frac{c}{f} = \frac{3×10^8}{900Mhz} = \frac{1}{3} m \)

\(\frac{P_r}{P_t} = \frac{G_t G_r \lambda^2}{(4\pi R)^2} = \frac{1.64×1.64×1/3^2}{(4\pi ×20)^2}\)

Pr=236.5μW

Right option is (b) 8.8mΩ

To elaborate: \(λ=\frac{C}{f}=\frac{3×10^8}{1×10^6}=300m\)

Radiation resistance of a HERTZIAN DIPOLE is given by \(R=80π^2(\frac{l}{\lambda})^2=80π^2 (\frac{1}{300})^2\)=0.0088Ω

The correct option is (a) Independent, independent

The best explanation: Directive gain is the ratio of POWER density to the average power radiated. \(G_d = \frac{P_{d(\theta,\emptyset)}}{P_{avg}}\)

So, the Directive gain is independent on both input power to ANTENNA and power DUE to ohmic losses.

Power gain is DEPENDENT on input power and ohmic losses to antenna.

The correct OPTION is (c) 5m

The EXPLANATION: Effective length of the RECEIVING antenna \(L_{EFF}=\frac{V_{OC}}{E_i}=\frac{1}{0.2}=5m.\)