300+ REAL TIME Antennas Objective Questions & Answers 2023

250+ TOP MCQs on Uniform Aperture Field and Answers

Antennas Multiple Choice Questions on “Uniform Aperture Field”.

1. For aperture antenna to be efficient and have high directivity, its area should be ____________
a) ≥ λ²
b) ≥ 1/λ
c) ≤ λ²
d) ≤ λ
Answer: a
Clarification: Antenna with an aperture at the end is known as aperture antenna. Example is waveguide. For aperture antenna to have high directivity its area should be ≥ λ². These antennas usually operated at UHF and above frequencies.

2. At which of the following frequencies aperture antennas are operated?
a) UHF and EHF
b) MF and HF
c) HF and UHF
d) LF and MF
Answer: a
Clarification: Antenna with an aperture at the end is known as aperture antenna. Example is waveguide. These antennas usually operated at UHF and EHF frequencies (300MHz to 300GHz).
LF – 30 kHz – 300 kHz
MF – 300k-3MHz
HF – 3MHz -30MHz

3. Which of the following does not belong to the aperture antenna?
a) Half-Dipole
b) Horn Antenna
c) Waveguide antenna
d) Slot antenna
Answer: a
Clarification: Antenna with an aperture at the end is known as aperture antenna. Horn antenna, Waveguide antenna and slot antenna are examples of aperture antenna. Half-dipole is a wire antenna.

4. Which of the following principle is used for analysis of aperture antennas?
a) Equivalence principle
b) Friss Equation
c) Reflectivity
d) Diffraction
Answer: a
Clarification: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the boundary conditions. So this is used in the analysis of the aperture antennas. Friss transmission is used to relate the distance and power radiation between the antennas. Reflectivity and diffraction are not the principles used for analysis of aperture antenna.

5. Equivalence Principle follows which f the following theorem?
a) Uniqueness Theorem
b) Poynting theorem
c) Friss Theorem
d) Gauss theorem
Answer: a
Clarification: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the boundary conditions. Uniqueness theorem is defined from the pointing theorem. Friss transmission is used to relate the distance and power radiation between the antennas. Gauss theorem states that total electric flux enclosed by charge is equal to net positive charge.

6. Which of the following condition is true for the electric conductor equivalent?
a) J_s=n×H=0;M_s=-n×E
b) J_s=n×H;M_s=-n×E=0
c) J_s=n×H≠0;M_s=-n×E
d) J_s=n×H≠0;M_s=-n×E=0
Answer: a
Clarification: The equivalent surface currents J_s, M_s radiates fields H, E respectively. The conditions for the electric conductor equivalent is J_s=n×H=0;M_s=-n×E.

7. Which of the following condition is true for the magnetic conductor equivalent?
a) J_s=n×H=0;M_s=-n×E
b) J_s=n×H;M_s=-n×E=0
c) J_s=n×H≠0;M_s=-n×E
d) J_s=n×H≠0;M_s=-n×E=0
Answer: b
Clarification: The equivalent surface currents J_s, M_s radiates fields H, E. The conditions for the electric conductor equivalent is J_s=n×H;M_s=-n×E=0. This is one of the equivalence principle modes.

8. Equivalence principle is mainly used for far filed analysis of the antenna in outer region.
a) True
b) False
Answer: a
Clarification: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the boundary conditions. So this is used in the far field analysis of the aperture antennas in the outer regions.

9. The gain of the aperture antenna increases with square of the frequency.
a) True
b) False
Answer: a
Clarification: One of the distinguishing features of the aperture antenna is the increase in gain with operating frequency. The gain of the aperture antenna increases with square of the frequency if its aperture efficiency is kept constant with respect to the frequency.

10. Huygens principle in mathematical form is referred to as equivalence principle for aperture antennas.
a) True
b) False
Answer: a
Clarification: The equivalence principle replaces the aperture antenna with surface currents and thereby fields. It is derived from the uniqueness theorem and provides a unique solution for the boundary conditions for the fields. This is mainly used for the far field analysis of the aperture antenna.

250+ TOP MCQs on Isotropic Radiators and Answers

Antennas Multiple Choice Questions on “Isotropic Radiators”.

1. An ideal source in which the power is radiated equally in all directions is called as ________ radiator.
a) Isotropic
b) Omni-directional
c) Directional
d) Transducer
Answer: a
Clarification: Isotropic radiators radiate power in all directions uniformly. Omni-directional antenna radiates only in one direction. Directional antenna radiates maximum power in particular direction. Transducer converts one form of energy to other.

2. In Isotropic radiation, which of the following vector component is absent in pointing vector?
a) (widehat{a_r})
b) (widehat{a_theta})
c) (widehat{a_emptyset})
d) Both (widehat{a_theta} and widehat{a_emptyset} )
Answer: d
Clarification: Isotropic radiators will radiate power equally in all directions. Due to this symmetrical distribution, components of θ, Φ get cancelled. So it will have only radial component.

3. What is the amount of Electric field present at a distance of 10km for an isotropic radiator with radiating power 3kW?
a) 30mV/m
b) 60mV/m
c) 15mV/m
d) 10mV/m
Answer: a
Clarification: For isotropic radiator, power per unit area = (frac{P_t}{4πr^2}) — (1)
From pointing theorem, (P=frac{mid E_{max}^2mid}{2eta}, where eta=120π. )—– (2)
Equating (1) & (2) and E_rms=E_max/√2, we get
(E_{rms}=frac{sqrt{30P_t}}{r}=frac{sqrt{30×3000}}{10000}=30mV/m.)

4. What is the radiation intensity for isotropic antenna having radiation power density (frac{3 sin⁡θ}{r^2}a_r W/m^2)?
a) 3sinθ a_r W/Steradian
b) 3cosθa_r W/Steradian
c) 6πsinθ a_r W/Steradian
d) 6πcosθ a_r W/steradian
Answer: a
Clarification: Radiation intensity is defined as the power radiated per unit solid angle.
U= Wr.r²= 3sinθ a_r W/Steradian.

5. For an isotropic source, Radiation intensity will be _____ on θ and ______ on Φ.
a) Dependent, independent
b) Independent, independent
c) Independent, dependent
d) Dependent, dependent
Answer: b
Clarification: Due to symmetric distribution in isotropic source, there will be no θ, ф components. Only radial component will be present. ( P_{total ,rad}=∯ Usinθdθd∅=4πU (For ,θ: 0 ,to ,π, phi: 0 ,to, 2π).)
∴ Radiation intensity (U=frac{P_{total ,rad}}{4π})

6. Find the effective length of a receiving antenna with open circuit voltage 1V and incident electric field 200mV/m?
a) 0.2m
b) 50m
c) 5m
d) 5cm
Answer: c
Clarification: Effective length of the receiving antenna (L_{eff}=frac{V_{oc}}{E_i}=frac{1}{0.2}=5m.)

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250+ TOP MCQs on N-element Linear Array and Answers

Antennas Multiple Choice Questions on “N-element Linear Array”.

1. Which of the following statement is true?
a) As the number of elements increase in array it becomes more directive
b) As the number of elements increase in array it becomes less directive
c) Point to point communication is not possible with more number of array elements
d) There is no uniform progressive phase shift in linear uniform array
Answer: a
Clarification: To get a single beam for the point to point communication more number of array elements is used. It increases the directivity of the antenna. An array is said to be uniform if the elements are excited equally and there is a uniform progressive phase shift.

2. Normalized array factor of N –element linear array is ________
a) (frac{sin(Nᴪ/2)}{Nᴪ/2} )
b) (frac{cos(Nᴪ/2)}{Nᴪ/2} )
c) (Nfrac{sin(ᴪ/2)}{ᴪ/2} )
d) (Nfrac{cos(Nᴪ/2)}{Nᴪ/2} )
Answer: a
Clarification: The N-element linear uniform array, having a constant phase difference will have the array factor (AF = ∑_{n=1}^Ne^{j(n-1)ᴪ})
Normalized array factor is given by (frac{sin(Nᴪ/2)}{Nᴪ/2}. )

3. Which of the following expression gives the nulls for the N- element linear array?
a) (θ_n=cos^{-1}⁡(frac{λ}{2πd}[-β±frac{2πn}{N}]))
b) (θ_n=sin^{-1}⁡(frac{λ}{2πd}[-β±frac{2πn}{N}]))
c) (θ_n=cos^{-1}⁡(frac{λ}{2πd}[-β±frac{πn}{N}]))
d) (θ_n=cos^{-1}⁡(frac{λ}{2πd}[β±frac{2πn}{N}]))
Answer: a
Clarification: To determine Null points the array factor is set equal to zero.
(frac{sin(frac{Nᴪ}{2})}{frac{Nᴪ}{2}}=0)
⇨ (sin(frac{Nᴪ}{2})=0 )
⇨ (frac{Nᴪ}{2}=±nπ )
⇨ (kdcosθ+β=±frac{2nπ}{N} )
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2πd}[-β±frac{2πn}{N}]).)

4. Maximum value of array factor for N-element linear array occurs at ______
a) (θ_m=cos^{-1}⁡(frac{λ}{2πd}[-β±2πm]))
b) (θ_m=cos^{-1}⁡(frac{λ}{2πd}[β±2πm]))
c) (θ_m=sin^{-1}⁡(frac{λ}{2πd}[-β±2πm]))
d) (θ_m=sin^{-1}(frac{λ}{2πd}[-β±2πm]))
Answer: a
Clarification: Normalized array factor is given by (frac{sin(Nᴪ/2)}{Nsin(frac{ᴪ}{2})}.)
To get maximum denominator is equal to zero.
⇨ (sin(frac{ᴪ}{2})=0 )
⇨ (frac{ᴪ}{2}=±nπ )
⇨ kdcosθ+β=±2πm
⇨ (θ_m=cos^{-1}⁡(frac{λ}{2πd}[-β±2πm]).)

5. Find the maximum value of array factor when elements are separated by a λ/4 and phase difference is 0?
a) θ_m=cos^-1⁡(4m)
b) θ_m=sin^-1⁡(4πm)
c) θ_m=cos^-1⁡(4πm)
d) θ_m=sin^-1(2m)
Answer: a
Clarification: The maximum value of array factor is (θ_m=cos^{-1}⁡(frac{λ}{2πd}[-β±2πm]).)
(θ_m=cos^{-1}⁡(frac{λ}{2πλ/d}[0±2πm]).)
θ_m=cos^-1⁡(4m).

6. Find the Nulls of the N-element array in which elements are separated by λ/4 and phase difference is 0?
a) (θ_n=cos^{-1}⁡(left[±frac{4n}{N}right]))
b) (θ_n=cos^{-1}⁡(left[±frac{2n}{N}right]))
c) (θ_n=sin⁡(left[±frac{4n}{N}right]))
d) (θ_n=sin^{-1}⁡(left[±frac{4n}{N}right]))
Answer: a
Clarification: The nulls of the N- element array is given by (θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2πλ/4} [0±frac{2πn}{N}]))
⇨ (θ_n=cos^{-1}⁡([±frac{4n}{N}]))

7. Find the Nulls of the 8-element array in which elements are separated by λ/4 and phase difference is 0?
a) (θ_n=cos^{-1}⁡(left[±frac{n}{2}right]))
b) (θ_n=cos^{-1}⁡(left[±frac{n}{4}right]))
c) (θ_n=sin^{-1}⁡(left[±frac{n}{2}right]))
d) (θ_n=sin^{-1}⁡(left[±frac{n}{4}right]))
Answer: a
Clarification: The nulls of the N- element array is given by (θ_n=cos^{-1}⁡(frac{λ}{2πd} left[-β±frac{2πn}{N}right]))
⇨ (θ_n=cos^{-1}⁡(frac{λ}{2πλ/4}left[0±frac{2πn}{N}right]))
⇨ (θ_n=cos^{-1}⁡(left[±frac{4n}{8}right]))
⇨ (θ_n=cos^{-1}⁡(left[±frac{n}{2}right]).)

8. The radiating pattern of single element multiplied by the array factor simply gives the ___________
a) Pattern multiplication
b) Normalized array factor
c) Beamwidth of the array
d) Field strength of the array
Answer: a
Clarification: The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. Array factor is the function of antenna positions in the array and its weights. Total array field is the field generated by the sum of the individual elements in array.

9. Condition for the half power width of the Array factor is given by ___________
a) (frac{Nᴪ}{2}=±1.391 )
b) (frac{Nᴪ}{2}=±3)
c) (frac{Nᴪ}{2}=±0.5)
d) (frac{Nᴪ}{2}=±1)
Answer: a
Clarification: Array factor is the function of antenna positions in the array and its weights. Half power beamwidth is also known as the 3 decibel points. The half power beam widths of the array factor will occur at (frac{Nᴪ}{2}=±1.391. )

10. The maximum of the first minor lobe of array factor occurs at 13.46 dB down the maximum major lobe.
a) True
b) False
Answer: a
Clarification: The maximum of 1st minor lobe occurs at (frac{Nᴪ}{2}=±3π/2)
⇨ (AF = frac{sin(frac{Nᴪ}{2})}{frac{Nᴪ}{2}}=0.212= -13.46dB.)

250+ TOP MCQs on Types and Answers

Antennas Multiple Choice Questions on “Types”.

1. For a horn antenna, in which flaring is done only in one direction is ________
a) Conical antenna
b) Sectoral antenna
c) Pyramidal horn antenna
d) Exponential horn antenna
Answer: b
Clarification: In Sectoral horn antenna, flaring is done only in one direction. Depending on the flaring direction with respect to field propagation, it is divided into E-plane or H-plane horn antenna. Conical horn flared cross section is in shape of a cone.Pyramidal horn flared cross section is in shape of a four-sided pyramid. In exponential horn, the separation of sides increases as a function of length.

2. If flaring is done in the boarder direction of the rectangular waveguide then it is called ______
a) E-plane horn
b) H-plane horn
c) Conical horn
d) Pyramidal horn
Answer: a
Clarification: If the flaring is done in the electric field that means in the boarder direction of the rectangular waveguide, then it is called E-plane horn antenna.

3. If flaring is done in the magnetic field direction of the rectangular waveguide then it is called ______
a) E-plane horn
b) H-plane horn
c) Conical horn
d) Pyramidal horn
Answer: b
Clarification: In Sectorial horn antenna, flaring is done only in one direction.If the flaring is done in the magnetic field direction, then it is called H-plane horn antenna.

4. Which of the following antenna has the parallel slots along the inside surface of the horn?
a) E-plane horn
b) Pyramidal horn
c) Exponential horn
d) Corrugated horn
Answer: d
Clarification: A Corrugated horn has parallel slots or grooves which are small in size compared to the wavelength. These are present along the inside surface of the horn and are transverse to the axis.These corrugated antennas have wider bandwidths and minimizes the side lobes.

5. In which of the following antennas the separation of sides increases as a function of length?
a) Conical horn
b) Exponential tapered pyramidal horn
c) Pyramidal horn
d) Sectoral horn
Answer: b
Clarification: In exponentially tapered pyramidal horn, the transition region is gradually tapered exponentially to minimize the reflections. These are mainly used when reflections in the waveguide are critical.

6. If the walls of the circular waveguide are flared out, then it is called _____
a) Pyramidal horn
b) E-plane horn
c) H-plane horn
d) Conical horn
Answer: d
Clarification: In conical horn, the walls of the circular waveguide are flared out. For pyramidal horn, the walls of the rectangular waveguide are flared out in both directions.

7. If the all the walls of rectangular waveguide are flared out, then it is called ______
a) Pyramidal horn
b) Conical horn
c) E-plane horn
d) H-plane horn
Answer: a
Clarification: For pyramidal horn, all the walls of the rectangular waveguide are flared out. E-plane and H-plane are Sectoral horn antennas which are flared in only one direction.

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250+ TOP MCQs on Directive Gain and Answers

Antennas Multiple Choice Questions on “Directive Gain”.

1. For an isotropic antenna, the average power P_av can be expressed in terms of radiated power P_r as ____
a) P_av=P_r/4π
b) P_av=P_r/2πr²
c) P_av=P_r/2π
d) P_av=P_r/4πr²
Answer: d
Clarification: Average power is the total power radiated in the unit area. Here for isotropic radiation, area is spherical (say with radius r) and the area is 4πr².
∴ P_av=P_r/4πr²

2. Directive gain is defined as a measure of concentration of power in a particular direction.
a) True
b) False
Answer: a
Clarification: Directive gain is the ratio of power density to the average power radiated.
(G_d = frac{P_{d(theta,emptyset)}}{P_{avg}})

3. What is the directive gain when the magnitude of radiation intensity equals to average radiation intensity?
a) 4π
b) ∞
c) 1
d) 0
Answer: c
Clarification: Directive gain (G_d = frac{P_{d(theta,emptyset)}}{P_{avg}} = frac{P_{d(theta,emptyset)}}{P_r/4pi r^2} = frac{P_{d(theta,emptyset)^{r^2}}}{P_r/4pi} = frac{U_{d(theta,emptyset)}}{U_{avg}} )
∴ (G_d = frac{U_{(theta,emptyset)}}{U_{avg}})=1.

4. Directive gain of antenna when radiation intensity is 5W/Steradian and radiated power 5W is ____
a) 4π
b) 1/4π
c) 25
d) 1
Answer: a
Clarification: Given U_d(θ,∅) = 5W/steradian , P_r=5W
Directive gain (G_d=frac{P_{d(theta,emptyset)^{r^2}}}{P_r/4pi} = frac{U_{d(theta,emptyset)}}{P_r/4pi}=4pi)

5. The Directive gain is ______ on input power to antenna and _____ on power due to ohmic losses.
a) Independent, independent
b) Dependent, independent
c) Independent, dependent
d) Dependent, dependent
Answer: a
Clarification: 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.

6. What is the maximum directive gain of antenna with radiation efficiency 98% and maximum power gain 1?
a) 0.98
b) 1.02
c) 1.98
d) 1
Answer: b
Clarification: G_pmax=η_r G_dmax where η_r is radiation efficiency
Therefore G_dmax=1/0.98=1.02

7. Which of the following expression is correct for radiation efficiency?
a) (eta_r=frac{R_r}{R_l})
b) (eta_r=frac{R_r}{R_r-R_l})
c) (eta_r=frac{R_r}{R_r+R_l})
d) (eta_r=frac{R_l}{R_r+R_l})
Answer: c
Clarification: Radiation efficiency is defined as the ratio of power radiated to the total input power to the antenna. Total input power is the sum of the radiated power P_r and the ohmic losses P_l.
(eta_r = frac{P_r}{P_{in}} = frac{P_r}{P_r+P_l} = frac{R_r I_{rms}^2}{I_{rms}^2 R_r+I_{rms}^2 R_l} = frac{R_r}{R_r+R_l} )

8. For a lossless antenna, maximum Power gain equals to the maximum directive gain.
a) True
b) False
Answer: a
Clarification: 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.

250+ TOP MCQs on Antenna Array Introduction and Answers

Antennas Multiple Choice Questions on “Introduction”.

1. Which of the following is false regarding Antenna array?
a) Directivity increases
b) Directivity decreases
c) Beam width decreases
d) Gain increases
Answer: b
Clarification: A single antenna provides low gain and less directivity. To increase the directivity antenna arrays are used. With the antenna arrays, directivity and gain increases and beam width decreases.

2. Electrical size of antenna is increased by which of the following?
a) Antenna Array
b) Decreasing the coverage area
c) Increasing the coverage area
d) Using a single antenna
Answer: a
Clarification: To increase the directivity antenna arrays are used. With the antenna arrays, directivity and gain increases and beam width decreases. The electrical size of the antenna is increased by placing an array antenna together to achieve high directivity.

3. For long distance communication, which of the property is mainly necessary for the antenna?
a) High directivity
b) Low directivity
c) Low gain
d) Broad beam width
Answer: a
Clarification: Long distance communication requires antenna with high directivity. To increase the directivity antenna arrays are used. With the antenna arrays, directivity and gain increases and beam width decreases.

4. Which of the following is false about the single antenna for long distance communication?
a) Enlarging may create side lobes
b) No side lobes
c) High directivity is required
d) High Gain is required
Answer: b
Clarification: High directive antennas are required for the long distance communications. The array of antennas is used to increase the directivity. The directivity can be increased by increasing the dimensions of antenna but it creates side lobes.

5. The electrical size of antenna is increased by antenna array to avoid size lobes compared to single antenna.
a) True
b) False
Answer: a
Clarification: Increasing the dimensions of antennas may lead to the appearance of the side lobes. So by placing a group of antennas together the electrical size of antenna can be increased. With the antenna arrays, directivity and gain increases and beam width decreases.

6. A uniform linear array contains _____________
a) N elements placed at equidistance and fed currents of equal magnitude and progressive phase shift
b) N elements at non-equidistance and fed currents of equal magnitude and progressive phase shift
c) N elements at equidistance and fed currents of unequal magnitude and progressive phase shift
d) N elements at equidistance and fed currents of unequal magnitude and equal phase shift
Answer: a
Clarification: An array is said to be linear if N elements are spaced equally long the line and is a uniform array if the current is fed with equal magnitude to all elements and progressive phase shift along the line. High directivity can be obtained by antenna array.

7. Total resultant field obtained by the antenna array is given by which of following?
a) Vector superposition of individual field from the element
b) Maximum field from individual sources in the array
c) Minimum field from individual sources in the array
d) Field from the individual source
Answer: a
Clarification: The total resultant field is obtained by adding all the fields obtained by the individual sources in the array. An Array containing N elements has the resultant field equal to the vector superposition of individual field from the elements.

8. If the progressive shift in antenna array is equal to zero then it is called _________
a) Broad side
b) End-fire
c) Yagi-uda
d) Fishbone antenna
Answer: a
Clarification: The total phase difference of the fields is given by Ѱ=kdcosθ+β
Here β is the progressive phase shift
⇨ β=0, array is a uniform broadside array
⇨ β=180, array is a uniform end-fire array
Yagi-uda antenna, fishbone antenna are end-fire antenna array.

9. What is the progressive phase shift of the end-fire array?
a) 0
b) 90
c) 180
d) 60
Answer: c
Clarification: The progressive phase shift of the end-fire array is 180°. It is a linear array whose direction of radiation is along the axis of the array. For a broadside array it is 0°.

10. Which of the following statement about antenna array is false?
a) Field pattern is the product of individual elements in array
b) Field pattern is the sum of individual elements in array
c) Resultant field is the vector superposition of the fields from individual elements in array
d) High directivity can be achieved for long distance communications
Answer: b
Clarification: The total resultant field is obtained by adding all the fields obtained by the individual sources in the array. Radiation pattern is obtained by multiplying the individual pattern of the element. Field pattern is the product of individual elements in array. Antenna arrays are used to get high directivity with less side lobes.