Category: Antennas Objective Questions

250+ TOP MCQs on Aperture Antenna – Beamwidths and Answers

Antennas written test Questions & Answers on “Aperture Antenna – Beamwidths”.

1. Half-power Beamwidth is given by ____
a) 70λ/D
b) 70D/λ
c) 35λ/D
d) 35D/λ
Answer: a
Clarification: The area the power is radiated is given by Beamwidth. Half power Beamwidth is the area at which the power is radiated 50% of peak power. The half-power beamwidth is given by70λ/D.

2. If the antenna dimension is two times the wavelength of the signal then the half power beam width will be _____
a) 35
b) 140
c) 70
d) 280
Answer: a
Clarification: The half-power beamwidth is given by70λ/D.
⇨ (frac{70lambda}{D}=frac{70}{2}=35.)

3. For a circular aperture the FNBW is ______
a) 140λ/D
b) 70λ/D
c) 140D/λ
d) 70D/λ
Answer: a
Clarification: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. For circular aperture the FNBW is given by 140λ/D. The half-power beam width is given by 70λ/D.

4. If the antenna dimension is two times the wavelength of the signal then the First null beam width will be _____
a) 35
b) 140
c) 70
d) 280
Answer: c
Clarification: The first null beam-width is given by 140λ/D.
⇨ (frac{140lambda}{D} = frac{140}{2}=70.)

5. For a rectangular aperture of a*b the first null in E-plane occur at _______
a) sin-1⁡(λ/b)
b) sec-1⁡(λ/b)
c) cos-1⁡(λ/a)
d) sin-1⁡(λ/a)
Answer: a
Clarification: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. (frac{kb}{2}) sinθ=nπ
⇨ θ= sin-1⁡(nλ/b)
⇨ For first null n=1 θ= sin-1⁡(λ/b).

6. The first null beam width in the E-plane of a rectangular aperture of a×b is given by _______________
a) 2sin-1⁡(λ/b)
b) sin-1⁡(λ/a)
c) 2sec-1⁡(λ/b)
d) 2cos-1⁡(λ/a)
Answer: a
Clarification: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. (frac{kb}{2}) sinθ=nπ
θ= sin-1(⁡(frac{nlambda}{b}))
Therefore, the FNBW in E-plane is given by FNBW=2 θ = sin-1⁡(⁡(frac{lambda}{b})).

7. Larger the size of the aperture, the narrower is the Beam-widths.
a) True
b) False
Answer: a
Clarification: The FNBW in E-plane is given by FNBW=2 θ = sin-1⁡(⁡(frac{nlambda}{b})). As the dimension of the antenna aperture increases, the FNBW will decrease. Thereby, beam-width becomes narrower.

8. Half-power Beam width in E-plane for a rectangular aperture antenna of a×b is given by ____
a) 0.886λ/b
b) 0.443λ/b
c) 0.5λ/b
d) λ/b
Answer: a
Clarification: By equating the field in E-plane to half power point
(frac{sin⁡(0.5kbsintheta)}{0.5kbsintheta} = frac{1}{sqrt 2} => theta = arcsin⁡(frac{0.443
lambda}{b}))
Now HPBW = 2 arcsin⁡((frac{0.443lambda}{b}))≈0.886λ/b.

9. Find the HPBW of the uniform rectangular aperture antenna with 4λ×2λ in the E-plane?
a) 0.443
b) 0.886
c) 0.25
d) 0.5
Answer: a
Clarification: HPBW = 2 arcsin⁡((frac{0.443lambda}{b}))≈0.886λ/b=0.886/2=0.443.

10. The value at which the second null occurs in H-plane of rectangular aperture of a*b is given by ____
a) sin-1⁡(2λ/a)
b) sin-1⁡(λ/a)
c) sin-1⁡(a/2λ)
d) sin-1⁡(a/λ)
Answer: a
Clarification: The area the power is radiated is given by Beam-width. The beam-width between the first nulls is the FNBW. For H-plane (frac{ka}{2}) sinθ=nπ
⇨ θ= sin-1⁡(nλ/a)
⇨ For null n=2 θ= sin-1⁡(2λ/a).

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250+ TOP MCQs on Directivity and Answers

Antennas Multiple Choice Questions on “Directivity”.

1. The ratio of maximum power density in the desired direction to the average power radiated from the antenna is called as _______
a) directivity
b) directive gain
c) power gain
d) partial directivity
Answer: a
Clarification: The ratio of maximum power density in the desired direction to the average power radiated from the antenna is called Directivity. The ratio of power radiated in a particular direction to the actual power input to antenna is called Power gain. Directive gain is the ratio of power radiated in desired direction to the average power radiated from the antenna. Maximum Directive gain is called as Directivity. Partial directivity is the part of radiation intensity in a particular polarization to radiation intensity in all directions.

2. What is the Beam area for Directivity to be 1 in Steradian?
a) 4π
b) 1/2π
c) 2π
d) 1/4π
Answer: a
Clarification: The Directivity in terms of the Beam area ΩA is given by D=(frac{4pi}{Omega_A})
⇨ (frac{4pi}{D})=4π steradians.

3. If directivity of antenna increases, then the coverage area _____
a) decreases
b) increases
c) increases and then decreases
d) remains unchanged
Answer: a
Clarification: As the directivity increases, the beam area decreases. So, the coverage area of beam decreases and takes more time to scan a target for target detection.

4. If half power beam width in one plane and other plane orthogonal to it are equal to π then the directivity is ____
a) π
b) 4π
c) 4/π
d) 2π
Answer: c
Clarification: Beam area ΩA≈ θ1r θ2r where θ1r, θ2r are half-power beam widths in radians.
Directivity (D=frac{4pi}{Omega_A}=frac{4pi}{theta_{1r}theta_{2r}}=frac{4pi}{pi.pi} = 4/pi)

5. Directive gain with maximum radiation intensity is called as Directivity.
a) True
b) False
Answer: a
Clarification: Directivity is defined as ratio of maximum power density in desired direction to the average power radiated in all directions. This is simply, maximum Directive gain. Maximum Directive gain is obtained if maximum radiation intensity is present in desired direction.

6. How the directivity and effective aperture related to each other?
a) Inversely proportional
b) Directly proportional
c) Independent
d) Proportionality depends on input power
Answer: b
Clarification: The directivity D is directly proportional to the effective aperture of antenna. (A_e=Dfrac{lambda^2}{4pi})

7. What is the directivity of half-wave dipole?
a) 1.64
b) 1.5
c) 1.43
d) 1.44
Answer: a
Clarification: Directivity (D=frac{U_{max}}{U_{av}} = frac{4pi U_{max}}{P_{rad}} )
Prad=36.54I02 Since the radiation resistance for Half-wave dipole is 36.54Ω.
Maximum radiation intensity is given by (U_{max}=P_{max}.r^2=frac{mid E_thetamid_{max}^2}{2eta}.r^2)
Where (E_theta=frac{jeta I_0 e^{-jkr}cos⁡(frac{pi costheta}{2})}{2pi rsintheta})
⇨ (D=frac{4pi}{1}frac{eta I_0^2}{8pi^2} frac{1}{36.54I_0^2}=1.64.)

8. What is the directivity of antenna having effective aperture 1 m2?
a) (frac{4pi}{lambda^2})
b) (frac{lambda^2}{4pi})
c) 1
d) 4π
Answer: a
Clarification: Directivity D=(frac{4pi}{lambda^2}A_e)
Given effective aperture Ae=1m2
⇨ D=(frac{4pi}{lambda^2}A_e = frac{4pi}{lambda^2} × 1 =frac{4pi}{lambda^2})

250+ TOP MCQs on Applications and Answers

Antennas Multiple Choice Questions on “Applications”.

1. In mobiles, which of the following antenna is widely used?
a) Microstrip antenna
b) Horn antenna
c) Yagi-Uda antenna
d) Lens antenna
Answer: a
Clarification: Microstrip antenna is a printed antenna. It is fabricated using photolithography technique. It is small in size and is used in mobile communication widely. Horn, Yagi-Uda and Lens antennas are large in size.

2. One of the advantages of the Microstrip antenna compared to conventional microwave antenna is _________
a) Small size
b) Low gain
c) No surface wave excitation
d) High gain
Answer: a
Clarification: Compared to conventional microwave antennas like wire, lens, reflector antennas, the Microstrip antenna is small in size. It has low gain and has surface wave excitation which is a disadvantage.

3. In Microstrip antennas, the feed line and matching networks cannot be fabricated separated separately
a) False
b) True
Answer: a
Clarification: Microstrip antenna is a printed antenna. It consists of a radiation patch on one side of dielectric and a ground plane on other side. The patch is connected to the feed line. In this at fabrication only, the feed line and matching networks are fabricated simultaneously.

4. Which of the following is the application of Microstrip antenna in a telemedicine industry?
a) Wireless Body Area Network
b) Detection of moving targets
c) Rectenna application
d) WiMax
Answer: a
Clarification: A 2.4 GHz Wearable Microstrip antenna is used for the Wireless Body Area Network. Detection of moving targets comes under radar application. Rectenna is a special rectifying antenna used to converts directly the microwave energy to the DC power. WiMax is an IEEE 802.16 standard used for communication.

5. Rectenna Application of Microstrip antenna converts _______________
a) Microwave energy to DC power
b) Microwave energy to AC power
c) Microwave energy to solar energy
d) Inductive to capacitive
Answer: a
Clarification: Rectenna is a special rectifying antenna used to converts directly the microwave energy to the DC power. This is used in the long distance links. It is a combination of antenna, rectification filters and rectifiers.

6. One of the disadvantages of the Microstrip antenna is excitation of surface waves
a) True
b) False
Answer: a
Clarification: Microstrip antenna is a printed antenna. It consists of a radiation patch on one side of dielectric and a ground plane on other side. It has low gain and has surface wave excitation which is a disadvantage.

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250+ TOP MCQs on Radiation from Rectangular Aperture and Answers

Antennas Question Paper on “Radiation from Rectangular Aperture”.

1. Which of the following antenna belongs to rectangular aperture?
a) Horn antenna
b) Helical antenna
c) Parabolic antenna
d) Conical antenna
Answer: a
Clarification: The aperture of antenna at the end determines its shape. Horn antenna has a rectangular aperture. Helical, Parabolic, conical antenna have circular apertures. Helical antenna belongs to frequency independent antenna.

2. The radiation pattern of rectangular is similar to line source integrated in two directions
a) True
b) False
Answer: a
Clarification: The radiation pattern for the rectangular distributions is similar to the line source distributions. In this the patterns is calculated by integration in two directions as rectangle have two different lengths in different direction.

3. The total pattern function for rectangular aperture f(x, y) if f(x) and f(y) are separable is given by ____
a) f(x, y)=f(x) f(y)
b) f(x, y)=f(x)+f(y)
c) f(x, y)=f(x)/f(y)
d) f(x, y)=f(x)-f(y)
Answer: a
Clarification: The radiation pattern for the rectangular aperture is likely relatable to the line source distributions. If the functions f(x) and f(y) are separable, then total pattern will be the product of the two functions. f(x, y)=f(x)f(y).

4. The first-level of the side lobe occurs at ______ dB for a uniform rectangular aperture antenna.
a) -13.26
b) -6.63
c) 3
d) 8.5
Answer: a
Clarification: The aperture of antenna at the end determines its shape. If the field is uniform in amplitude and phase along the rectangular aperture then it is called a uniform rectangular aperture antenna. The first side-lobe occurs at -13.26dB.

5. A rectangular aperture a ×b is placed in xy-plane, The HPBW in H-plane is given by _____
a) 0.886λ/a
b) 0.443λ/a
c) 0.5λ/b
d) λ/b
Answer: a
Clarification: By equating the field in H-plane to half power point
(frac{sin⁡(0.5kasintheta)}{0.5kasintheta} = frac{1}{sqrt{2}} => theta =arcsin⁡(frac{0.443lambda}{a}))
Now HPBW = (2 arcsin⁡(frac{0.443lambda}{a}) approx 0.886lambda/a.)

6. The relation between directivity and the effective aperture of the uniform aperture antenna is given by _____
a) (D = frac{4pi}{lambda^2}A_{eff})
b) (A_{eff} = frac{4pi}{lambda^2}D)
c) (A_{eff} = frac{4pilambda^2}{D})
d) (D = frac{4pilambda^2}{A_{eff}})
Answer: a
Clarification:For a uniform aperture antenna, the physical and effective apertures are equal.
The relation between directivity and the effective aperture of the antenna is given by
(D = frac{4pi}{lambda^2}A_{eff}).

7. Which of the following is used to reduce side lobe levels in aperture antenna?
a) Tapering
b) Increasing the power
c) Using repeaters
d) Reducing power
Answer: a
Clarification: The uniform aperture produces the high SLL under a constant phase amplitude excitation. To reduce this SLL effect, tapering is done. Tapering is done maximum at center and reduces to zero at the edges for anequivalent source distribution.

8. The principle plane pattern of the E-plane of rectangular aperture of a×b is given by F(θ) = ______
a) (frac{sin⁡(0.5kbsintheta)}{0.5kbsintheta})
b) (frac{cos⁡(0.5kbcostheta)}{0.5kbsintheta})
c) (frac{sec⁡(0.5kbsintheta)}{0.5kbsintheta})
d) (frac{1-sin⁡(0.5kbsintheta)}{0.5kbsintheta})
Answer: a
Clarification: The principle plane patterns for the uniform rectangular aperture antenna is given by
F(θ) = (frac{sin⁡(0.5kbsintheta)}{0.5kbsintheta}) and main lobe occurs when θ =0.

9. Tapering is done in order to reduce the side lobe level.
a) True
b) False
Answer: a
Clarification: The uniform aperture produces the high SLL under a constant phase amplitude excitation. To reduce this SLL effect, tapering is done. Tapering is done maximum at center and reduces to zero at the edges for an equivalent source distribution.

10. If the aperture antenna is tapered only in H-plane then which of the following is true compared to uniform non-tapered aperture antenna?
a) Principle patterns in E-plane and H-plane are same in both cases
b) Principle patterns in E-plane and H-plane are different in both cases
c) Principle patterns in E-plane is same and H-plane is different
d) Principle patterns in E-plane is different and H-plane is same
Answer: c
Clarification: Tapering is done to reduce the SLL effect. Since the E-plane is not tapered, its principle pattern is same but in H-plane as the aperture is tapered principle pattern will be different from the uniform aperture antenna.

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250+ TOP MCQs on Effective Aperture and Answers

Antennas Multiple Choice Questions on “Effective Aperture”.

1. Effective aperture is the ability of antenna to extract energy from the electromagnetic wave.
a) True
b) False
Answer: a
Clarification: Effective aperture is defined as the ratio of power received from load to the average power density produced at that point. So it is the ability of antenna to extract energy from EM wave.

2. Which of the following best describes the condition for Maximum effective aperture?
a) Load impedance must be equal to the antenna impedance
b) Load impedance must be equal complex conjugate to the antenna impedance
c) Receiver power should be minimum
d) Transmitter power should be minimum
Answer: b
Clarification: Foe effective aperture to be maximum, the receiver power should be maximum. Maximum power transfer states that the maximum received power is obtained when the impedance of network matches with the complex conjugate of the load impedance. Hence, Load impedance must be equal to the complex conjugate of the antenna impedance.

3. What is the effective aperture of Hertzian dipole antenna operating at frequency 100 MHz?
a) 1.07m2
b) 0.17m2
c) 1.7m2
d) 1.2m2
Answer: a
Clarification: Effective aperture for a Hertzian dipole is given by (A_e=1.5frac{lambda^2}{4pi} )
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}{4pi} =1.5frac{3^2}{4pi} =1.07m^2)

4. If physical aperture of antenna is 0.02m2 and aperture efficiency is 0.5, then what is the value of effective aperture?
a) 0.0004m2
b) 0.001m2
c) 0.01m2
d) 25m2
Answer: c
Clarification: Effective aperture Aem=Aeη=0.02×0.5=0.01 m2

5. Expression for aperture efficiency in terms of physical aperture Ae and effective aperture Aem is ____
a) (frac{A_e}{A_{em}})
b) (frac{A_{em}}{A_e})
c) (frac{A_e+A_{em}}{A_e-A_{em}})
d) (frac{A_e-A_{em}}{A_e+A_{em}})
Answer: b
Clarification: The aperture efficiency is defined as the ratio of effective aperture to physical aperture of antenna. So aperture efficiency (eta=frac{A_{em}}{A_e}.)

6. What is the effective aperture of a Half-wave dipole operating at 100MHz?
a) 1.07m2
b) 1.17m2
c) 1.27m2
d) 1.77m2
Answer: b
Clarification: The directivity of half-wave dipole is 1.64
The effective aperture of half-wave dipole is (A_e=1.64frac{lambda^2}{4pi})
(lambda=frac{c}{f}=frac{3×10^8}{100×10^6}=3m)
(A_e=1.64 frac{lambda^2}{4pi} = 1.64 frac{3^2}{4pi})=1.17m2.

7. What is the relation between effective length and Effective aperture of antenna?
a) (A_e = frac{dL^2eta}{4R_{rad}})
b) (A_e = frac{dL^2}{4eta R_{rad}})
c) (A_e = frac{dL^2 R_{rad}}{4eta})
d) (A_e = frac{dL^2 eta ^2}{4R_{rad}})
Answer: a
Clarification: Maximum effective aperture is ratio of maximum power received to the average power density. (A_e=frac{P_{Rmax}}{P_{avg}})
The received power is maximum when load equal to complex conjugate of network resistance.
⇨ (I_{total}=frac{V_{oc}}{2R_{rad}})
⇨ (P_{Rmax}=I_{rms}^2 R_{rad}= (frac{I_{total}}{sqrt 2})^2 R_{rad}=frac{V_{oc}^2}{8R_{rad}}=frac{mid E_thetamid^2 dL^2}{8R_{rad}} and P_{avg}=frac{mid E_thetamid^2}{2eta} )
(A_e=frac{P_{Rmax}}{P_{avg}}=frac{dL^2eta}{4R_{rad}})

8. The physical aperture of an isotropic radiator is _______
a) (frac{4pi eta}{lambda^2})
b) (frac{4pi}{lambda^2 eta})
c) (frac{lambda^2}{4pi eta})
d) (frac{lambda^2eta}{4pi})
Answer: c
Clarification: For isotropic radiator, directivity is 1. So the effective aperture is given by (A_{em}=D frac{lambda^2}{4pi} = frac{lambda^2}{4pi})
Then physical aperture =(frac{Effective, aperture}{Aperture, efficiency}=frac{lambda^2}{4pi eta})

250+ TOP MCQs on Feed Methods and Answers

Antennas Multiple Choice Questions on “Feed Methods”.

1. Which of the following is used to excite to radiate the antenna?
a) Feed line
b) Ground plane
c) Patch
d) Substrate
Answer: a
Clarification: The feedline is sued to excite to radiate the antenna in direct or indirect way. There are mainly four different types of feeding in Microstrip antenna: Microstrip line feed, coaxial probe, aperture coupling, proximity coupling.

2. Which of the following fed line is connected to patch just like a conducting strip?
a) Microstrip line feed
b) Coaxial feed
c) Aperture coupling
d) Proximity Coupling
Answer: a
Clarification: Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane. In Aperture coupling, two different substrates are separated by ground plane. In Proximity coupling, the length of stub and length to width ratio of patch controls the match.

3. Which of the following feed line contains two different substrates separated by a ground plane?
a) Microstrip line feed
b) Coaxial feed
c) Aperture coupling
d) Proximity Coupling
Answer: c
Clarification: In Aperture coupling, the ground plane separates two different substrates. In Microstripline feed, the feed is connected to the patch like a conducting strip. The coaxial feed consists an inner conductor attached to the patch and ground plane to the outer conductor. The length of stub and length to width ratio of patch is used for matching in proximity coupling.

4. Which of the following is true regarding the coaxial coupling feed in Microstrip antenna?
a) Inner conductor is connected to patch and outer conductor to ground plane
b) Outer conductor is connected to patch and inner conductor to ground plane
c) It is connected to patch just like a conducting strip
d) It contains two different substrates separated by a ground plane
Answer: a
Clarification: A coaxial feed consists of inner and outer conductors to which the patch and ground planes are attached respectively. In a Microstrip feed line the patch is attached to the feed like a conducting strip. Different substrates are used in aperture coupling.

5. Which of the following feed line uses the length of stub and L/W ratio to control the match?
a) Microstrip line feed
b) Coaxial feed
c) Aperture coupling
d) Proximity Coupling
Answer: d
Clarification: In Proximity coupling, the length of stub and length to width ratio of patch controls the match. Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane. In Aperture coupling, ground plane separates the different substrates.

6. Which of the following is the disadvantage of Microstrip line feeding?
a) Spurious feed radiation increases with increase in substrate thickness
b) Spurious feed radiation decreases with increase in substrate thickness
c) There is no Bandwidth limit
d) Low spurious radiation
Answer: a
Clarification: Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. So the substrate thickness may increases leading to the increase in spurious radiation and this limits the bandwidth.

7. Which of the following is the disadvantage of the coaxial feeding?
a) Fabrication is very difficult
b) Low spurious radiation
c) Narrow Bandwidth
d) No generation of higher order modes
Answer: c
Clarification: In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane. Coaxial feeding fabrication is simple and has low spurious radiation which is its advantage. Due to asymmetries, higher modes are generated and produces cross polarization radiation.

8. Which of the following allows independent optimization of feed mechanism?
a) Microstrip line feed
b) Coaxial feed
c) Aperture coupling
d) Proximity Coupling
Answer: c
Clarification: In Aperture coupling, two different substrates are separated by ground plane. This type of arrangement optimizes feed mechanism and radiating element.In Proximity coupling, the length of stub and length to width ratio of patch controls the match. Microstrip line feed acts as an extent to patch and is connected to patch just like a conducting strip. In coaxial feed, inner conductor is attached to patch and outer conductor to ground plane.

9. Which of the following feeding has largest bandwidth?
a) Microstrip line feed
b) Coaxial feed
c) Aperture coupling
d) Proximity Coupling
Answer: d
Clarification: Proximity coupling feed provides largest bandwidth. Microstrip line feed limits the bandwidth due to increase in substrate thickness. Coaxial feed has narrow bandwidth.

10. In aperture coupling, ground plane is used to minimize the inference due to spurious radiation.
a) True
b) False
Answer: a
Clarification: In Aperture coupling, two different substrates are separated by ground plane. This type of arrangement optimizes feed mechanism and radiating element. The ground plane is between the substrates and isolates feed from the radiating element and minimizes the interference due to spurious radiation.