300+ REAL TIME Antennas Objective Questions & Answers 2023

250+ TOP MCQs on Polarization Mismatch and Answers

Antennas Multiple Choice Questions on “Polarization Mismatch”.

1. What is the polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅?
a) tan⁡∅
b) cos⁡∅
c) cos²∅
d) cos^-1⁡∅
Answer: c
Clarification: The polarization loss factor describes the power loss due to polarization mismatch. The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF =cos²∅.

2. What is the polarization loss factor when a linear polarized antenna receives a circular polarized?
a) 0.5
b) 1
c) 0
d) 2
Answer: a
Clarification: Circular polarization is combination of two orthogonal linear polarized waves at 90° phase difference. The linear component just selects one in-phase component from this so ∅=45
⇨ PLF = cos²∅.=1/2=0.5

3. What is the polarization loss factor when two antennas are rotated by an angle 30°?
a) 0.75
b) 0.25
c) 1
d) 0
Answer: a
Clarification: The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos²∅
⇨ PLF = cos²30=(frac{3}{4}=0.75)

4. Which of the following is best described by the polarization loss factor?
a) Power loss due to mismatch
b) Height of antenna
c) Directivity
d) Power gain
Answer: a
Clarification: The polarization loss factor describes the power loss due to polarization mismatch. The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF =cos²∅.

5. The polarization of received antenna is not same as incident wave polarization is termed as ___
a) polarization mismatch
b) polarization loss factor
c) directivity
d) transmitter
Answer: a
Clarification: The polarization of received antenna is not same as incident wave polarization is termed as Polarization mismatch. The polarization loss factor describes the power loss due to polarization mismatch. PLF =cos²∅.

6. Which of the following is the condition for no power loss?
a) PLF=1
b) PLF=0
c) PLF=0.5
d) PLF=0.75
Answer: a
Clarification: Polarization loss factor describes the power loss. When PLF = 1 = 0dB indicates total power incident is received by the antenna. Therefore the condition for the no power loss is PLF=1

7. Which of the following limit is correct for the polarization loss factor?
a) 0≤PLF≤1
b) -1≤PLF≤1
c) -1≤PLF≤0
d) 0≤PLF≤∞
Answer: a
Clarification: The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos²∅.
⇨ Max value of PLF = 1
⇨ Min value of PLF = 0
⇨ 0≤PLF≤1

8. Which of the following holds true for complete polarization mismatch?
a) PLF=1
b) PLF=0
c) PLF=0.5
d) PLF=0.75
Answer: b
Clarification: Polarization loss factor describes the power loss. When PLF=0=∞ dB, the receiver antenna doesn’t receive any incident power so there will be a complete mismatch.

9. Polarization loss factor describes the amount of power loss due to polarization mismatch at the receiving antenna from incident wave.
a) True
b) False
Answer: a
Clarification: Polarization loss factor describes the amount of power loss due to polarization mismatch at the receiving antenna from incident wave. The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos²∅.

10. Polarization loss factor in dB when two linear polarized antennas are rotated by an angle 45° is ___
a) 3dB
b) -3dB
c) 0.5dB
d) -0.5dB
Answer: b
Clarification: The polarization loss factor due to mismatch if the two linear polarized antennas are rotated by an angle ∅ is given by PLF = cos²∅.
⇨ PLF = cos²∅=cos² 45=0.5=-3dB.

250+ TOP MCQs on Antenna Noise Temperature and Answers

Antennas Multiple Choice Questions on “Antenna Noise Temperature”.

1. Relation between brightness temperature T_B and physical body temperatureT_p is ____
a) T_B=((1-midGamma_s mid^2) T_p)
b) T_B=(T_p/(1-midGamma_s mid^2))
c) T_B=((1-midGamma_s mid)T_p)
d) T_B=((1-midGamma_s mid)^2 T_p)
Answer: a
Clarification: The relation between brightness temperature and the physical body temperature is given by T_B=((1-midGamma_s mid^2) T_p)
Here Γ_s is the reflection coefficient for a given polarization and emissivity =(1-midGamma_s mid^2.)

2. If the reflection co-efficient is ½ then emissivity is ___
a) 3/4
b) 1/4
c) 1/2
d) 3/2
Answer: a
Clarification: Emissivity in terms of reflection coefficient is given by (epsilon=1-midGamma_smid^2=1-frac{1}{4}=frac{3}{4}.)

3. Overall receiver noise temperature expression if T₁, T₂… are amplifier 1, 2, and so on noise Temperature and G₁, G₂, and so on are their gain respectively is_____
a) T = (T_1+frac{T_2}{G_1}+frac{T_3}{G_1 G_2}+⋯ )
b) T = T₁+T₂ (1-G₁)+T₃(1-G₁G₂)+⋯
c) T = (T_1+frac{T_2}{(1-G_1)}+frac{T_3}{(1-G_1 G_2)}+⋯)
d) T = T₁+T₂ (G₁)+T₃(G₁G₂)+⋯
Answer: a
Clarification: Overall receiver noise temperature expression is given by T = (T_1+frac{T_2}{G_1}+frac{T_3}{G_1 G_2}+⋯ )
System Temperature is one of the important factors to determine the antenna sensitivity and SNR.

4. Total noise power of the system is P=_____
a) k(T_A+T_R)B
b) k(T_A+T_R)/B
c) k(T_R)B
d) kB/T_sys
Answer: a
Clarification: The overall noise temperature of the system is the sum of noise temperature of antenna T_A and the receiver surrounding T_R.
⇨ Total noise power of the system is P= k(T_A+T_R)B
⇨ K is Boltzmann’s constant and B is the bandwidth

5. What is the relation between noise temperature introduced by beam T_B and the antenna temperature T_A when the solid angle obtained by the noise source is greater than antenna solid angle?
a) T_A= T_B
b) T_A > T_B
c) T_A < T_B
d) T_A « T_B
Answer: a
Clarification: When the solid angle obtained by the noise source Ω_B is greater than antenna solid angle Ω_A, then relation between noise temperature introduced by beam T_B and the antenna temperatureT_A is given by
T_A= T_B (If Lossless antenna).
For radio astronomy, Ω_B<Ω_A and T_A≠ T_B; ΔT_A=(frac{Omega_B}{Omega_A}T_B)

6. Which expression suits best when the solid angle obtained by the noise source is less than antenna solid angle?
a) P_A Ω_A=P_B Ω_B and ΔT_A=(frac{Omega_B}{Omega_A} T_B)
b) P_A Ω_B=P_B Ω_A and ΔT_A=(frac{Omega_B}{Omega_A} T_B)
c) ΔT_A=(frac{Omega_A}{Omega_B} T_B) and P_A Ω_B=P_B Ω_A
d) ΔT_A=(frac{Omega_A}{Omega_B} T_B) and P_A Ω_A=P_B Ω_B
Answer: a
Clarification: When the solid angle obtained by the noise sourceΩ_B is greater than antenna solid angle Ω_A, then relation between noise temperature introduced by beam T_B and the antenna temperatureT_A is given by
T_A= T_B (If Lossless antenna).
For radio astronomy, Ω_B<Ω_A and T_A≠ T_B; ΔT_A=(frac{Omega_B}{Omega_A} T_B) and P_A Ω_A=P_B Ω_B

7. Expression for noise figure F related to the effective noise temperature T_e is ____
a) (F=1+frac{T_e}{T_o})
b) (F=1+frac{T_0}{T_e})
c) (F=1-frac{T_e}{T_o})
d) (F=1-frac{T_0}{T_e})
Answer: a
Clarification: The noise introduced by antenna is known as the effective noise temperature. The relation between noise figure and effective noise temperature is given by
(F=1+frac{T_e}{T_o}, T_o) is the room temperature.

8. Effective noise temperature T_e in terms of noise figure is ____
a) T_e=T_o (F-1)
b) T_e=T_o/(F-1)
c) T_e=T_o/(F+1)
d) T_e=T_o (F+1)
Answer: a
Clarification: The relation between noise figure and effective noise temperature is given by F=1+(frac{T_e}{T_o})
⇨ F-1=(frac{T_e}{T_o})
⇨ T_e=T_o (F-1)

9. Which of the following statement is false?
a) Noise power of antenna depends on the antenna temperature as well as the noise due to the receiver surroundings
b) Noise figure value lies between 0 and 1
c) Any object with physical temperature greater than 0K radiates energy
d) Noise power per unit bandwidth is kT_A W/Hz
Answer: b
Clarification: The relation between noise figure and effective noise temperature is given by F=1+(frac{T_e}{T_o})
And object with physical temperature greater than 0K radiates energy. So (frac{T_e}{T_o}) > 0 and F > 1
Noise power per unit bandwidth of antenna is kT_A W/Hz while noise power of antenna is kT_AB W

10. Find the effective noise temperature if noise figure is 3 at room temperature (290K)?
a) 290K
b) 580K
c) 289K
d) 195K
Answer: b
Clarification: Room temperature T_o=290K
Noise figure F=1+(frac{T_e}{T_o})
T_e=T_o(F-1)=290(3-1)=580K.

11. What should be the noise figure value at which the effective noise temperature equals to room temperature?
a) 2
b) 1
c) 0
d) 1/T_(o )
Answer: a
Clarification: Noise figure F=1+(frac{T_e}{T_o})
T_e=T_o(F-1)
F-1=1
F=2.

250+ TOP MCQs on Aperture Blockage and Answers

Reflector Antenna online quiz on “Aperture Blockage”.

1. Which of the following is used to avoid the aperture blockage due to secondary reflector?
a) Offset feed system
b) Cassegrain
c) Plane reflector
d) Parabolic antenna
Answer: a
Clarification: Offset feed system is used to avoid the aperture blocking effect due to the dependence of secondary reflector. In this a feed radiator is placed at the focus in an inclined position such that all rays are collimated without formation of blockage region. In Cassegrain, plane reflector there is a blocking effect.

2. Which of the following causes aperture blockage in a Cassegrain antenna?
a) Main reflector
b) Sub-reflector
c) Feed radiator
d) Noise from surroundings
Answer: b
Clarification: The sub-reflector causes the aperture blockage in a Cassegrain antenna. It can be reduced by reducing the size of the sub-reflector.

3. In which of the following the secondary reflector faces the primary reflector?
a) Cassegrain
b) Gregorian
c) Corner reflector
d) Plane reflector
Answer: b
Clarification: In Gregorian antenna, sub-reflector is ellipsoidal and has its concave side faces the primary reflector. In Cassegrain, sub-reflector is hyperboloid with its convex side facing the feed. Corner and plane reflectors are normal reflector antennas while Cassegrain and Gregorian are dual reflector antennas.

4. Which of the following statement is true regarding a Gregorian antenna?
a) Sub-reflector ellipsoidal lies closer to the focus of the paraboloid
b) Sub-reflector is a hyperboloid with its convex facing feed
c) Sub-reflector is an ellipsoidal with its concave facing reflector
d) Sub-reflector is an ellipsoidal with its concave facing feed
Answer: c
Clarification: Gregorian antenna is a dual reflector antenna having a concave secondary reflector. The secondary reflector also known as sub-reflector is a ellipsoidal. It lies beyond the focus of the paraboloid.

5. In which of the following the aperture blocking is reduced by permitting antenna to operate in a single polarization?
a) Truncated reflector
b) Cassegrain
c) Polarization-twist reflector
d) Plane reflector
Answer: c
Clarification: Aperture blocking is reduced by permitting the antenna to operate with a single polarization in Polarization-twist reflector. Truncated reflector is a type of parabolic reflector. Cassegrain is a dual reflector.

6. In which of the following a sub-reflector called trans-reflector is present?
a) Polarization-twist reflector
b) Plane reflector
c) Cassegrain
d) Gregorian
Answer: a
Clarification: It is present in Polarization-twist reflector. Sub-reflector consists of a horizontal grating of wires. Cassegrain contains a hyperboloid sub-reflector. Gregorian contains an ellipsoidal sub-reflector.

7. Which of the following is not used for aperture blocking?
a) Reducing size of sub-reflector
b) Offset feed system
c) High directive feed
d) Reducing spillover
Answer: c
Clarification: A high directive feed means a large feed. It partially shadows the primary reflector and becomes an obstacle for blockage. Aperture blocking can be avoided by reducing size of sub-reflector, thereby spillover and by using Offset feed system.

8. Which of the following condition holds good for minimum total aperture blocking in Cassegrain?
a) Area of sub-reflector and projected area of feed are equal
b) Area of primary reflector and projected area of feed are equal
c) Area of sub-reflector greater than area of primary reflector
d) Area of sub-reflector is less than projected area of feed
Answer: a
Clarification: Minimum total aperture blocking occurs when the area of sub-reflector and projected area of feed are approximately equal. In other conditions, the main ray back radiated ray gets added up with collinear rays from the primary reflector.

9. In Polarization-twist reflector, sub-reflector consists of a horizontal grating of wires.
a) True
b) False
Answer: a
Clarification: In Polarization-twist reflector, sub-reflector consists of a horizontal grating of wires called trans-reflector. In this the antenna is permitted to operate only at a single polarization. The polarized radiation is rotated by 90° by sub-reflector at the primary reflector with the help of twist reflector.

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250+ TOP MCQs on Types of Polarization and Answers

Antennas Multiple Choice Questions on “Types of Polarization”.

1. Which of the following polarization is used in monopole antenna?
a) Right-hand Circular
b) Linear
c) Depends on the feed
d) Left-hand Circular
Answer: b
Clarification: The linear polarization takes place in the monopole dipole. In the Dipole also a linear polarization takes place. Parabolic reflectors take the polarization of the feed. In helical, circular loop antennas we can find the circular polarization.

2. Which type of polarization is found in parabolic reflector?
a) Right-hand Circular
b) Linear
c) Depends on the feed
d) Left-hand Circular
Answer: c
Clarification: The parabolic antennas take the polarization of the feed. The linear polarization takes place in the monopole dipole. In the Dipole also a linear polarization takes place. In helical, circular loop antennas we can find the circular polarization.

3. In which of the following polarization the electric field components are perpendicular to each other and have equal magnitude?
a) Linear
b) Vertical
c) Circular
d) Elliptical
Answer: c
Clarification: In linear components are in the same plane. Vertical and horizontal are types in linear polarization. Circular polarization has electric field components are perpendicular to each other and have equal magnitude. In elliptic, the electric field components are perpendicular to each other and have unequal magnitude.

4. In which of the following polarization the electric field components are perpendicular to each other and have unequal magnitudes?
a) Linear
b) Vertical
c) Circular
d) Elliptical
Answer: d
Clarification: In linear components are in the same plane. Vertical and horizontal are types in linear polarization. Circular polarization has electric field components are perpendicular to each other and have equal magnitude. In elliptic, the electric field components are perpendicular to each other and have unequal magnitude.

5. Tilt angle of the elliptic polarization with respect to horizontal is ____________
a) τ=(tan^{-1}(frac{a_{vertical}}{a_{horizontal}}))
b) τ=(frac{a_{horizontal}}{a_{vertical}})
c) τ=((frac{a_{vertical}}{a_{horizontal}}))
d) τ=(tan^{-1}(frac{a_{horizontal}}{a_{vertical}}))
Answer: a
Clarification: The two properties of ellipse that relates to the polarization are — eccentricity and tilt or inclination angle with respect to horizontal. Tilt angle of the elliptic polarization is given by
τ=(tan^{-1}(frac{a_{vertical}}{a_{horizontal}}))

6. The locus traced by the extremity of the time-varying field vector at a fixed observational point is called ________
a) polarization
b) gain
c) directivity
d) height
Answer: a
Clarification: The locus traced by the extremity of the time-varying field vector at a fixed observational point is called polarization. Gain is the output to input power ratio. Directivity is the amount of power radiated in the desired direction.

7. Which of the following is true for the circular polarization?
a) E_x=E_y, and ∅=π/2
b) E_x≠E_y, and ∅=π/2
c) E_x≠E_y, and ∅=π/4
d) E_x=E_y, and ∅=π/4
Answer: a
Clarification: The locus traced by the extremity of the time-varying field vector at a fixed observational point is called polarization. Circular polarization has electric field components are perpendicular to each other and have equal magnitude. E_x=E_y, and ∅=π/2.

8. Which of the following is true for the elliptical polarization?
a) E_x=E_y, and ∅=π/2
b) E_x≠E_y, and ∅=π/2
c) E_x≠E_y, and ∅=π/4
d) E_x=E_y, and ∅=π/4
Answer: b
Clarification: The locus traced by the extremity of the time-varying field vector at a fixed observational point is called polarization. In elliptical polarization, the electric field components are perpendicular to each other and have unequal magnitude.
E_x≠E_y,and ∅=π/2

9. The transmission mode polarization vector and receiving mode polarization vector of antenna polarization are ___________
a) always equal
b) conjugate to each other
c) negative of conjugate of other
d) inverse of Conjugate of other vector
Answer: b
Clarification: The antenna polarization is defined by the polarization vector it transmits. In a common coordinate system, the transmission mode polarization vector is the conjugate of its receiving mode polarization vector.

10. The linear and circular polarizations are special cases of elliptical polarization.
a) True
b) False
Answer: a
Clarification: In Elliptical polarization, if the amplitude is made equal, then it becomes a circular polarization. If the phase difference of the two linear components is zero or nπ then it is becomes a linear polarization. Therefore linear and circular polarization is a special case of elliptical polarization.

250+ TOP MCQs on Power Radiation from Half Wave Dipole and Answers

Antenna Parameters Interview Questions and Answers for Experienced people on “Power Radiation from Half Wave Dipole”.

1. If the current input to the antenna is 100mA, then find the average power radiated from the half-wave dipole antenna?
a) 365mW
b) 0.356mW
c) 0.365mW
d) 356mW
Answer: a
Clarification: Average Power radiated from the half-wave dipole P_avg=(I_{rms}^2 R=(frac{I_m}{sqrt 2})^2 R )
Radiation resistance of a half-wave dipole is 73Ω.
Given I_m=100mA => P_avg=((frac{100×10^{-3}}{sqrt 2})^2×73=365mW.)

2. The average radiated power of half-wave dipole is given by ______
a) (73I_{rms}^2)
b) (36.5I_{rms}^2)
c) (13.25I_{rms}^2)
d) (146I_{rms}^2)
Answer: a
Clarification: Radiation resistance of a half-wave dipole is 73Ω.
Average Power radiated from the half-wave dipole (P_{avg}=I_{rms}^2 R=73I_{rms}^2)
Radiation resistance of a quarter-wave monopole is 36.5Ω.

3. If the power radiated by a quarter-wave monopole is 100mW then power radiated by a half wave dipole under same current input is _____
a) 100W
b) 100mW
c) 200W
d) 200mW
Answer: d
Clarification: Average Power radiated from the half-wave dipole (P_{avg-hlf}=I_{rms}^2 R_{hlf})
⇨ (frac{P_{avg-hlf}}{P_{avg, mono}} = frac{R_{hlf}}{R_{mono}} = frac{73}{36.5}=2) (under same current)
⇨ P_avg-hlf = 2P_{avg mono} = 2*100mW=200mW

4. Power radiated by a half wave dipole is how many times the power radiated by a quarter wave monopole under same input current to antennas?
a) 2
b) 3
c) 4
d) 1
Answer: a
Clarification: Average Power radiated from the half-wave dipole (P_{avg-hlf}=I_{rms}^2 R_{hlf})
⇨ (frac{P_{avg-hlf}}{P_{avg, mono}} = frac{R_{hlf}}{R_{mono}} = frac{73}{36.5}=2) (under same current)

5. Find the magnetic field if the electric field radiated by the half-wave dipole is 60mV/m?
a) 159μA/m
b) 195μA/m
c) 159mA/m
d) 195mA/m
Answer: a
Clarification: η=(frac{E}{H})
⇨ 120π=60m/H
⇨ H = 159μA/m

6. In which of the following the power is radiated through a complete spherical surface?
a) Half-wave dipole
b) Quarter-wave Monopole
c) Both Half-wave dipole & Quarter-wave Monopole
d) Neither Half-wave dipole nor Quarter-wave Monopole
Answer: a
Clarification: In a half-wave dipole the power is radiated in the entire spherical surface and in quarter wave monopole the power is radiated only through a hemispherical surface. Hence its radiation resistance is also twice that of the quarter wave monopole.

7. Find the power radiated from the half wave dipole at 2km away with magnetic field at point (theta=frac{pi}{2}) is 10μA/m ?
a) 0.576mW
b) 0.576W
c) 0.756W
d) 0.675W
Answer: b
Clarification: Magnetic field strength (H=frac{I_m}{2pi r}(frac{cos⁡(frac{pi}{2}costheta)}{sintheta}))
⇨ 10×10^-6=(frac{I_m}{2pi×2×10^3}(frac{cos⁡(frac{pi}{2}cosfrac{pi}{2})}{sinfrac{pi}{2}}))
⇨ I_m=0.125A
Now Average power radiated (P_{avg}=I_{rms}^2 R=(frac{I_m}{sqrt 2})^2 R=(frac{0.125}{sqrt 2})^2 ×73×=0.576W)

8. For the same current, the power radiated by half-wave dipole is four times that of the radiation by quarter wave monopole.
a) True
b) False
Answer: b
Clarification: The radiation resistance of a half wave dipole is 73Ω and that of a quarter wave monopole is 36.5Ω. So the power radiated by half-wave dipole is two times that of the radiation by quarter wave monopole.

9. If the power radiated by a half wave dipole is 100mW then power radiated by a quarter wave monopole under same current input is _____
a) 50mW
b) 200mW
c) 100mW
d) 50W
Answer: a
Clarification: Average Power radiated from the half-wave dipole (P_{avg-hlf}=I_{rms}^2 R_{hlf})
⇨ (frac{P_{avg-hlf}}{P_{avg, mono}} = frac{R_{hlf}}{R_{mono}} = frac{73}{36.5}=2) (under same current)
(P_{avg ,mono}=frac{P_{avg-hlf}}{2}=frac{100mW}{2}=50mW.)

10. If the power radiated by a quarter wave monopole is 100mW, then power radiated by a half wave dipole with doubled current is ______
a) 800mW
b) 400mW
c) 200mW
d) 100mW
Answer: a
Clarification: Average Power radiated from the half-wave dipole (P_{avg-hlf}=I_{rms}^2 R_{hlf}=4I_{rms}^2 R_{hlf})
⇨(frac{P_{avg-hlf}}{P_{avg, mono}}=frac{4I_{rms}^2 R_{hlf}}{I_{rms}^2 R_{mono}}=4*frac{73}{36.5}=8) (under same current)
P_avg-hlf=8(P_{avg mono})=800mW

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250+ TOP MCQs on Feeding Systems and Answers

Antennas Multiple Choice Questions on “Feeding Systems”.

1. Which of the following is called as the source placed at the focus?
a) Feed radiator
b) Reflector
c) Secondary radiator
d) Primary / reflector
Answer: a
Clarification: The source placed at the focus is called as feed radiator. It is also known as primary radiator. The reflector is called as the secondary radiator. A reflector antenna contains primary and secondary radiators.

2. In which of the following cases, a parabolic reflector primary radiator is said to be ideal feed?
a) The entire reflector is illuminated with no radiation in unwanted direction when feed radiated entire energy towards it
b) Some part of reflector is illuminated with no radiation in unwanted direction when feed radiated entire energy towards it
c) Some part of reflector is illuminated with radiation in unwanted direction when feed radiated entire energy towards it
d) The entire reflector is illuminated with small radiation in unwanted direction when feed radiated entire energy towards it
Answer: a
Clarification: A parabolic reflector primary radiator is said to be ideal feed, when the feed radiates entire energy, the reflector is fully illuminated and there is no energy radiation in unwanted direction.

3. To obtain maximum beam pattern, the primary radiator is placed ______ in a reflector antenna.
a) between Focus and Directrix
b) after the focus
c) at the focus point
d) can be placed anywhere
Answer: c
Clarification: Maximum beam pattern is obtained only when the feed is placed at the focus of the parabola. Reflector antenna is a high gain antenna. So this is one of the important points to obtain the maximized gain.

4. When the feed is moved along the main axis in a reflector antenna what happens to the beam pattern?
a) It broadens
b) It deteriorates
c) Beam remains unchanged
d) Side lobes are increased
Answer: a
Clarification: Maximum beam pattern is obtained only when the feed is placed at the focus of the parabola. When the feed is moved along the main axis, beam gets broadened.

5. The beam gets deteriorated when it is moved along a line perpendicular to the main axis passing through focus?
a) True
b) False
Answer: a
Clarification: Beam pattern is maximum only when the feed is placed at the focus of the parabola. When the feed is moved along the main axis, beam gets broadened. When it is moved along perpendicular line beam gets deteriorated.

6. In a Cassegrain feed system; the feed is placed at _____
a) Focus
b) Vertex
c) Directrix
d) Anywhere between vertex and focus
Answer: b
Clarification: Cassegrain is a dual reflector antenna. In this the feed is placed at the vertex of the parabolic reflector. The focus of the sub-reflector coincides with the focus of parabolic reflector and it illuminates the reflector.

7. Which of the following coincides with the focus of the parabolic reflector in a Cassegrain antenna?
a) Feed
b) Parabolic reflector
c) Focus of hyperboloid reflector
d) Focus of primary radiator
Answer: c
Clarification: The focus of the sub-reflector coincides with the focus of parabolic reflector and it illuminates the reflector. Since it is a Cassegrain, the sub-reflector is a hyperboloid reflector.

8. Which of the following regarding the Cassegrain feed system is false?
a) Spill over is reduced
b) It is a dual reflector antenna
c) Minor lobe radiation increases
d) A convex sub-reflector is used
Answer: c
Clarification: Spill over is the power loss when the reflector fails to redirect the energy. With the use of sub-reflector, this is reduced. Thus the minor lobe radiation also decreased. A convex hyperboloid is used as a sub-reflector. It is a dual reflector antenna with primary and two secondary radiators.

9. When a dipole with a parasitic reflector is used as a feed system, the distance between them is _____
a) 0.125λ
b) 0.4λ
c) 1λ
d) 0.625λ
Answer: a
Clarification: In a parabolic reflector antenna, a dipole along a parasitic reflector serves well as a feed radiator. The distance between parasitic reflector and the dipole is 0.125λ. The distance between dipole to dipole is approximately 0.4λ.

10. Which of the following uses a concave sub-reflector in a dual reflector antenna?
a) Cassegrain
b) Gregorian
c) Both Gregorian &Cassegrain
d) Neither Gregorian nor Cassegrain
Answer: b
Clarification: Cassegrain and Gregorian are dual reflector antennas. Gregorian antenna has a concave sub-reflector which is elliptic. Cassegrain uses a hyperboloid reflector which is convex towards the feed. The sub-reflectors are used to reduce the spillover efficiency.