Category: Microwave Engineering Objective Questions

250+ TOP MCQs on System Aspects of Antennas and Answers

Microwave Engineering Multiple Choice Questions on “System Aspects of Antennas”.

1. If an antenna has a directivity of 16 and radiation efficiency of 0.9, then the gain of the antenna is:
A. 16.2
B. 14.8
C. 12.5
D. 19.3
Answer: A
Clarification: Gain of an antenna is given by the product of radiation efficiency of the antenna and the directivity of the antenna. Product of directivity and efficiency thus gives the gain of the antenna to be 16.2.

2. Gain of an antenna is always greater than the directivity of the antenna.
A. True
B. False
Answer: B
Clarification: Gain of an antenna is always smaller than the directivity of an antenna. Gain is given by the product of directivity and radiation efficiency. Radiation efficiency can never be greater than one. So gain is always less than or equal to directivity.

3. A rectangular horn antenna has an aperture area of 3λ × 2λ. Then the maximum directivity that can be achieved by this rectangular horn antenna is:
A. 24 dB
B. 4 dB
C. 19 dB
D. Insufficient data
Answer: C
Clarification: Given the aperture dimensions of an antenna, the maximum directivity that can be achieved is 4π A/λ2, where A is the aperture area and λ is the operating wavelength. Substituting the given values in the above equation, the maximum directivity achieved is 19 dB.

4. A rectangular horn antenna has an aperture area of 3λ × 2λ. If the aperture efficiency of an antenna is 90%, then the directivity of the antenna is:
A. 19 dB
B. 17.1 dB
C. 13 dB
D. 21.1 dB
Answer: B
Clarification: Given the aperture dimensions of an antenna, the directivity that can be achieved is ap4π A/λ2, where A is the aperture area and λ is the operating wavelength, ap is the aperture efficiency. Substituting the given values in the above equation, the directivity achieved is 17.1 dB.

5. If an antenna has a directivity of 16 and is operating at a wavelength of λ, then the maximum effective aperture efficiency is:
A. 1.27λ2
B. 2.56λ2
C. 0.87λ2
D. None of the mentioned
Answer: A
Clarification: Maximum effective aperture efficiency of an antenna is given by D λ2/4π, D is the directivity of the antenna. Substituting in the equation the given values, the maximum effective aperture is 1.27λ2.

6. A resistor is operated at a temperature of 300 K, with a system bandwidth of 1 MHz then the noise power produced by the resistor is:
A. 3.13×10-23 watts
B. 4.14×10-15 watts
C. 6.14×10-15 watts
D. None of the mentioned
Answer: B
Clarification: For a resistor noise power produced is given by kTB, where T is the system temperature and B is the bandwidth. Substituting in the above expression, the noise power produced is 4.14×10-15 watts.

7. With an increase in operating frequency, the background noise temperature:
A. Increases
B. Decreases
C. Remains constant
D. Remains unaffected
Answer: A
Clarification: The plot of frequency v/s background noise temperature shows that with the increase of the signal frequency, the background noise temperature increases. Also, with the increase of the elevation angle from the horizon, background noise temperature increases.

8. The noise temperature of an antenna is given by the expression:
A. radTb + (1-raD. Tp
B. (1-raD. TP
C. radTb
D. None of the mentioned
Answer: A
Clarification: The noise temperature of an antenna is given by the expression radTb + (1-raD. Tp. here, Tb is the brightness temperature and Tp is the physical temperature of the system. rad is the radiation efficiency. Noise temperature of a system depends on these factors.

9. Low is the G/T ratio of an antenna, higher is its efficiency.
A. True
B. False
Answer: B
Clarification: In the G/T ratio of an antenna, G is the gain of an antenna and T is the antenna noise temperature. Higher the G/T ratio of an antenna better is the performance of the antenna.

10._________ has a constant power spectral density.
A. White noise
B. Gaussian noise
C. Thermal noise
D. Shot noise
Answer: A
Clarification: Thermal noise has a power spectral density for a wide range of frequencies. Its plot of frequency v/s noise power is a straight line parallel to Y axis.


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250+ TOP MCQs on Rectangular Waveguide and Answers

Microwave Engineering Multiple Choice Questions on “Rectangular Waveguide”.

1. The modes of propagation supported by a rectangular wave guide is:
A. TM, TEM, TE modes
B. TM, TE
C. TM, TEM
D. TE, TEM
Answer: B
Clarification: A hollow rectangular waveguide can propagate TE and TM modes. Since only a single conductor is present, it does not support TEM mode of propagation.

2. A hollow rectangular waveguide cannot propagate TEM waves because:
A. Of the existence of only one conductor
B. Of the losses caused
C. It is dependent on the type of the material used
D. None of the mentioned
Answer: A
Clarification: A rectangular hollow waveguide can propagate both TE and TM modes of propagation. But due the presence of only one conductor, rectangular waveguide does not support the propagation of TEM mode.

3. For any mode of propagation in a rectangular waveguide, propagation occurs:
A. Above the cut off frequency
B. Below the cut off frequency
C. Only at the cut-off frequency
D. Depends on the dimension of the waveguide
Answer: A
Clarification: Both TE and TM modes of propagation in rectangular waveguide have certain separate and specific cut off frequencies below which propagation is not possible. Hence propagation of signal occurs above the cut off frequency.

4. In TE mode of wave propagation in a rectangular waveguide, what is the equation that has to be satisfied?
A. (∂2/ ∂x2 + ∂2/ ∂y2+ kC2).HZ(x, y) =0
B. (∂2/ ∂x2 + ∂2/ ∂y2– kC2).HZ(x, y) =0
C. (∂2/ ∂x2 – ∂2/ ∂y2+ kC2).HZ(x, y) =0
D. None of the mentioned
Answer: A
Clarification: For TE mode of propagation in a rectangular waveguide, electric field along the direction of propagation is 0. Hence for propagation, the above partial differential equation in terms of magnetic field along Z direction has to be satisfied.

5. Dominant mode is defined as:
A. Mode with the lowest cut off frequency
B. Mode with the highest cut off frequency
C. Any TEM mode is called a dominant mode
D. None of the mentioned
Answer: A
Clarification: Among the various modes of propagation in a rectangular waveguide, the mode of propagation having the lowest cutoff frequency or the highest wavelength of propagation among the other propagating modes is called dominant mode.

6. For TE1ₒ mode, if the waveguide is filled with air and the broader dimension of the waveguide is 2 cm, then the cutoff frequency is:
A. 5 MHz
B. 7.5 MHz
C. 7.5 GHz
D. 5 GHz
Answer: C
Clarification: The cutoff frequency for TE 10 mode of propagation in a rectangular waveguide is 1/2a√(∈μ) where ‘a’ is the broader dimension of the waveguide. Substituting for the given value and 1/√(∈μ)=3*108. The cutoff frequency is 7.5 GHz.

7. TEₒₒ mode for a rectangular waveguide:
A. Exists
B. Exists but defined only under special cases
C. Does not exist
D. Cannot be determined
Answer: C
Clarification: The field expressions for TEₒₒ mode disappears or becomes zero theoretically. Hence, TEₒₒ mode does not exist.

8. For dominant mode propagation in TE mode, if the rectangular waveguide has a broader dimension of 31.14 mm , then the cutoff wave number:
A. 100
B. 500
C. 50
D. 1000
Answer: A
Clarification: The cutoff wave number for the dominant mode of a rectangular waveguide is given by π/a where ‘a’ is the broader dimension of the waveguide, substituting the given values, the wave number 100.

9. The lowest mode of TM wave propagation is:
A. TM10 mode
B. TM01 mode
C. TM11 mode
D. TM12 mode
Answer: C
Clarification: The field components for other lower modes of propagation in TM mode disappear for other lower modes of propagation. Hence, the lowest mode of propagation is TM11 mode.

10. The cutoff frequency for the dominant mode in TM mode propagation for a rectangular waveguide of dimension of 30mm*40mm is:
A. 2 GHz
B. 1 GHz
C. 2 MHz
D. 4 MHz
Answer: A
Clarification: The cutoff frequency of dominant mode in TM mode is √((π/A.2 + (π/B.2). Here, ‘a’ and ‘b’ are the dimensions of the waveguide. Substituting the corresponding values, the cutoff frequency is 2 GHz.

11. In TE10 mode of wave propagation in a rectangular waveguide, if the broader dimension of the waveguide is 40 cm, then the cutoff wavelength for that mode is:
A. 8 cm
B. 6 cm
C. 4 cm
D. 2 cm
Answer: A
Clarification: In TE10 mode of propagation in a rectangular waveguide, the cutoff wavelength of the waveguide is given by 2a where ‘a’ is the broader dimension of the waveguide. Substituting, the cutoff wavelength is 8 cm.

12. In TE01 mode of wave propagation in a rectangular waveguide, if the smaller dimension of the waveguide is 2 cm, then the cutoff wavelength for that mode is:
A. 4 cm
B. 8 cm
C. 1 cm
D. 2 cm
Answer: A
Clarification: For TE01 mode of wave propagation in a rectangular wave guide, if the smaller dimension of the wave guide is 2 cm, then the cut off wavelength is 2b where b is the smaller dimension of the waveguide. substituting, the cutoff wavelength is 4 cm.


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250+ TOP MCQs on Chebyshev Multi-section Matching Transformers and Answers

Microwave Engineering Questions and Answers for Experienced people on “Chebyshev Multi-Section Matching Transformers”.

1. The major disadvantage of binomial multi section transformer is higher bandwidth cannot be achieved.
A. true
B. false
Answer: A
Clarification: In some applications, a flat curve in the operating frequency is a major requirement. This requirement can be satisfied using a binomial transformer. But the disadvantage is that a higher bandwidth can be achieved.

2. Advantage of chebyshev matching transformers over binomial transformers is:
A. higher gain
B. low power losses
C. higher roll-off in the characteristic curve
D. higher bandwidth
Answer: D
Clarification: Chebyshev transformers when designed to operate at a certain frequency called center frequency, the reflection co-efficient is low for a large frequency range implying that they have a higher operating range. This is the major advantage of chebyshev filters.

3. There are passband ripples present in the chebyshev characteristic curve.
A. true
B. false
Answer: A
Clarification: This is a major difference between chebyshev and binomial transformer. Binomial transformers have a flat curve in the passband while chebyshev transformers have ripples in the transformer passband.

4. Chebyshev matching transformers can be universally used for impedance matching in any of the microwave networks.
A. true
B. false
Answer: B
Clarification: Chebyshev transformers have passband ripples in the characteristic curve. In some critical applications, these ripples are not tolerable in the operating bandwidth. Hence, chebyshev transformers cannot be used for all the microwave networks for impedance matching.

5. The 4th order chebyshev polynomial is:
A. 8x4-8x2+1
B. 4x3-4x2+1
C. 4x3-3x
D. none of the mentioned
Answer: A
Clarification: nth order polynomial for a chebyshev polynomial is generated using lower polynomials by the expression Tn (x) = 2xTn-1(x) – Tn-2(x). T2(x) = 2x2-1, T3(x)= 4x3-3x. Substituting the lower level polynomials in the given expression, T4(x) = 8x4-8x2+1.

6. Chebyshev polynomials do not obey the equal-ripple property.
A. true
B. false
Answer: B
Clarification: For -1≤x≤1,│T(x)│≤ 1. In this range, the chebyshev polynomials oscillate between±1. This is the equal ripple property. Chebyshev polynomials obey the equal-ripple property.

7. Chebyshev polynomial can be expressed in trigonometric functions as:
A. Tn(cos θ)=cos nθ
B. Tn(sin θ)= sin nθ
C. Tn(cos θ)=cos nθ.sin nθ
D. none of the mentioned
Answer: A
Clarification: If the chebyshev polynomial variable x is equated to a trigonometric variable cos θ, then the higher order chebyshev polynomials can be defined in terms of the same function with multiples of θ. This can be theoretically proved and function generation becomes simpler.

8. For values of x greater than 1, the chebyshev polynomial in its trigonometric form cannot be determined.
A. true
B. false
Answer: B
Clarification: Since cosine function is defined for values of x between -1 and +1, for x values greater than 1, hyperbolic function is used to define the chebyshev polynomial. Tn(x)=cosh (n cosh-1x).

9. Reflection co-efficient Гn in terms of Zn and Zn+1, successive impedances of successive sections in the matching network are:
A. 0.5 ln (Zn+1/Zn)
B. 0.5 ln (Zn/Zn+1)
C. ln (Zn+1/Zn)
D. ln (Zn/Zn+1)
Answer: A
Clarification: When multiple sections are used in the chebyshev matching network, the reflection co-efficient of the nth matching section, given the impedances at the ends of the section, reflection co-efficient can be obtained using the expression 0.5 ln (Zn+1/Zn).

10. In a 3 section multisection chebyshev matching network, if Z3 = 100Ω, and Z2=50Ω, then the reflection co-efficient Г2 is:
A. 0.154
B. 0.3465
C. 0.564
D. none of the mentioned
Answer: B
Clarification: Гn for ‘n’ section matching chebyshev network is given by Гn=0.5 ln (Zn+1/Zn). substituting the given values in the expression, Г2 is 0.3465.

11. If Г3=0.2 and Z3=50Ω, then the impedance of the next stage in the multi-section transformer is:
A. 100Ω
B. 50Ω
C. 74.6Ω
D. 22.3Ω
Answer: C
Clarification: Гn for ‘n’ section matching chebyshev network is given by Гn=0.5 ln (Zn+1/Zn). Substituting the given values in the expression, the impedance of the next stage is Z4=74.6Ω.


for Experienced people,

250+ TOP MCQs on Properties of Ferrimagnetic Materials and Answers

Microwave Engineering Multiple Choice Questions on “Properties of Ferrimagnetic Materials”.

1. Example of a non reciprocal device:
A. Branch line coupler
B. Wilkinson coupler
C. Magic-T hybrid
D. Circulator
Answer: D
Clarification: Non reciprocal device is the one in which the response between any two ports I and j of a component depends on the direction of signal flow. Circulator is a device that allows power flow either in clockwise direction or counter clockwise direction.

2. A microwave network can be called non reciprocal only if it contains anisotropic materials like ferrite materials.
A. True
B. False
Answer: B
Clarification: A microwave network consisting of active non linear devices like transistor amplifiers, ferrite phase shifters and more. Presence of active devices or anisotropic materials can make a microwave network non reciprocal.

3. This is not an example of anisotropic material:
A. Yttrium aluminum garnet
B. Aluminum
C. Cobalt
D. Silicon
Answer: D
Clarification: Yttrium aluminum garnet is a ferromagnetic compound. Aluminum and cobalt are iron oxides that are anisotropic. Silicon is a non metal that is isotropic in nature.

4. The magnetic properties of a material are due to the existence of ___________
A. Electrons in atoms
B. Electric dipole moment
C. Magnetic dipole moment
D. None of the mentioned
Answer: C
Clarification: The magnetic properties of ferromagnetic materials are due to the existence of magnetic dipole moments, which arise primarily from electron spin. The magnetic dipole moment of an electron is 9.27×10-24 A-m2.

5. ___________ is a measure of the relative contributions of the orbital moment and the spin moment to the total magnetic moment.
A. Lande’s factor
B. Gibbs factor
C. Newton’s ratio
D. None of the mentioned
Answer: A
Clarification: An electron in orbit around a nucleus gives rise to an effective current loop and thus an additional magnetic moment, but this effect is negligible compared to the magnetic moment due to spin. Lande’s factor is a relative measure of these orbital moments.

6. Lande’s factor for all ferromagnetic materials is in the range of 0 to 1.
A. True
B. False
Answer: B
Clarification: Lande’s factor (g) is one when the moment is only due to orbital motion and 2 when the moment is only due to spin. For most microwave ferrite materials, g lies between 1.98 and 2.01.

7. The variation of magnetic moment of a ferromagnetic material with applied bias field is linear.
A. True
B. False
Answer: B
Clarification: With the increase in the applied bias field to a ferromagnetic material, the magnetic moment increases exponentially initially, after a certain applied bias field magnetic moment remains a constant.

8. A permanent magnet is made by placing the magnetic material in a strong magnetic field.
A. True
B. False
Answer: A
Clarification: A permanent magnet is made by placing the magnetic material in a strong magnetic field and then removing the field to leave the material magnetized in a remanent state.

9. The operating point of a permanent magnet is in the:
A. First quadrant
B. Second quadrant
C. Third quadrant
D. Fourth quadrant
Answer: B
Clarification: Unless the magnet shape forms a closed path, the demagnetization factors at the magnet ends will cause a slightly negative H field to be induced in the magnet. Thus the “operating point “of a permanent magnet will be in the second quadrant. This portion of the curve is called demagnetization curve.

10. After demagnetization of a magnetic material, the residual magnetization retained in the magnetic material is called:
A. Remanence
B. Residue
C. Retardation
D. None of the mentioned
Answer: A
Clarification: The residual magnetization called remanence characterizes the strength of the magnet, so magnetic material with large remanence is chosen.


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250+ TOP MCQs on Hybrid Microwave integrated Circuits and Answers

Microwave Engineering Questions and Answers for Campus interviews on “Hybrid Microwave integrated Circuits”.

1. In the course of development of microwave circuits, two distinct types of microwave integrated circuits have been developed according to the application requirements.
A. true
B. false
Answer: A
Clarification: There are two distinct types of microwave integrated circuits fabricated. They are hybrid microwave integrated circuits and monolithic microwave integrated circuits. They differ in the method of fabrication in the layers of metallization done.

2. __________ is an important consideration for a hybrid integrated circuit.
A. material selection
B. processing units
C. design complexity
D. active sources
Answer: A
Clarification: Material selection is an important consideration for a hybrid integrated circuit. Characteristics such as electrical conductivity, dielectric constant, loss tangent, thermal transfer and manufacturing compatibility of the material to be used for hybrid microwave circuits are evaluated first.

3. To fabricate a low frequency circuit using the hybrid microwave IC methodology, the material with _______ is preferred.
A. high dielectric constant
B. low dielectric constant
C. high resistivity
D. low resistivity
Answer: A
Clarification: At low frequency applications, a high dielectric constant is desirable because it results in smaller circuit size. At higher frequencies, however the substrate thickness must be decreased to prevent radiation loss and other spurious effects.

4. The mask in a hybrid microwave circuit is made of:
A. rubylith
B. silicon
C. quartz
D. arsenic
Answer: A
Clarification: The mask in hybrid microwave integrated circuits is made of Rubylith, a soft mylar film usually at a magnified scale for high accuracy. Then an actual size mask is made on a thin sheet of glass or quartz.

5. The metalized substrate is coated with __________ covered with the mast and exposed to light source.
A. photoresist
B. GaAs
C. germanium liquid
D. none of the mentioned
Answer: A
Clarification: The metalized substrate is coated with photoresist, covered with the mast and exposed to light source. The substrate can be etched to remove the unwanted areas of the metal.

6. Commonly used software packages for CAD of hybrid microwave integrated circuits are:
A. CADENCE
B. ADS
C. DESIGNER
D. all of the mentioned
Answer: D
Clarification: Before any microwave circuit design is implemented on the hardware, it is economical to simulate the same designs in software and check for the expected theoretical results. A few such software that provide such an environment is CADENCE, ADS, DESIGNER to name a few.

7. In hybrid microwave integrated circuits, the various components of the circuit are etched in the substrate.
A. true
B. false
Answer: B
Clarification: In hybrid integrated circuit design, after all initial design steps are completed, the discrete components are soldered or wire bonded to the conductors This can be done manually or through automated computer-controlled pick and place machines.

8. Once the circuit components are designed and fabricated for certain specific values, they cannot be changed as per the requirement later.
A. true
B. false
Answer: B
Clarification: IN HIC, provision is made for variations in component values and other circuit tolerances by providing tuning or trimming stubs that can be manually trimmed for each circuit. This increase circuit yield but also increases the cost of manufacture.

9.________ is a micromachining technique where suspended structures are formed on silicon substrates.
A. MMIC
B. HIC
C. RF MEMS
D. none of the mentioned
Answer: C
Clarification: RF MEMS switch technology is a micro machining technique where suspended structures are formed in silicon substrates. These can be used in microwave resonators, antennas and switches.

10. Depending on the single path (capacitive or direct contact) and the attenuation mechanism MEMS switch can be used for various configurations for various devices.
A. true
B. false
Answer: A
Clarification: A MEMS switch can be used in several different configurations depending on the single path, actuation mechanism, pull-back mechanism and the type of structure. One such example is switching the capacitance of a single path between high and low values by moving a flexible conductive membrane through the application of DC controlled voltage.


for Campus Interviews,

250+ TOP MCQs on Antenna Gain and Efficiency and Answers

Microwave Engineering Multiple Choice Questions on “Antenna Gain and Efficiency”.

1. A __________ is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space.
A. Transmitting antenna
B. Receiving antenna
C. Radar
D. Mixer
Answer: A
Clarification: A transmitting antenna is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space. It appears as an electrical circuit on one side, provides an interface with a propagating plane wave.

2. Antennas are bidirectional devices.
A. True
B. False
Answer: A
Clarification: Antennas can be used both as transmitters and receivers. As transmitters they radiate energy to free space and as receivers they receive signal from free space. Hence, they are called bidirectional devices as they are used at both transmitting end and receiving end.

3. Dipole antennas are an example for:
A. Wire antennas
B. Aperture antennas
C. Array antennas
D. None of the mentioned
Answer: A
Clarification: Dipoles, monopoles, oops, Yagi-Uda arrays are all examples for wire antennas. These antennas have low gains, and are mostly used at lower frequencies.

4. _________ antennas consist of a regular arrangement of antenna elements with a feed network
A. Aperture antennas
B. Array antennas
C. Printed antennas
D. Wire antennas
Answer: B
Clarification: Array antennas consist of a regular arrangement of antenna elements with a feed network. Pattern characteristics such as beam pointing angle and side lobe levels can be controlled by adjusting the amplitude and phase excitation of array elements.

5. A parabolic reflector used for reception with the direct broadcast system is 18 inches in diameter and operates at 12.4 GHz. The far-field distance for this antenna is:
A. 18 m
B. 13 m
C. 16.4 m
D. 17.3 m
Answer: D
Clarification: Far field distance for a reflector antenna is given by 2D2/λ. D is the diameter and λ is the operating signal wavelength. Substituting in the above expression, far field distance is 17.3 m.

6._________ of an antenna is a plot of the magnitude of the far field strength versus position around the antenna.
A. Radiation pattern
B. Directivity
C. Beam width
D. None of the mentioned
Answer: A
Clarification: Radiation pattern of an antenna is a plot of the magnitude of the far field strength versus position around the antenna. This plot gives the detail regarding the region where most of the energy of antenna is radiated, side lobes and beam width of an antenna.

7. Antennas having a constant pattern in the azimuthal plane are called _____________
A. High gain antenna
B. Omni directional antenna
C. Unidirectional antenna
D. Low gain antenna
Answer: B
Clarification: Omni directional antennas radiate EM waves in all direction. If the radiation pattern for this type of antenna is plotted, the pattern is a constant signifying that the radiated power is constant measured at any point around the antenna.

8. Beamwidth and directivity are both measures of the focusing ability of an antenna.
A. True
B. False
Answer: A
Clarification: Beamwidth and directivity are both measures of the focusing ability of an antenna. An antenna with a narrow main beam will have high directivity, while a pattern with low beam will have low directivity.

9. If the beam width of an antenna in two orthogonal planes are 300 and 600. Then the directivity of the antenna is:
A. 24
B. 18
C. 36
D. 12
Answer: B
Clarification: Given the beam width of the antenna in 2 planes, the directivity is given by 32400/θ*∅, where θ,∅ are the beam widths in the two orthogonal planes. Substituting in the equation, directivity of the antenna is 18.

10. If the power input to an antenna is 100 mW and if the radiated power is measured to be 90 mW, then the efficiency of the antenna is:
A. 75 %
B. 80 %
C. 90 %
D. Insufficient data
Answer: C
Clarification: Antenna efficiency is defined as the ratio of radiated power to the input power to the antenna. Substituting the given data in the efficiency equation, the efficiency of the antenna is 90%.


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