Category: Microwave Engineering Objective Questions

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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250+ TOP MCQs on Circular Waveguide and Answers

Microwave Engineering Multiple Choice Questions on “Circular Waveguide”.

1. In TE mode of a circular waveguide, EZ=0. The wave equation is:
A. ∇2HZ+k2HZ=0
B. ∇2HZ-k2HZ=0
C. ∇2HZ-HZ=0
D. ∇2HZ+HZ=0
Answer: A
Clarification: In TE mode, EZ=0. Hence, when we substitute it in the wave equation, we get ∇2HZ+k2HZ=0.

2. Bessel’s differential equation for a circular waveguide is:
A. ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0
B. n2(d2R/ dρ2) + n(dR/dρ) + (ρ2kC2– n2) R=0
C. d2R/ dρ2 + dR/dρ + (ρ2kC2– n2) R=0
D. None of the mentioned
Answer: A
Clarification: After solving the wave equation ∇2HZ+k2HZ=0 in TE mode by making suitable assumptions and making appropriate substitutions, the final equation obtained is ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0.

3. The lowest mode of TE mode propagation in a circular waveguide is:
A. TE10 mode
B. TE00 mode
C. TE01 mode
D. TE11 mode
Answer: C
Clarification: A circular waveguide can support various modes of propagation. Among these, the lowest mode of propagation supported by the waveguide is TE10 mode of propagation.

4. What is the cutoff frequency for TE₁₁ mode in a circular waveguide of radius 2 cm with P’₁₁= 1.841?
A. 5.5 GHz
B. 4.3 GHz
C. 7.7 GHz
D. 8.1 GHz
Answer: B
Clarification: The cutoff frequency for TE11 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.3 GHz.

5. In a circular waveguide, if the propagation is in TE21 mode with P21=3.054, with a diameter of 60 mm, then the cutoff frequency for the mode is:
A. 5.6 GHz
B. 6.4 GHz
C. 3.5 GHz
D. 4.8 GHz
Answer: D
Clarification: The cutoff frequency for TE21 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.8 GHz.

6. For a circular waveguide in TM11 mode of propagation with inner radius of 30mm, and the phase constant being equal to 0.3, then the wave impedance is equal to:
A. 0.16 Ω
B. 0.15 Ω
C. 0.5 Ω
D. 0.4 Ω
Answer: A
Clarification: For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.16 Ω.

7. For TM mode. The wave equation in cylindrical co ordinates is:
A. (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0
B. ∂2E2/∂ρ2 + 1/ρ ( ∂E/∂ρ)=0
C. ∂2E2/∂ρ2 + 1/ρ2 (∂2E2/∂∅2 ) = 0
D. None of the mentioned
Answer: A
Clarification: The wave propagation in a cylindrical waveguide in TM mode of propagation is governed by the equation (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0. This is a second order differential equation.

8. In TM mode, what is the first propagating mode?
A. TM01 mode
B. TM11 mode
C. TM12 mode
D. TM10 mode
Answer: A
Clarification: TM mode in a circular waveguide supports various modes of propagation. Among these modes of propagation, the first or the lowest mode of propagation is TM01 mode.

9. For TM01 mode of propagation in a circular waveguide with P01=2.405, with the inner diameter of the circular waveguide being equal to 25 mm. What is the cut off frequency for this mode of propagation?
A. 2.8 GHz
B. 6 GHz
C. 3.06 GHz
D. 4 GHz.
Answer: C
Clarification: The cutoff frequency for TM01 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 3.06 GHz.

10. If β is 0.3 for a circular wave guide operating in TM12 mode with P21=5.315, with the radius of the circular waveguide being equal to 25 mm, then the intrinsic impedance of the wave is:
A. 0.55 Ω
B. 0.4 Ω
C. 0.3 Ω
D. 1.2 Ω
Answer: A
Clarification: For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.55 Ω.

11. The cutoff frequencies of the first two propagating modes of a Teflon on a filled circular waveguide with a=0.5 with ∈r=2.08 is:
A. 12.19 GHz, 15.92 GHz
B. 10 GHz, 12 GHz
C. 12 GHz, 15 GHz
D. 15 GHz, 12 GHz
Answer: A
Clarification: The cutoff frequencies are given by the expression p*C/2πa√∈. Substituting the given values in the above expression, the cutoff frequencies are 12.19 GHz, 15.92 GHz.


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250+ TOP MCQs on Tapered Lines and Answers

Microwave Engineering Multiple Choice Questions on “Tapered Lines”.

1. A single section tapered line is more efficient in impedance matching than a multisection tapered line for impedance matching.
A. True
B. False
Answer: B
Clarification: As the number N of discrete transformer sections increases, the step changes in the characteristic impedance between the sections become smaller, and the transformer geometry approaches a continuous tapered line. Thus multisection are preferred over single section for impedance matching.

2. Passband characteristics of tapered lines differ from one type of taper to another.
A. True
B. False
Answer: A
Clarification: The impedance of the tapered line varies along the line depending on the type of the tapering done. Thus impedance is a function of the type of taper. Hence passband characteristics depend on the type of taper.

3. For a continually tapered line, the incremental reflection co-efficient is:
A. ∆Z/2Z
B. 2Z/∆Z
C. ∆Z0/2Z0
D. None of the mentioned
Answer: A
Clarification: The incremental reflection co-efficient ∆Г is a function of distance. If a step change in impedance occurs for z and z+∆z, then the incremental reflection co-efficient is given by ∆Z/2Z.

4. The variation of impedance of an exponentially tapered line along the length of the line is given by:
A. Z(z)=Z0eaz
B. Z(z)=Z0e-az
C. Z(z)=Z0e2az
D. Z(z)=Z0e-2az
Answer: A
Clarification: The variation of impedance along the transmission line is a positive growing curve and is given by Z(z)=Z0eaz. The constant ‘a’ is defined as L-1 ln(ZL/Z0).

5. The value of constant ‘a’ for an exponentially tapered line of length 5 cm with load impedance being 100Ω and characteristic impedance of the line is 50Ω is:
A. 0.1386
B. 0.265
C. 0.5
D. 0.2
Answer: A
Clarification: The constant ‘a’ for a tapered transmission line is given by L-1 ln(ZL/Z0). ‘a’ is a function of the tapered length, load and characteristic impedance. Substituting the given values in the above expression, ‘a’ has the value 0.1386.

6. Reflection co-efficient magnitude response is an exponential curve for tapered line.
A. True
B. False
Answer: B
Clarification: The reflection co-efficient magnitude response of a exponential tapered line resembles only positive valued sinc function or can be called as a function with multiple peaks.

7. Triangular taper is the best solution for any impedance matching requirement.
A. True
B. False
Answer: B
Clarification: Klopfenstein taper is the best and most optimized solution for impedance matching because reflection co-efficient has minimum value in the passband.

8. The maximum passband ripple in a Klopfenstein taper matching section is:
A. Г0/cos h A
B. Г0/sin h A
C. Г0/ tan h A
D. None of the mentioned
Answer: A
Clarification: The maximum passband ripple in a Klopfenstein taper matching section is Г0/cos h A. Here, Г0 is the reflection co-efficient at zero frequency. A is a trigonometric function relating reflection co-efficient at zero frequency and maximum ripple in the passband.

9. For any load impedance, perfect match can be obtained and the minimum reflection co-efficient achieved can be zero.
A. True
B. False
Answer: B
Clarification: From Bode-Fano criterion, there is a theoretical limit on the minimum achievable reflection co-efficient for a given load impedance. Hence, perfect match cannot be obtained.

10. For a given load (a fixed RC product), a broader bandwidth can be achieved with a low reflection co-efficient in the passband.
A. True
B. False
Answer: B
Clarification: Based on the theoretical results of Bode-Fano criterion, a broader bandwidth can be achieved only at the expense of a higher reflection coefficient in the passband.

11. A perfect match can be obtained in the passband for any impedance matching circuit around the center frequency for which it is defined.
A. True
B. False
Answer: B
Clarification: The passband reflection co-efficient cannot be zero unless the bandwidth is zero. Thus a perfect match can be obtained only at a finite number of discrete frequencies.


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