300+ REAL TIME Microwave Engineering Objective Questions & Answers 2023

250+ TOP MCQs on Circular Waveguide and Answers

Microwave Engineering Multiple Choice Questions on “Circular Waveguide”.

1. In TE mode of a circular waveguide, E_Z=0. The wave equation is:
A. ∇²H_Z+k²H_Z=0
B. ∇²H_Z-k²H_Z=0
C. ∇²H_Z-H_Z=0
D. ∇²H_Z+H_Z=0
Answer: A
Clarification: In TE mode, E_Z=0. Hence, when we substitute it in the wave equation, we get ∇²H_Z+k²H_Z=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 ∇²H_Z+k²H_Z=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 P_nm/ 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 P_nm/ 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 P_nm/ 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.

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)=Z₀e^az
B. Z(z)=Z₀e^-az
C. Z(z)=Z₀e^2az
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)=Z₀e^az. The constant ‘a’ is defined as L-1 ln(Z_L/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.

250+ TOP MCQs on Ferrite Isolators and Answers

Microwave Engineering Multiple Choice Questions on “Ferrite Isolators”.

1. Ferrite isolators are ____ port microwave devices.
A. Two
B. Three
C. Four
D. None of the mentioned

Answer: A
Clarification: Ferrite isolators are two port devices having unidirectional transmission characteristics. Isolator provides isolation between the two ports and the power flow occurs only in one direction.

2. The matrix of an ideal isolator is not ______
A. Unitary
B. Symmetric
C. Lossless
D. None of the mentioned

Answer: A
Clarification: Ferrite isolator allows wave propagation in only one direction and attenuates propagation in the other direction. So the isolator is lossy. Since the isolator is lossy, the scattering matrix of isolator is not unity.

3. A very common application of isolator is to provide isolation between a low power source and the load.
A. True
B. False

Answer: B
Clarification: Isolators have a wide variety of applications. The most common among them is the use of an isolator between a high-power source and a load to prevent the possible reflections from damaging the source. An isolator can be used in place of a matching network, but it should be realized that any power reflected from the source is absorbed by the isolator.

4. The attenuation of a ________ is very large near the gyro magnetic resonance of the ferrite.
A. Linearly polarized wave
B. Circularly polarized wave
C. Left polarized wave
D. Right polarized wave

Answer: B
Clarification: The attenuation of the circularly polarized wave is very large near the gyro magnetic resonance of the ferrite, while the attenuation of the wave propagating in the opposite direction is very small.

5. The isolators constructed using ferrite materials must operate at:
A. Gyro magnetic resonance
B. Magnetic resonance
C. Isolator resonance
D. None of the mentioned

Answer: A
Clarification: The attenuation of the circularly polarized wave is very large near the gyro magnetic resonance of the ferrite, while the attenuation of the wave propagating in the opposite direction is very small. Isolator constructed using ferrite must hence operate at gyro magnetic resonance.

6. Forward attenuation provided by a resonance ferrite isolator is:
A. Zero
B. Low
C. High
D. None of the mentioned

Answer: B
Clarification: Zero forward attenuation cannot be obtained in resonance isolators because the internal magnetic field is not truly circularly polarized. Because of this, there is some amount of forward attenuation in the isolator.

7. An isolator has a very large operating bandwidth and independent of any isolator parameter.
A. True
B. False

Answer: B
Clarification: The bandwidth of an isolator is relatively narrow, dictated essentially by the line width ∆H of the ferrite material.

8. The length of a ferrite slab required operating with a minimum forward insertion loss and 30 dB reverse attenuation and the reverse attenuation at this point is:
A. 3 cm
B. 2.4 cm
C. 4 cm
D. 3.6 cm

Answer: B
Clarification: Length of the ferrite slab required is equal to the ratio of the minimum forward insertion loss to the reverse attenuation at the point. Substituting the given values in the above equation, length of the ferrite slab is 2.4 cm.

9. The electric field distribution of the forward and reverse waves in a ferrite slab-loaded waveguide is quite different. This property is used in:
A. Field displacement resonator
B. Resonance isolator
C. Waveguide isolator
D. None of the mentioned

Answer: A
Clarification: In a field displacement isolator, electric field distribution of the forward and reverse waves in a ferrite slab-loaded waveguide is different. The electric field for the forward wave can be made to vanish at the side of the ferrite slab.

10. Field displacement isolators require higher bias field than resonance isolators.
A. True
B. False

Answer: B
Clarification: Field displacement isolators have much smaller bias field requirement since it operates well below gyro magnetic resonance. This property of field displacement isolator make it more preferred than resonance isolators.

250+ TOP MCQs on Monolithic Microwave Integrated Circuits and Answers

Microwave Engineering Multiple Choice Questions on “Monolithic Microwave Integrated Circuits”.

1. Progress in ________ and other related semiconductors material processing led to the feasibility of monolithic microwave integrated circuits.
A. GaAs
B. Silicon
C. Germanium
D. GaAlAs
Answer: A
Clarification: Progress in GaAs and other related semiconductor material processing led to the feasibility of MMIC, where all the passive and active components required for a given circuit can be grown or implanted in a substrate.

2. MMICs are high cost devices that involve complex fabrication methods and contain multiple layers to contain even small circuits.
A. True
B. False
Answer: B
Clarification: Min MIC can be made at low cost because the labor involved with fabricating hybrid MIC’s is reduced. In addition, a single wafer can contain a large number of circuits, all of which can be processed and fabricated simultaneously.

3. The substrate of an MMIC must be a _____________ to accommodate the fabrication of all the type of devices.
A. Semiconductor
B. Insulator
C. Partial conductors
D. Metals operable at high frequencies
Answer: A
Clarification: Substrate of MMIC must be a semiconductor material to accommodate the fabrication of active devices. The type of devices and the frequency range dictate the type of substrate material. One such material is GaAs MESFET.

4. GaAs MESFETs are versatile device because it finds application in:
A. Low-noise amplifiers
B. High gain amplifiers
C. Mixers
D. All of the mentioned
Answer: D
Clarification: GaAs MESFET find application in low noise amplifiers, high gain amplifiers, broadband amplifiers, mixers, oscillators, phase shifters, and switches. These are the mostly used and cost effective substrates.

5. Transmission lines and other conductors in microwave devices are usually made with ___________
A. Gold metallization
B. Silver metallization
C. Copper metallization
D. Zinc metallization
Answer: A
Clarification: Transmission lines and conductors at microwave operation are usually made with gold metallization. To improve the adhesion of gold to the substrate, a thin layer of chromium or titanium may be deposited first since these metals are relatively lossy.

6. For the capacitors used in MMICs, the insulating dielectric films used are:
A. Air
B. SiO
C. Titanium
D. GaAs
Answer: B
Clarification: Capacitor and overlaying lines require insulating dielectric films, such as SiO, SiO2, SiN4 and Ta2O5. These materials have high dielectric constants and low loss and are compatible with integrated circuit processing.

7. Resistors used at normal operating frequencies can be directly used at microwave frequencies in MMIc.
A. True
B. False
Answer: B
Clarification: Resistors used at normal operating frequencies cannot be used directly in MMICs. Resistors require the deposition of the lossy films like NiCr, Ta, Ti, and doped GaAs commonly used.

8. Processing in MMICs is done by __________
A. Ion implantation
B. Net list generation
C. Floor planning
D. None of the mentioned
Answer: A
Clarification: Processing begins by forming an active layer in the semiconductor substrate for the necessary active devices. This is done by ion implantation or epitaxial techniques.

9. MMICs are the best microwave integrated circuit fabrication methodologies without any drawbacks in it.
A. True
B. False
Answer: B
Clarification: Major drawback of MMIC is that they tend to waste large area of relatively expensive semiconductor substrate for components such as transmission lines and hybrids.

10. MMICs have higher circuit flexibility as compared to other microwave integrated fabrication methods.
A. True
B. False
Answer: A
Clarification: Since the fabrication of additional FETs in an MMIC design is much easy, the circuit flexibility and performance can be enhanced with only little additional cost and the requirement for the fabrication of the entire device is prevented.

250+ TOP MCQs on Wireless Communication and Answers

Microwave Engineering Multiple Choice Questions on “Wireless Communication”.

1. Most of the wireless systems today operate at a frequency of about:
A. 800 MHz
B. 100 MHz
C. 80 MHz
D. None of the mentioned
Answer: A
Clarification: With all advancement in wireless communication today, the need of the hour is higher data rates of transmission and reception. These higher data rates can be achieved only at microwave frequency range and in giga hertz frequency range.

2. Point to point communication systems use low gain antennas for communication.
A. True
B. False
Answer: B
Clarification: In point to point communication a single transmitter communicates with a single receiver. Such systems use high gain antennas to maximize received power and minimize interference with other radios.

3. In this method of wireless communication, communication happens only in one direction:
A. Simplex
B. Duplex
C. Half duplex
D. None of the mentioned
Answer: A
Clarification: In simplex systems, communication happens only in one direction that is from the transmitter to the receiver. Examples for this type of communication include radio, television and paging systems.

4. The power density radiated by an isotropic antenna is given by the relation:
A. P_t/4πR²
B. P_t/4R²
C. P_t/R²
D. None of the mentioned
Answer: A
Clarification: An isotropic antenna radiates energy equally in all the directions. Hence, the power density radiated at a distance R is given by the relation P_t/4πR².

5. The power received by a receiving antenna given that P_t is the transmitted power is:
A. G_rG_tλ²p_t/ (4πR)²
B. G_tλ²p_t/ (4πR)²
C. G_rλ²p_t/ (4πR)²
D. None of the mentioned
Answer: A
Clarification: The power received by a receiving antenna given that P_t is the transmitted power is G_rG_tλ²pt/ (4πR)². Here G_r is the gain of the receiving antenna; G_t is the gain of the transmitting antenna. R is the distance between the transmitting and receiving antenna.

6. If the distance between a transmitting station and receiving station is 1 Km and if the antennas are operating at a wavelength of 5 cm, then the path loss is:
A. 108 dB
B. 12 dB
C. 45 dB
D. 48 dB
Answer: A
Clarification: Path loss is given by the expression 20 log (4πR/λ) in db. Substituting the given values in the above expression, the path loss is 108 dB.

7. The amount of power by which the received power must be greater than the threshold level required to maintain a minimum quality of service is called _______
A. Line loss
B. Link budget
C. Link margin
D. None of the mentioned
Answer: C
Clarification: Link margin is the amount of power by which the received power must be greater than the threshold level required to maintain a minimum quality of service. Link margin signifies the minimum amount of power required to sustain communication maintaining a minimum quality of service.

8. Link margin that is used to account for fading effects is called fade margin.
A. True
B. False
Answer: A
Clarification: Link margin that is used to account for fading effects is called fade margin. Satellite links operating at frequencies of above 10 GHz require a fading margin of about 20dB or more to account for attenuation during heavy rain.

9. One of the most important requirements of a radio receiver is high gain.
A. True
B. False
Answer: A
Clarification: Radio receivers must have very high gain of about 100 dB in order to detect the very low power level of the received signal to a level near its original baseband value.

10. A radio receiver operating at microwave frequencies must have very high selectivity.
A. True
B. False
Answer: A
Clarification: Today, most of the applications use wireless communication at microwave frequency. Hence space is a sea of EM waves. In order to receive only the desired signal in the desired range of frequencies, the radio receiver must have high sensitivity.

250+ TOP MCQs on Co-axial Lines and Answers

Microwave Engineering Multiple Choice Questions on “Co-axial Lines”.

1. What are the modes of propagation that a co axial line supports?
A. TM, TE mode
B. TM, TE, TEM mode
C. TM, TEM mode
D. TE, TEM mode
Answer: B
Clarification: Certain propagating media support specific modes of propagation. Coaxial cables support all the three modes of propagation. They are TM, TE, and TEM modes.

2. The dominant waveguide mode of a co axial line is:
A. TE11 mode
B. TE01 mode
C. TM01 mode
D. TEM mode
Answer: A
Clarification: Co-axial cable many modes of propagation. Among those supported modes of propagation, the dominant mode, the mode with lowest propagating wavelength is TE11 mode.

3. In a co axial line with inner and outer diameters of 0.0645 and 0.0215 inches and a Teflon di electric with ∈r=2.2. The highest usable frequency before the TE11 waveguide mode starts to propagate is:
A. 16.8 GHz
B. 117.7 GHz
C. 15.3 GHz
D. 8.4 GHz
Answer: A
Clarification: Before the wave propagation starts, the unstable frequency is given by the expression Ck/2π√∈. Here, C is the velocity of light. Substituting the given values in the above expression, the frequency is 16.8 GHz.

4. If a wave guide has he inner and outer conductor diameters of 0.0645 and 0.0215 inches respectively for a co axial lines then the cut off wave number is:
A. 298
B. 300
C. 285
D. 123
Answer: A
Clarification: The cutoff wave number for wave propagation is given by 2/ (a + B.. a, b are the inner and outer diameter respectively. Substituting in the above expression, cut off wave number is 298.

5. The commercially used co axial cable and connectors used has a characteristic impedance is:
A. 50Ω
B. 100Ω
C. 33.34Ω
D. 66.6Ω
Answer: A
Clarification: All commercial manufacturer s of coaxial cables and connectors have set a standard for all the manufactured products. The standard value is 50 Ω.

6. In television systems the characteristic impedance of the cables used is:
A. 75Ω
B. 150Ω
C. 100Ω
D. 50Ω
Answer: A
Clarification: All cable manufacturers of the television system follow a set of standards. As per these set standards, the characteristic impedance of the line is 75 Ω.

7. SWR standing wave ratio has to be ________for co axial connector.
A. Low
B. High
C. Infinite
D. Cannot be calculated
Answer: A
Clarification: Higher the value of the standing wave ratio more is the reflection which implies mismatch. Hence, standing wave ratio has to be low.

8. What are the connectors used in pairs called?
A. Jack and plug
B. Male and female connectors
C. Both the mentioned
D. None of the mentioned
Answer: C
Clarification: There are certain special connectors that can be used to connect two devices operating devices. These special connectors are to be used in pairs and are with both the names.

9. The frequency range for N type co axial connector is:
A. 8-12 GHz
B. 11-18 GHz
C. 14-20 GHz
D. 2-8 GHz
Answer: B
Clarification: N type connectors are coaxial connectors used at microwave frequency range. These type of connectors can be used in the frequency range of 11-18 Hz.

10. The frequency range of SMA co-axial connector used most commonly is:
A. 18-25 GHz
B. 25-50 GHz
C. 11-18 GHz
D. 8-12 GHz
Answer: A
Clarification: SMA coaxial cable connectors are designed to operate at high frequencies. The frequency ranges from 18-25 GHz.