Microwave Engineering Quiz! Exam

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| By Dr. Dinesh
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Dr. Dinesh
Community Contributor
Quizzes Created: 1 | Total Attempts: 451
Questions: 19 | Attempts: 451

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Microwave Engineering Quiz! Exam - Quiz

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Questions and Answers
  • 1. 

    A waveguide section in a microwave circuit acts as:

    • A.

      LP filter

    • B.

      All pass filter

    • C.

      HP filter

    • D.

      Band stop filter

    Correct Answer
    C. HP filter
    Explanation
    A waveguide section in a microwave circuit acts as a high pass (HP) filter because it allows high-frequency signals to pass through while attenuating or blocking low-frequency signals. The physical dimensions of the waveguide section are designed to support the propagation of higher-frequency waves, effectively filtering out lower-frequency components. This allows the waveguide section to selectively transmit high-frequency signals and reject or attenuate low-frequency signals, making it function as a high pass filter.

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  • 2. 

    Which of the following devices uses a slow wave structure?

    • A.

      Klystron two cavity amplifier

    • B.

      Klystron multicavity amplifier

    • C.

      Travelling wave tube

    • D.

      Reflex klystron oscillator

    Correct Answer
    C. Travelling wave tube
    Explanation
    A travelling wave tube uses a slow wave structure. A slow wave structure is a device that slows down the propagation of electromagnetic waves, allowing for efficient energy transfer between the electrons and the waves. The travelling wave tube achieves this by using a helix or a coupled cavity structure, which creates a longer interaction path for the electrons and the waves. This slow wave structure enhances the amplification capabilities of the travelling wave tube, making it suitable for high-power applications such as satellite communication and radar systems.

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  • 3. 

    A line has Z0 = 300 ∠ 0° Ω. If ZL = 150 ∠ 0° Ω, reflection coefficient is

    • A.

      0.5

    • B.

      0.3333

    • C.

      -0.5

    • D.

      -0.3333

    Correct Answer
    D. -0.3333
    Explanation
    The reflection coefficient can be calculated using the formula (ZL - Z0) / (ZL + Z0). In this case, substituting the given values, we get (150 - 300) / (150 + 300) = -0.3333.

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  • 4. 

    In a directional coupler:

    • A.

      Isolation (dB) equals coupling plus directivity

    • B.

      Coupling (dB) equals isolation plus directivity

    • C.

      Directivity (dB) equals isolation plus coupling

    • D.

      isolation (dB) equals (coupling) (directivity)

    Correct Answer
    A. Isolation (dB) equals coupling plus directivity
    Explanation
    The correct answer is that in a directional coupler, isolation (dB) equals coupling plus directivity. This means that the amount of isolation between the input and output ports of the coupler is determined by both the coupling factor (which determines how much power is transferred from the input to the output port) and the directivity (which determines how much power is reflected back towards the input port). By adding these two factors together, we can determine the overall level of isolation provided by the coupler.

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  • 5. 

    The reflection coefficient on a line is 0.2 ∠45°. The SWR is:

    • A.

      0.8

    • B.

      1.1

    • C.

      1.2

    • D.

      1.5

    Correct Answer
    D. 1.5
    Explanation
    The reflection coefficient on a line is given by the formula SWR = (1 + |Γ|) / (1 - |Γ|), where Γ is the reflection coefficient. In this case, the reflection coefficient is 0.2 ∠45°. The magnitude of the reflection coefficient is 0.2, so |Γ| = 0.2. Plugging this value into the formula, we get SWR = (1 + 0.2) / (1 - 0.2) = 1.2 / 0.8 = 1.5. Therefore, the SWR is 1.5.

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  • 6. 

    The velocity factor of a transmission line depends on:

    • A.

      Temperature

    • B.

      Skin effect

    • C.

      Relative permittivity of dielectric

    • D.

      None of the above

    Correct Answer
    C. Relative permittivity of dielectric
    Explanation
    The velocity factor of a transmission line depends on the relative permittivity of the dielectric. The relative permittivity, also known as the dielectric constant, is a measure of how much the electric field in the dielectric material is reduced compared to the electric field in a vacuum. It affects the speed at which electromagnetic waves propagate through the transmission line. Different dielectric materials have different relative permittivity values, which in turn affect the velocity factor of the transmission line.

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  • 7. 

    A frequency at which microwave ovens operate is:

    • A.

      50 GHz

    • B.

      8 GHz

    • C.

      2.45 GHz

    • D.

      3.5 GHz

    Correct Answer
    C. 2.45 GHz
    Explanation
    Microwave ovens operate at a frequency of 2.45 GHz. This frequency is commonly used because it allows for efficient heating of food and is also the frequency at which water molecules in food absorb the most energy. As a result, the microwave radiation is able to generate heat and cook the food effectively.

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  • 8. 

    Which of the following materials are generally preferred for waveguides?

    • A.

      Cast iron and steel

    • B.

      Nonmetallic solids including plastic

    • C.

      High carbon steel and vanadium steel

    • D.

      Brass and aluminium

    Correct Answer
    D. Brass and aluminium
    Explanation
    Brass and aluminium are generally preferred for waveguides because they have low electrical conductivity, which allows for the transmission of electromagnetic waves without significant loss. Additionally, both materials have good mechanical properties, such as high strength and durability, making them suitable for use in waveguide applications.

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  • 9. 

    In a magnetron why do the electrons travel in a cycloidal path?

    • A.

      The cathode is positive

    • B.

      Strong field is supplied by the permanent magnet

    • C.

      The anode is negative

    • D.

      The cavities are resonant

    Correct Answer
    B. Strong field is supplied by the permanent magnet
    Explanation
    The strong field supplied by the permanent magnet in a magnetron causes the electrons to travel in a cycloidal path. This is because the magnetic field exerts a force on the electrons, causing them to move in a curved path. The cycloidal path is a result of the combination of the magnetic force and the electric field between the anode and cathode. The electrons are accelerated towards the anode by the electric field, but the magnetic field deflects their path, resulting in a cycloidal motion.

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  • 10. 

    Which of the following is wrong for a magic used to tee?

    • A.

      E and H arms are decoupled

    • B.

      Coplanar arms are coupled

    • C.

      All ports are perfectly matched

    • D.

      A signal into coplanar arm splits equally between E and H arms

    Correct Answer
    D. A signal into coplanar arm splits equally between E and H arms
    Explanation
    The given statement states that a signal into the coplanar arm splits equally between the E and H arms. However, this is incorrect because in a magic tee, the signal splits equally between the E and H arms only when the arms are decoupled, which is mentioned as a wrong statement in the question. Therefore, the correct answer is that a signal into the coplanar arm does not split equally between the E and H arms.

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  • 11. 

    Which of the following bands that not comes under Microwave Band?

    • A.

      C band

    • B.

      X band

    • C.

      Ku band

    • D.

      F band

    Correct Answer
    D. F band
    Explanation
    The F band does not come under the Microwave Band because it is a frequency range used for satellite communications and has a lower frequency compared to the microwave bands. Microwave bands typically range from 1 GHz to 300 GHz, while the F band operates at around 4 GHz. Therefore, the F band is not considered a part of the microwave band.

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  • 12. 

    Which of the following is the main advantage of the microwave?

    • A.

      Highly directive

    • B.

      Moves at the speed of light

    • C.

      Greater S/N ratio

    • D.

      High penetration power

    Correct Answer
    A. Highly directive
    Explanation
    The main advantage of the microwave is its highly directive nature. This means that the microwave can be focused and concentrated in a specific direction, allowing for precise targeting and transmission of signals. This makes it ideal for applications such as long-distance communication, radar systems, and satellite communication. The highly directive nature of microwaves also helps in minimizing interference and maximizing signal strength, making it a preferred choice in various industries.

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  • 13. 

    _______ is the best medium for handling large microwave power.

    • A.

      Coaxial line

    • B.

      Rectangular wave guide

    • C.

      Strip line

    • D.

      Copper wires

    Correct Answer
    B. Rectangular wave guide
    Explanation
    A rectangular waveguide is the best medium for handling large microwave power because it has a larger cross-sectional area compared to other options like coaxial lines and strip lines. This allows for better power handling capabilities and lower transmission losses. The rectangular shape of the waveguide also helps in guiding the microwave signals effectively without significant signal degradation. Additionally, the waveguide's design allows for easy integration with other microwave components and systems.

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  • 14. 

    Which of the following is used for amplification of microwave energy?

    • A.

      Reflex Klystron

    • B.

      Megnetron

    • C.

      Travelling wave tube

    • D.

      Gunn Diode

    Correct Answer
    C. Travelling wave tube
    Explanation
    The travelling wave tube is used for amplification of microwave energy. It is a specialized vacuum tube that utilizes electromagnetic waves traveling along a helix to amplify microwave signals. As the signal passes through the helix, it interacts with an electron beam, causing the signal to be amplified. This technology is commonly used in applications such as satellite communications, radar systems, and microwave ovens.

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  • 15. 

    Which of the following microwave tube amplifier uses an axial magnetic field & radial electric field?

    • A.

      Reflex Klystron

    • B.

      Coaxial Magnetron

    • C.

      Travelling Wave Tube

    • D.

      Crossed field Megnetron

    Correct Answer
    D. Crossed field Megnetron
    Explanation
    The crossed field magnetron is the microwave tube amplifier that uses an axial magnetic field and a radial electric field. In this type of magnetron, the magnetic field is applied along the axis of the tube, while the electric field is applied radially. This configuration allows for the generation of high-power microwaves by the interaction of the electron beam with the crossed electromagnetic fields. The crossed field magnetron is commonly used in radar systems and microwave ovens.

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  • 16. 

    The fabrication of microstrip line is done by?

    • A.

      Etching

    • B.

      Printed circuit Board technique

    • C.

      Oxidation

    • D.

      Cladding

    Correct Answer
    B. Printed circuit Board technique
    Explanation
    The fabrication of microstrip line is done using the printed circuit board (PCB) technique. This technique involves creating a thin conductive trace on a PCB substrate, which serves as the microstrip line. The PCB technique is commonly used for fabricating microstrip lines due to its simplicity, cost-effectiveness, and ability to produce accurate and precise results. By using this technique, the desired dimensions and characteristics of the microstrip line can be achieved, making it a preferred method for manufacturing microstrip lines.

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  • 17. 

    For TE10 mode a rectangular wave guide has dimensions are 7 * 3.5 cm2. its cut off frequency is:

    • A.

      2.14 GHz

    • B.

      2.5 GHz

    • C.

      1.5 GHz

    • D.

      1.9 GHz

    Correct Answer
    A. 2.14 GHz
    Explanation
    The cutoff frequency of a waveguide is the frequency below which the waveguide cannot support any propagating modes. In the case of the TE10 mode, the cutoff frequency is determined by the dimensions of the waveguide. A rectangular waveguide with dimensions of 7 * 3.5 cm^2 will have a cutoff frequency of 2.14 GHz.

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  • 18. 

    For the case of "No Propagation" the propagation constant must be:

    • A.

      Real

    • B.

      Imaginary

    • C.

      Complex

    • D.

      None of Above

    Correct Answer
    A. Real
    Explanation
    For the case of "No Propagation", the propagation constant must be real. This means that there is no change in the amplitude or phase of the signal as it propagates through the medium. In other words, the signal does not experience any attenuation or distortion. A real propagation constant indicates that the medium is lossless and the signal travels without any changes in its characteristics.

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  • 19. 

    The range of VSWR given as:

    • A.

      0 to 1

    • B.

      1 to 2

    • C.

      1 to infinite

    • D.

      -1 to 1

    Correct Answer
    C. 1 to infinite
    Explanation
    The correct answer is "1 to infinite" because VSWR (Voltage Standing Wave Ratio) is a measure of how well a transmission line is matched to the impedance of the connected device. A VSWR of 1 indicates a perfect match, meaning all the power is transmitted without any reflections. As the VSWR increases beyond 1, it indicates an imperfect match and the amount of power reflected back increases. A VSWR of infinite means that all the power is reflected back, indicating a complete mismatch. Therefore, the range of VSWR can be from 1 to infinite.

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