Antennas And Microwave Engineering Set 6

Electrical Engineering

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This set of Antennas and Microwave Engineering Multiple Choice Questions & Answers (MCQs) focuses on Antennas And Microwave Engineering Set 6

Q1 | The klystron tube used in a klystron amplifier is a                    type beam amplifier.
  • linear beam
  • crossed field
  • parallel field
  • none of the mentioned
Answer: linear beam
Explaination:
in klystron amplifier, the electron beam passes through two or more resonant cavities. the first cavity accepts an rf input and modulates the electron beam by bunching it into high density and low density regions.
Q2 | In crossed field tubes, the electron beam traverses the length of the tube and is parallel to the electric field.
  • true
  • false
Answer: false
Explaination:
in a crossed field or ‘m’ type tubes, the focusing field is perpendicular to the accelerating electric field. since the focusing field and accelerating fields are perpendicular to each other, they are called crossed field tubes.
Q3 |                  is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity.
  • backward wave oscillator
  • reflex klystron
  • travelling wave tube
  • magnetrons
Answer: reflex klystron
Explaination:
reflex klystron is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity to provide positive feedback via the electron beam. it can be tuned by mechanically adjusting the cavity size.
Q4 | A major disadvantage of klystron amplifier is:
  • low power gain
  • low bandwidth
  • high source power
  • design complexity
Answer: low bandwidth
Explaination:
klystron amplifier offers a very narrow operating bandwidth. this is overcome in travelling wave tube (twt).
Q5 | In a                    oscillator, the RF wave travels along the helix from the collector towards the electron gun.
  • interaction oscillator
  • backward wave oscillator
  • magnetrons
  • none o the mentioned
Answer: backward wave oscillator
Explaination:
in a backward wave oscillator, the rf wave travels along the helix from the collector towards the electron gun. thus the signal for oscillation is provided by the bunched electron beam itself and oscillation occurs.
Q6 | Magnetrons are microwave devices that offer very high efficiencies of about 80%.
  • true
  • false
Answer: true
Explaination:
magnetrons are capable of very high power outputs, on the order of several kilowatts, and with efficiencies of 80% or more. but disadvantage of magnetron is that they are very noisy and cannot maintain frequency or phase coherence when operated in pulse mode.
Q7 | Klystron amplifiers have high noise output as compared to crossed field amplifiers.
  • true
  • false
Answer: false
Explaination:
crossed filed amplifiers have very good efficiencies – up to 80%, but the gain is limited to 10-15 db) in addition, the cfa has a noisier output than either a klystron amplifier or twt. its bandwidth can be up to 40%.
Q8 |                          is a microwave device in which the frequency of operation is determined by the biasing field strength.
  • vtm
  • gyratron
  • helix bwo
  • none of the mentioned
Answer: gyratron
Explaination:
gyratron is a microwave device in which the frequency of operation is determined by the biasing field strength and the electron velocity, as opposed to the dimensions of the tube itself. this makes the gyrator especially useful for microwave frequencies.
Q9 | 2 IMPEDANCE MATCHING
  • swr, first voltage minimum
  • swr, first voltage maximum
  • characteristic impedance, first voltage minimum
  • characteristic impedance, first voltage maximum
Answer: swr, first voltage minimum
Explaination:
with a slotted line, swr and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.
Q10 | A modern device that replaces a slotted line is:
  • digital cro
  • generators
  • network analyzers
  • computers
Answer: network analyzers
Explaination:
although slotted lines used to be the principal way of measuring unknown impedance at microwave frequencies, they have largely been superseded by the modern network analyzer in terms of accuracy, versatility and convenience.
Q11 | If SWR=1.5 with a wavelength of 4 cm and the distance between load and first minima is 1.48cm, then the reflection coefficient is:
  • 0.0126+j0.1996
  • 0.0128
  • 0.26+j0.16
  • none of the mentioned
Answer: 0.0126+j0.1996
Explaination:
┌= (swr-1)/ (swr+1). substituting for swr in the above equation for reflection co-efficient, magnitude of the reflection co-efficient is 0.2. to find θ, θ=π+2βlmin, substituting lmin as 1.48cm, θ=86.4⁰. hence converting the polar form of the reflection co-efficient into rectangular co- ordinates, reflection co-efficient is 0.0126+j0.1996.
Q12 | High gain is not achievable at microwave frequencies using BJT amplifiers because:
  • device construction
  • complex architecture
  • ports are not matched at high frequencies
  • none of the mentioned
Answer: ports are not matched at high frequencies
Explaination:
at higher frequencies, if higher bandwidth is desired, a compromise on maximum achievable gain is made. but at these higher frequencies, the ports of the amplifier are not matched to 50 Ω.
Q13 | To flatten the gain response of a transistor:
  • biasing current has to be increased
  • input signal level has to increased
  • increase the operational bandwidth
  • give negative feedback to the amplifier
Answer: give negative feedback to the amplifier
Explaination:
negative feedback can be used to increase the gain response of the transistor, improve the input and output match, and increase the stability of the device.
Q14 | In conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. But this drawback can be overcome by using:
  • balanced amplifiers
  • distributed amplifiers
  • differential amplifiers
  • none of the mentioned
Answer: balanced amplifiers
Explaination:
in conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. but this drawback can be overcome by using balanced amplifiers. this is overcome by using two 900 couplers to cancel input and output reflections from two identical amplifiers.
Q15 | Bandwidth of balanced amplifier can be an octave or more, but is limited by the bandwidth of the coupler.
  • true
  • false
Answer: true
Explaination:
in order to achieve flat gain response, balanced amplifiers use couplers to minimize reflections. but this in turn reduces the bandwidth of the amplifier to the coupler bandwidth.
Q16 | Coupler that is mostly used in balanced amplifiers to achieve the required performance is:
  • branch line coupler
  • wilkinson coupler
  • lange coupler
  • waveguide coupler
Answer: lange coupler
Explaination:
lange couplers are broadband couplers and are compact in size. since the bandwidth of a balanced amplifiers depends on the bandwidth of the coupler used. lange coupler is thus preferred over couplers.
Q17 | Distributed amplifiers offer very high
  • gain
  • bandwidth
  • attenuation
  • none of the mentioned
Answer: bandwidth
Explaination:
distributed amplifiers offer very high bandwidth of about 10 decade. but higher gain cannot be achieved using distributed amplifiers and matching at the ports is very important to achieve higher bandwidth.
Q18 | In distributed amplifiers, all the FET stages in the amplifier are connected in series to one another.
  • true
  • false
Answer: false
Explaination:
in distributed amplifiers, cascade of n identical fets have their gates connected to a transmission line having a characteristic impedance of zg with a spacing of lg while the drains are connected to a transmission line of characteristic impedance zd, with a spacing ld.
Q19 |                          uses balanced input and output, meaning that there are 2 signal lines, with opposite polarity at each port.
  • differential amplifier
  • distributed amplifier
  • balanced amplifier
  • none of the mentioned
Answer: differential amplifier
Explaination:
differential amplifier uses balanced input and outputs, meaning that there are 2 signal lines, with opposite polarity at each port. it has two input ports and one output port. the difference of the 2 input signals is amplified.
Q20 | A major advantage of differential amplifiers is:
  • high gain
  • low input impedance
  • higher output voltage swing
  • none of the mentioned
Answer: higher output voltage swing
Explaination:
differential amplifiers can provide higher voltage swings that are approximately double that obtained with single ended amplifier.
Q21 | Along with a differential amplifier, 1800 hybrid is used both at the input and output.
  • true
  • false
Answer: true
Explaination:
a differential amplifier can be constructed using two single-ended amplifiers
Q22 |                    is a non linear circuit that converts DC power to an AC waveform of desired frequency based on the oscillator design.
  • attenuator
  • amplifier
  • oscillator
  • none of the mentioned
Answer: oscillator
Explaination:
oscillator is a non linear circuit that converts dc power to an ac waveform. most rc oscillators provide sinusoidal outputs, which minimizes undesired harmonics and noise sidebands.
Q23 | The transfer function of an RF oscillator is given by:
  • a/ (1-ah (ω))
  • a/ (1+ah (ω))
  • a/ (-1+ah (ω))
  • 1/ (1-ah (ω))
Answer: a/ (1-ah (ω))
Explaination:
transfer function of an rf oscillator is given by a/ (1-ah (ω)). here, a is the gain of the transistor multiplier used.
Q24 | The criterion on which oscillations are produced in the oscillator circuit is called:
  • shannon’s criteria
  • barkhausen criteria
  • colpitts criteria
  • none of the mentioned
Answer: barkhausen criteria
Explaination:
when the condition 1-ah (ω)
Q25 | The necessary condition for oscillation in a Colpitts oscillator is:
  • c2/c1=gm/gi
  • c1/c2=gm/gi
  • c2/c1= gm*gi
  • none of the mentioned
Answer: c2/c1=gm/gi
Explaination:
the condition for sustained oscillation in a colpitts oscillator is c2/c1 = gm/gi. here c1 and c2 are the capacitance in the feedback network, gm is the transconductance of the transistor and gi is the input admittance.
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