# Multiple choice question for engineering

## Set 1

1. ______ is a device that produces a phase shift of a required amount of the input wave.
a) Phase shifter
b) Attenuator
c) Resonator
d) None of the mentioned

Answer: a [Reason:] Ferrite phase shifter is a two port component that provides a variable phase shift by changing the bias field of the ferrite. Microwave diodes and FETs can also be used to implement phase shifters.

2. Phase shifters are used in _______ where the antenna beam can be steered in space by electronically controlled phase shifters.
a) Phased array antennas
b) Dipole array antennas
c) Slot antennas
d) Patch antennas

Answer: a [Reason:] Phase shifters are used in phased array antennas where the antenna beam can be steered in space by electronically controlled phase shifters. Phase shifters can be of two types, Reciprocal phase shifters and non reciprocal phase shifters. These are chosen as per the application requirement.

3. Reciprocal phase shifters give different phase shift in different direction.
a) True
b) False

Answer: b [Reason:] Reciprocal phase shifters are those devices which give the same phase shift in either direction. That is, either if port 1 or port 2 of the phase shifter is used as input port, the phase shifts at the output remains the same.

4. If a ferrite slab provides a phase shift of 48⁰/ cm, then the length of the ferrite slab required to produce a phase shift of 180⁰ is:
a) 4 cm
b) 3.75 cm
c) 4.5 cm
d) 3.5 cm

Answer: b [Reason:] The given ferrite slab provides a phase shift of the 48⁰/ cm. hence the length of the required ferrite slab is 180/45, the required length is 3.75 cm.

5. If a ferrite slab provides a phase shift of 48⁰/ cm, then the length of the ferrite slab required to produce a phase shift of 90⁰ is:
a) 2.44 cm
b) 1.88 cm
c) 4.5 cm
d) 3.5 cm

Answer: b [Reason:] The given ferrite slab provides a phase shift of the 48⁰/ cm. hence the length of the required ferrite slab is 90/45, the required length is 1.88 cm.

6. Gyrator is a device that produces a phase shift of ____ between the input and output.
a) 90⁰
b) 180⁰
c) 45⁰
d) None of the mentioned

Answer: b [Reason:] Gyrator is a device that produces a phase shift of 180⁰ between the input and output of the gyrator. This is a special case of the ferrite phase shifter which gives a constant phase shift and cannot be changed.

7. The scattering matrix of a gyrator is:
a) Symmetric
b) Skew symmetric
c) Identity matrix
d) Null matrix

Answer: b [Reason:] The scattering matrix of a gyrator is Skew symmetric. This is because of the 180⁰ phase shift that occurs in the device.

8. Ferrite phase shifters have more advantages over FETs and diodes in using them in microwave integrated circuits.
a) True
b) False

Answer: Even though PIN diode and FET circuits offer a less bulky and more integratable alternative to ferrite components, ferrite phase shifters are cost effective; have high power handling capacity and power requirements.

9. A gyrator can be made a passive device by certain design methods so that they do not affect the power levels of the circuit in which they are used.
a) True
b) False

Answer: a [Reason:] The gyrator can be implemented as a phase shifter with a 180⁰ phase shift; bias can be provided with a permanent magnet, making the gyrator a passive device.

10. If a ferrite slab produces a phase shift of 0.836 rad/ cm, then the length of the slab required to produce a phase shift of 135⁰ is:
a) 2.81 cm
b) 3 cm
c) 2 cm
d) 3.4 cm

Answer: a [Reason:] Converting the given phase shift from radian scale to degree scale, the produced phase shift is 48⁰/ cm. To produce a phase shift of 135⁰, the required length is 135/45 this is equal to 2.81 cm.

## Set 2

1. Field effect transistors are different from BJTs in that they are _________
a) monopolar devices
b) bipolar devices
c) bidirectional device
d) none of the mentioned

Answer: a [Reason:] FETs are called monopolar devices, with only one carrier type, either electrons or holes providing current flow through the device. N-channel FETs employ electrons while p-channel FETs employ holes as source of current.

2. GaAs MESFET –metal semiconductor field effect transistor are one of the widely used categories of FETs.
a) true
b) false

Answer: a [Reason:] One of the most important developments in microwave technology has been the GaAs metal semiconductor field effect transistor, as this device permitted the first practical solid-state implementation of amplifiers and oscillators.

3. At frequencies above 10GHz, MESFET are not suitable for microwave applications due to parasitic effects.
a) true
b) false

Answer: b [Reason:] GaAs MESFETs can be used at frequencies well into the millimeter wave range, with high gain and low noise figure often making them the device of choice in monolithic microwave integrated circuits above frequencies of 10 GHz.

4. Advantage of using GaAs in MESFET as compared to use of silicon is:
a) GaAs are cost effective
b) they have higher mobility
c) they have high resistance for flow of current in the reverse direction
d) none of the mentioned

Answer: b [Reason:] The desired high gain and noise features of this transistor are a result of high electron mobility of GaAs compared to silicon and the absence of shot noise in them.

5. In MESFET, an applied signal at the gate modulates the electron carriers; this produces _______ in the FET.
a) voltage amplification
b) voltage attenuation
c) electron multiplication
d) electron recombination

Answer: a [Reason:] In operation, the electrons are drawn from the source to drain by a positive voltage applied to the source and drain. These carriers are modulated by the voltage applied to the gate hence resulting in voltage amplification.

6. The frequency of operation of an FET is limited by:
a) drain to source voltage
b) gate to source voltage
c) gate length
d) effective area of an FET

Answer: c [Reason:] The frequency of operation of an FET is given by the gate length. Present FETs have a gate length of the order of 0.2-0.6 µm, with corresponding upper frequency limits of 100-50 GHz.

7. The S21 parameter for a MESFET is lesser than 1.
a) true
b) false

Answer: b [Reason:] From the analysis of the small signal equivalent circuit of MESFE, S21 parameter of the transistor was found to be greater than one under normal operating conditions. Here port 1 is at the gate and port 2 is at the drain.

8. The expression for short circuit current gain of an FET is given by:
a) gm/ ωCgs
b) Ig/gmVc
c) ωCgs/ gm
d) none of the mentioned

Answer: a [Reason:] Short circuit current gain of an FET is defined as the ratio of drain current to gate current when the output is short circuited. This is expressed as ID/IG. This ratio in simplified form is given as gm/ ωCgs.

9. The upper threshold frequency of an FET, where short circuit gain is unity is given by:
a) gm/2πCgs
b) gm/Cgs
c) gm/ 2π
d) none of the mentioned

Answer: a [Reason:] The upper threshold frequency is dependent on the factor gm, associated with the current generator of the small signal equivalent circuit. Cgs is the capacitance measured between the gate and source terminals.

10. The scattering parameter S11 for an FET __________ with increase in the frequency of operation of the transistor.
a) increases
b) decreases
c) remains constant
d) none of the mentioned

Answer: b [Reason:] S11 parameter of an FET decreases with the increase in the frequency of operation of an FET. The measured values are 0.97 at 1 GHz, and 0.49 at 12 GHz.

11. The curve of IDS v/s VDS of an FET does not vary with the gate to source voltage applied.
a) true
b) false

Answer: b [Reason:] Curve of IDS v/s VDS of an FET varies with the gate to source voltage applied. As the gate to source voltage applied becomes more positive, the drain to source current goes on increasing for an applied constant gate to source voltage.

12. High-power circuits generally use higher values of:
a) gate to source current
b) drain to source current
c) drain current
d) gate to source voltage

Answer: c [Reason:] In order to achieve high drain current for high power applications, DC bias voltage must be applied to both gate and the drain, without disturbing the RF signal paths.

13. High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply.
a) true
b) false

Answer: a [Reason:] High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply. The RF chokes provide a very low DC resistance for biasing, and a very high impedance at RF frequencies to isolate the signal from the bias supply.

## Set 3

1. In a two wire transmission line, if the distance between the lines is 20 mm and the radii is 5 mm then the inductance of the line is:
a) 0.1 µH
b) 0.526 µH
c) 0.9 µH
d) 1 µH

Answer: b [Reason:] The inductance of a two wire transmission line is given by the equation µ*ln(b/a)/2π. Substituting the given values in the above equation, inductance is 0.526 µH.

2. In a two wire transmission line , if the distance between the lines is 60mm and the radii is 10 mm , then the capacitive reactance of the line when operated at 12.5 GHz is
a) 20 pF
b) 21.13 pF
c) 23 pF
d) 12 pF

Answer: b [Reason:] The capacitive reactance of a two wire transmission line is π∈/cosh-1(D/2a). substituting the given values in the above expression, the capacitive reactance is 21.13 pF.

3. For a parallel plate type of a transmission line, then expression for conductance of the line is:
a) ∈(ω)w/d
b) 2Rx/x
c) μ/2π ln⁡(w/d)
d) μ/π cosh-1(w/2d)

Answer: a [Reason:] The conductance of a parallel plate waveguide is dependent on the complex value of the permittivity, width of the waveguide and the distance between the waveguide plates.

4. One of the Maxwell‘s curl equation that is satisfied inside a coaxial line is:
a) ∇×E =-jωµ (vector H)
b) ∇×E =-jωμ(vector E)
c) ∇×H=-jωμ(vector H)
d) ∇×H=jωμ(vector H)

Answer: a [Reason:] ∇×E = -jωµ (vector H).This is the Maxwell’s equation satisfied by the electric and magnetic fields inside a waveguide.

5. The wave impedance of air for a wave propagating in it is:
a) 377 Ω
b) 345 Ω
c) Insufficient data
d) None of the mentioned

Answer: a [Reason:] Intrinsic impedance is the impedance offered by air for a wave propagating in it. This is a standard value and is 377 Ω.

6. Wave impedance of a wave travelling in a medium of a relative permittivity 2 and permeability 4 is
a) 188.5 Ω
b) 200 Ω
c) 300 Ω
d) None of the mentioned

Answer: a [Reason:] Intrinsic impedance of a medium is given by the expression √μ/ϵ. Substituting the given values in the above expression, the wave impedance is 188.5 Ω.

7. For a parallel plate transmission line, if w= 12 mm and the distance between the plates is 2 mm, then the inductance of the transmission line is:
a) 0.2 µH
b) 0.1 µH
c) 0.3 µH
d) 0.4 µH

Answer: a [Reason:] The inductance of a parallel plate transmission line is given by µd/W. substituting the given values in the above expression, the inductance is 0.2 µH.

8. Expression for capacitance of a two wire transmission line is
a) ∈’*π/cosh⁡-1(D/2a)
b) μ/π*cosh-1(D/2a)
c) 2π∈/ln⁡(D/2a)
d) ∈”*πω/cosh-1(D/2a)

Answer: a [Reason:] The expression for capacitance of a two wire transmission line is ∈’ π/cosh-1(D/2a). Capacitance of a two wire transmission line is dependent on the distance between the two lines and the radius of the line.

9. If the distance between the 2 wires in a 2 wire transmission line is 10 mm and the radii 2 mm, then the inductance of the transmission line is:
a) 0.62 µH
b) 1 µH
c) 2 µH
d) None of the mentioned

Answer: a [Reason:] The inductance of a two wire transmission line is given by the expression µcosh-1(D/2a)/π. Substituting the given values in the above expression, the inductance is 0.62 µH.

10. For a parallel plate transmission line, if the complex part of permittivity is 2.5, if the width is 100 mm and the distance between the plates is 10 mm, then the conductance of the transmission line is:
a) 25 Ʊ
b) 30 Ʊ
c) 45 Ʊ
d) None of the mentioned

Answer: a [Reason:] Conductance of a parallel plate transmission line is ∈W/d. substituting the given values in the above expression, the conductance of the transmission line is 25 Ʊ.

11. For a parallel plate transmission line, if the series resistance is 10 mΩ/m, and the width is 100 mm, then the resistance of the transmission line is:
a) 0.2 Ω
b) 1 Ω
c) 2 Ω
d) 5 Ω

Answer: a [Reason:] For a parallel plate transmission line, the series resistance is given by the expression 2RS/W. substituting the given values in the above expression, the series resistance is 0.2 Ω.

## Set 4

1. Oscillators operating at millimeter wavelength are difficult to realize and are also less efficient.
a) true
b) false

Answer: a [Reason:] As frequency increases to the millimeter wave range, it becomes increasingly difficult to build fundamental frequency oscillators with good power, stability and noise characteristics. An alternative approach is to produce a harmonic of a low frequency oscillator through the use of frequency multiplier.

2. __________ is an example for a frequency multiplier.
a) resistor
b) inductor
c) capacitor
d) transistor

Answer: d [Reason:] A non linear device has the ability to generate the harmonics of the input sinusoidal signal. Transistor and diodes are non linear devices and hence can be used as a frequency multiplier.

3. The major drawback of frequency multipliers is that they have:
a) higher attenuation
b) complex construction methods
c) complex design
d) none of the mentioned

Answer: c [Reason:] Designing a good quality frequency multiplier is more difficult since it non-linear analysis, matching at multiple frequencies, stability analysis and thermal considerations. Considering all these issues for designing a multiplier makes it very complex.

4. A reactive diode multiplier uses _______ as the key electronic component for frequency multiplication.
a) zener diode
b) light emitting diode
c) varactor diode
d) Gunn diode

Answer: c [Reason:] Reactive diode multipliers use either a varactor diode or step recovery diode biased to present a non linear junction capacitance. Since losses in these diodes are low, the fraction of RF power converted to the desired harmonic is relatively high.

5. A major disadvantage of frequency multipliers is that they multiply the noise factor along with frequency.
a) true
b) false

Answer: a [Reason:] A disadvantage of frequency multipliers is that noise levels are also increased by the multiplication factor. Frequency multiplication process is a phase multiplication process as well, so phase noise variations get multiplied by the same factor as the frequency gets multiplied.

6. If a frequency multiplier has a multiplication factor of 10, then the increase in noise level due to frequency multiplication is:
a) 10 dB
b) 20 dB
c) 25 db
d) 15 dB

Answer: b [Reason:] For a frequency multiplier, the increase in noise power is given by 20 log n, where n is the multiplication factor of the multiplier. Substituting in the below equation, increase in noise level is 20 dB.

7. In a diode frequency multiplier, an input signal of frequency fo applied to the diode is terminated with_________ at all frequencies other than required harmonic.
a) real impedances
b) reactive impedance
c) complex impedance
d) none of the mentioned

Answer: b [Reason:] In a diode frequency multiplier, an input signal of frequency fo applied to the diode is terminated with reactive impedance at all frequencies other than required harmonic nfo. if the diode junction capacitance has a square –law I-V characteristic , it is necessary to terminate unwanted harmonics with short circuit.

8. Resistive multipliers are more efficient as compared to reactive multipliers.
a) true
b) false

Answer: b [Reason:] Resistive multipliers generally use forward biased Schottky-barrier diodes to provide non linear characteristic. Resistive multipliers have low efficiency but have better bandwidth.

9. Reactive multipliers have a disadvantage that they cannot be used at very high frequencies and they become less efficient.
a) true
b) false

Answer: a [Reason:] At millimeter frequencies, varactor diode exhibits resistive property. Hence, at high frequency the multiplier becomes lossy and also does not offer high bandwidth, which is a major disadvantage.

10. For a resistive frequency multiplier of multiplication factor 2, the maximum theoretical conversion efficiency is:
a) 50 %
b) 25 %
c) 75 %
d) 12.5 %

Answer: b [Reason:] For a resistive frequency multiplier of multiplication factor 2, the maximum theoretical conversion efficiency is given by 1/m2 where m is the multiplication factor. For a factor 2 multiplier, maximum theoretical conversion efficiency is 25 %.

## Set 5

1. Surface waves are typical by a field that decays ______away from the dielectric surface, with most of the field contained in or near the dielectric.
a) Linearly
b) Exponentially
c) Cubical
d) Field remains a constant

Answer: b [Reason:] Surface waves are typified by a field that decays exponentially away from the dielectric surface, with most of the field contained in or near the dielectric. At higher frequencies, the field generally becomes more tightly bound to the dielectric, making such waveguides practical.

2. Because of the presence of the dielectric, the phase velocity of a surface wave is:
a) Greater than that in vacuum
b) Lesser than that in vacuum
c) Independent of the presence of dielectric
d) Insufficient data

Answer: b [Reason:] The fields are stronger and concentrated near the dielectric, and hence because of the presence of the dielectric, the phase velocity of a surface wave is lesser than that in vacuum.

3. For wave propagation on grounded dielectric sheet, the equation to be satisfied by Ez , in the region of presence of dielectric 0≤x≤d for the propagation to be in Z direction
a) (∂2/∂x2 + ∈rk02– β2) eZ(x,y)=0
b) (∂2/∂x2 + k02– β2) eZ(x,y)=0
c) (∂2/∂x2 – k022) eZ(x,y)=0
d) (∂2/∂x2 + ∈rk02) eZ(x,y)=0

Answer: a [Reason:] The equation describes the variation of the electric field along the direction of propagation that is the Z direction. In the equation, it is clear that the relative permittivity of the dielectric is also a part of the second term of the equation.

4. The cut off wavenumber for the region of dielectric in a grounded dielectric sheet is:
a) kC2= ∈rk022
b) kC2= ∈rk022
c) h2= -k022
d) kC2= k2 + β2

Answer: a [Reason:] Cutoff wave number signifies the minimum threshold wave number required for propagation. Here the expression kC2= ∈rk022 gives the cutoff wave number for the propagation of waves on a grounded dielectric sheet.

5. For surface waves on a dielectric sheet the cutoff frequency of the TM mode can be given as:
a) fC = nC/2d√(ϵr-1)
b) fC = C/2nd√(ϵr-1)
c) fC = C/2nd√ϵr
d) fC = 2C/nd√ϵr

Answer: a [Reason:] Grounded dielectric sheets allow TM mode of propagation on them. The cut off frequency for the propagation of TM mode is given by the expression fC = nC/2d√(ϵr-1).

6. The cutoff frequency in TM1 mode for the propagation of EM waves on dielectric slab of relative permittivity 2.6 and thickness 20 mm is:
a) 6.5 GHz
b) 5.92 GHz
c) 4 GHz
d) 2 GHz

Answer: b [Reason:] The expression for cutoff frequency for wave propagation in TMn mode is fC = nC/2d√(ϵr-1).here n represents the mode. Substituting the given values, cutoff frequency is 5.92 GHz.

7. In TE of propagation, HZ must obey the below equation for wave propagation in the region of presence of dielectric:
a) (∂2/∂x2 + kc2) hZ(x,y)=0
b) (∂2/∂x2– h2)hZ(x,y)=0
c) (∂2/∂x2 – Kc2)hZ(x,y)=0
d) (∂2/∂x2 +h2)hZ(x,y)=0

Answer: a [Reason:] In TE mode of propagation,, electric field does not exist in the direction of wave propagation. Hence only magnetic field exists in the direction of wave propagation. This magnetic field must obey the equation (∂2/∂x2 + kc2) hZ(x,y)=0

8. Cutoff frequency fC for TEM mode of propagation is:
a) Fc= (2n-1)c/4d√εr -1
b) Fc= (2n-c)/2d(√εr-1)
c) Fc= (2n-1)/4d(√εr)
d) Fc= (2n-1)/8d√εr – 1

Answer: a [Reason:] Grounded dielectric sheets allow TE mode of propagation on them. The cut off frequency for the propagation of TM mode is given by the expression (2n-1)c/4d√εr -1 .

9. What is the cutoff frequency of TE₁ mode of propagation if the relative permittivity of the slab is 3.2 and the thickness of the slab is 45 mm?
a) 2.24 GHz
b) 4 GHz
c) 1.12 GHz
d) 8 GHz

Answer: c [Reason:] The expression for cutoff frequency for wave propagation in TEN mode is (2n-1)c/4d√(εr -1). substituting the given values in the above expression, the cutoff frequency for TE₁ mode of propagation is 1.12 GHz.

10. The first propagating mode on a grounded dielectric is:
a) TMO mode
b) TM1 mode
c) TM2 mode
d) TM3 mode 