Category: Microwave Engineering Objective Questions

250+ TOP MCQs on Ferrite Phase Shifters and Answers

Microwave Engineering Multiple Choice Questions on “Ferrite Phase Shifters”.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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.


Microwave Engineering,

250+ TOP MCQs on Micro-wave Tubes and Answers

Microwave Engineering Multiple Choice Questions on “Micro-wave Tubes”.

1. The production of power at higher frequencies is much simpler than production of power at low frequencies.
A. True
B. False
Answer: B
Clarification: As frequency increases to the millimeter and sub millimeter ranges, it becomes increasingly more difficult to produce even moderate power with solid state devices, so microwave tubes become more useful at these higher frequencies.

2. Microwave tubes are power sources themselves at higher frequencies and can be used independently without any other devices.
A. True
B. False
Answer: B
Clarification: Microwave tubes are not actually sources by themselves, but are high power amplifiers. These tubes are in conjunction with low power sources and this combination is referred to as microwave power module.

3. Microwave tubes are grouped into two categories depending on the type of:
A. Electron beam field interaction
B. Amplification method
C. Power gain achieved
D. Construction methods
D. None of the mentioned
Answer: A
Clarification: Microwave tubes are grouped into two categories depending on the type of electron beam field interaction. They are linear or ‘O’ beam and crossed field or the m type tube. Microwave tubes can also be classified as oscillators and amplifiers.

4. The klystron tube used in a klystron amplifier is a _________ type beam amplifier.
A. Linear beam
B. Crossed field
C. Parallel field
D. None of the mentioned
Answer: A
Clarification: 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.

5. In crossed field tubes, the electron beam traverses the length of the tube and is parallel to the electric field.
A. True
B. False
Answer: B
Clarification: 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.

6. ________ is a single cavity klystron tube that operates as on oscillator by using a reflector electrode after the cavity.
A. Backward wave oscillator
B. Reflex klystron
C. Travelling wave tube
D. Magnetrons
Answer: B
Clarification: 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.

7. A major disadvantage of klystron amplifier is:
A. Low power gain
B. Low bandwidth
C. High source power
D. Design complexity
Answer: B
Clarification: Klystron amplifier offers a very narrow operating bandwidth. This is overcome in travelling wave tube (TWT). TWT is a linear beam amplifier that uses an electron gun and a focusing magnet to accelerate beam of electrons through an interaction region.

8. In a _________ oscillator, the RF wave travels along the helix from the collector towards the electron gun.
A. Interaction oscillator
B. Backward wave oscillator
C. Magnetrons
D. None o the mentioned
Answer: B
Clarification: 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.

9. Extended interaction oscillator is a ________ beam oscillator that is similar to klystron.
A. Linear beam
B. Crossed beam
C. Parallel beam
D. M beam
Answer: A
Clarification: Extended interaction oscillator is a linear beam oscillator that uses an interaction region consisting of several cavities coupled together, with positive feedback to support oscillation.

10. Magnetrons are microwave devices that offer very high efficiencies of about 80%.
A. True
B. False
Answer: A
Clarification: 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.

11. Klystron amplifiers have high noise output as compared to crossed field amplifiers.
A. True
B. False
Answer: B
Clarification: 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%.

12. ____________ is a microwave device in which the frequency of operation is determined by the biasing field strength.
A. VTM
B. Gyratron
C. Helix BWO
D. None of the mentioned
Answer: B
Clarification: 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.


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250+ TOP MCQs on Digital Modulation and Bit Error Rate and Answers

Microwave Engineering Question Bank on “Digital Modulation and Bit Error Rate”.

1. All modern wireless communication systems rely on digital modulation methods due to:
A. superior performance
B. low power requirements
C. sustainability
D. all of the mentioned
Answer: D
Clarification: Major advantages of digital modulation over analog modulation is that they have superior performance in the presence of noise and signal fading and the power requirement is very as compared to analog communication.

2. The probability of error depends on the ratio of bit energy to noise power density.
A. true
B. false
Answer: A
Clarification: If the noise power is higher than the energy transmitted per bit, then the probability of error is high. Suppose signal 0 is transmitted, if the noise level is greater than the threshold to detect signal 1, bit 0 is interpreted as 1.

3. Probability of bit error is greater for ASK as compared to FSK.
A. true
B. false
Answer: A
Clarification: Frequency shift keying method of modulation is more efficient than amplitude shift keying where the carrier is varied with respect to the amplitude of the message signal. Hence FSK has a lower bit error rate as compared to ASK.

4. Probability of bit error rate is greater for QPSK as compared to FSK.
A. true
B. false
Answer: B
Clarification: In QPSK (Quadrature phase shift keying), two data bits are used to select one of the four possible phase states of the received data sequence. Hence, this has a low error probability as compared to FSK.

5. Global positioning system uses _____ satellites in medium earth orbits to provide accurate position information.
A. 12
B. 24
C. 36
D. 48
Answer: B
Clarification: Global positioning system uses 24 satellites in medium earth orbits to provide accurate position information to users on land, air and sea. GPS has become one of the most pervasive applications of wireless technology to consumers throughout the world.

6. GPS operates at a single frequency band.
A. true
B. false
Answer: B
Clarification: GPS operates at two frequency bands. L1 at 1575.42 MHz and L2 at 1227.60 MHz, transmitting spread spectrum signals with BPSK modulation.

7. WLAN is used for providing connection between a host computer and satellite for communication.
A. true
B. false
Answer: B
Clarification: WLANs are used to provide connections between a computer and peripherals over a short distance. These types of networks are used in airports, college campuses and many more.

8. There is no standard to be formed by commercial WLAN products.
A. true
B. false
Answer: B
Clarification: Most commercial WLANs follow IEEE 802.11 standards (Wi-Fi). They operate at either 2.4 or 5.7 GHz in the industrial, scientific and medical frequency bands and use either frequency hopping or direct sequence spread spectrum techniques.

9. Bluetooth devices operate at a frequency of :
A. 2.4 GHz
B. 4.5 GHz
C. 5.7 GHz
D. none of the mentioned
Answer: A
Clarification: Bluetooth operates at a frequency range of 2.4 GHz and RF power range of 1-100 mW and corresponding operating ranges of 1-100 m.

10. The modulation technique in which both amplitude and phase of the carrier are varied simultaneously is:
A. ASK
B. BPSK
C. QAM
D. QPSK
Answer: C
Clarification: In Quadrature amplitude modulation, both amplitude and phase of the carrier is varied with respect to the message signal. This is one of the most advanced modulation techniques which is used in advanced applications.


Microwave Engineering Question Bank,

250+ TOP MCQs on Surface Waves on Grounded Di-electric Sheet and Answers

Microwave Engineering Questions and Answers for Freshers on “Surface Waves on Grounded Di-electric Sheet”.

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
Clarification: 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
Clarification: 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 – k02+β2) eZ(x,y)=0
D. (∂2/∂x2 + ∈rk02) eZ(x,y)=0
Answer: A
Clarification: 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= ∈rk02-β2
B. kC2= ∈rk02+β2
C. h2= -k02+β2
D. kC2= k2 + β2
Answer: A
Clarification: Cutoff wave number signifies the minimum threshold wave number required for propagation. Here the expression kC2= ∈rk02-β2 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Answer: B
Clarification: Since for TMO mode of propagation on a dielectric sheet the cutoff frequency is 0, it is not practically possible for propagation. Hence, TM1 mode is the first propagating mode.


for Freshers,

250+ TOP MCQs on Transmission Line Resonators and Answers

Microwave Engineering Multiple Choice Questions on “Transmission Line Resonators”.

1. Lumped elements can be used to make resonators that rare to be operated at microwave frequencies.
A. True
B. False
Answer: B
Clarification: Lumped elements cannot be used at microwave frequencies since their behavior is not deterministic at these frequencies and the required response cannot be achieved.

2. Short circuited λ/2 transmission line has a quality factor of:
A. β/2α
B. 2β/α
C. β/α
D. Z0/ZL
Answer: A
Clarification: Quality factor of a short circuited transmission line is a function of attenuation constant and phase constant of the transmission line. Higher is the attenuation in the transmission line, lower is the quality factor of the transmission line.

3. Quality factor of a coaxial cable transmission line is independent of the medium between the wires of the transmission line.
A. True
B. False
Answer: B
Clarification: Quality factor is dependent on the permeability of the medium between the inner and outer conductor of the co-axial cable. For example, air has twice the quality factor as that of Teflon filled co-axial fiber.

4. A coaxial cable is air filled with air as dielectric with inner and outer radius equal to 1 mm and 4 mm. If the surface resistivity is 1.84*10-2Ω,then the attenuation due to conductor loss is:
A. 0.011
B. 0.022
C. 0.11
D. 0.22
Answer: A
Clarification: Conductor loss in a coaxial cable is given by Rs(a-1+b-1)/2ln (b/A.. Here ‘a’ and ‘b’ are the inner and outer radii of the coaxial cable. is the intrinsic impedance of the medium, for air is 377Ω. Substituting the given values in the equation, conductor loss is 0.022 Np/m.

5. An air coaxial cable has attenuation of 0.022 and phase constant of 104.7, then the quality factor of a λ/2 short circuited resonator made out of this material is:
A. 2380
B. 1218
C. 1416
D. Insufficient data
Answer: A
Clarification: Quality factor of a λ/2 short circuited transmission line is β/2α. β is the phase constant and α is the attenuation constant of the line, substituting the given values, the quality factor of the transmission line is 2380.

6. The equivalent resistance of a short circuited λ/4 transmission line is independent of the characteristic impedance of the transmission line.
A. True
B. False
Answer: B
Clarification: The equivalent resistance of a short circuited λ/4 transmission line is dependent of the characteristic impedance of the transmission line. The expression for equivalent resistance is Z0/αl. Resistance of a short circuited line is directly proportional to the characteristic impedance of the transmission line.

7. A microstrip patch antenna has a width of 5.08mm and surface resistivity of 1.84*10-2. Then the attenuation due to conductor loss is:
A. 0.0724
B. 0.034
C. 0.054
D. None of the mentioned
Answer: A
Clarification: Attenuation due to conductor loss of a microstrip line is given by Rs/Z0W. Substituting the given values, attenuation due to conductor loss is 0.0724 Np/m.

8. If the attenuation due to dielectric loss and attenuation due to conductor loss in a microstrip transmission line is 0.024Np/m and 0.0724 Np/m, then the unloaded quality factor if the propagation constant is 151 is:
A. 150
B. 783
C. 587
D. 234
Answer: B
Clarification: Unloaded Q for a microstrip line is given by β/2α. Α is the sum of attenuation due to conductor loss and dielectric loss. Substituting the given values the equation, unloaded Q is 783.

9. The equivalent capacitance of a short circuited λ/4 transmission line is dependent on the characteristic impedance of the transmission line.
A. True
B. False
Answer: A
Clarification: Equivalent capacitance of a short circuited λ/4 transmission line is dependent on the characteristic impedance of the transmission line. It is inversely proportional to the characteristic impedance of the transmission line. Equivalent capacitance is π/4ω0Z0.

10. Inductance of an open circuited λ/2 transmission line is dependent on the characteristic impedance of the transmission line.
A. True
B. False
Answer: A
Clarification: Inductance of an open circuited λ/2 transmission line is dependent on the characteristic impedance of the transmission line. Expression for inductance is 1/ω02c, C is the equivalent capacitance of the open circuited line. C has the expression π/4ω0Z0.


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250+ TOP MCQs on Ferrite Circulators and Answers

Microwave Engineering Multiple Choice Questions on “Ferrite Circulators”.

1. ________ is a three-port microwave device that can be lossless and matched at all spots.
A. Hybrid junction
B. Magic Tee
C. Circulator
D. Isolator
Answer: C
Clarification: A circulator is a three-port microwave device that can be lossless and matched at all ports; by using the unitary properties of scattering matrix it is proved that such a device must be non-reciprocal.

2. The total number of ones in the scattering matrix of an ideal circulator is:
A. 4
B. 3
C. 2
D. 5
Answer: B
Clarification: Since the circulator is matched at all the ports, the diagonal elements are zero. As the circulator allows power flow in only one direction, only one of the elements in each row has a 1 in the matrix. With three rows, there are three ones.

3. There is no method in which the scattering matrix of the opposite circularity can be obtained from the matrix we have.
A. True
B. False
Answer: B
Clarification: By transposing the port indices of the existing matrix, the opposite circulatory can be obtained. For example, if S13 is a 1 in the given circulator, then S31 is automatically one in the opposite circulator.

4. Practically, opposite circulatory in a ferrite circulator can be obtained by:
A. Changing the order of port operation
B. Impedance matching the input ports
C. Changing the polarity of the magnetic bias field
D. None of the mentioned
Answer: C
Clarification: For a ferrite circulator, opposite circulatory can be produced by changing the polarity of the magnetic bias field. This change in polarity causes power to flow in opposite direction but only in one direction.

5. A circulator device can also used as an isolator with a few modifications.
A. True
B. False
Answer: A
Clarification: A circulator can be used as an isolator by terminating one of the circulator ports with known impedance so that the remaining two ports are used for operation. As power flow occurs only in one direction in these two ports, they can be used as isolators.

6. In the scattering matrix representation of a non-ideal circulator, the diagonal elements of the matrix are:
A. Zero
B. One
C. Reflection coefficient Г
D. None of the mentioned
Answer: C
Clarification: In a non-ideal circulator, the three ports of the circulator are not properly matched and hence there will be some reflection back to the same ports. This impedance mismatch can be represented by the reflection co-efficient Г.

7. In a stripline junction circulator, the ferrite material is present in the form of a:
A. Slab
B. Ferrite disk
C. Ferrite material is not used in a microstrip circulator
D. Ferrite cubes
Answer: B
Clarification: In a stripline junction circulator, two ferrite disks fill the space between the center metallic disk and the ground planes of the stripline. Three striplines are attached to the periphery of the center disk and the ground plane on the stripline.

8. The dielectric resonator in the circulator has a single highest order resonant mode.
A. True
B. False
Answer: B
Clarification: In operation of a microstrip circulator, the ferrite disks form a dielectric resonator; in the absence of the bias field this resonator has a single lowest order resonant mode with a cos φ dependence.

9. In the plot of the magnitude of electric field around the periphery of the junction circulator, the curve has:
A. Three peaks
B. Two peaks
C. Four peaks
D. None of the mentioned
Answer: A
Clarification: In the plot of the magnitude of electric field around the periphery of the junction circulator, the curve has three peaks. These three peaks are due to the three ports of the circulator where the field measured is maximum.

10. In the non-ideal scattering matrix representation of the circulator, the attenuation constant and phase constant α, β respectively are approximated as 1.
A. True
B. False
Answer: B
Clarification: In the non-ideal scattering matrix representation of the circulator, the attenuation constant and phase constant α, β respectively are approximated in terms of the reflection coefficient which represents the impedance mismatch in the network. Α is approximated as 1-Г2 and β is approximated as Г.


Microwave Engineering,