Category: Microwave Engineering Objective Questions

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,

250+ TOP MCQs on Two Port Power Gains and Answers

Microwave Engineering Multiple Choice Questions on “Two Port Power Gains”.

1. The power gain G of a two port network is independent of the source impedance of the two port network.
A. True
B. False
Answer: A
Clarification: Power gain G is the ratio of power dissipated in the load ZL to the power delivered to the input of the two port network. This gain is independent of ZS although the characteristic of some active devices is dependent on ZS.

2. __________ is defined as the ratio of power available from the two port network to the power available from the source.
A. Transducer power gain
B. Available power gain
C. Power gain
D. None of the mentioned
Answer: B
Clarification: Available power gain is defined as the ratio of power available from the two port network to the power available from the source. This assumes conjugate matching of both source and the load and depends on ZS, not ZL.

3. Transducer power gain of a two port network is dependent on :
A. ZS and ZL
B. ZS
C. ZL
D. Independent of both the impedances
Answer: A
Clarification: Transducer power gain of a two port network is the ratio of the power delivered to the load to the power available from the source. This depends on both ZS and ZL.

4. For a two port network the voltage reflection coefficient seen looking towards the load, ГS is:
A. (ZS –Z0)/ (ZS –Z0)
B. (ZS +Z0)/ (Z0 – Z0)
C. ZS / (ZS –Z0)
D. Z0/ (ZS –Z0)
Answer: A
Clarification: For a two port network, the reflection coefficient ГS seen looking towards the load is (ZS –Z0)/ (ZS –Z0). Here ZS is the input impedance of the transmission line and Z0 is the characteristic impedance of the transmission line.

5. In a two port network, the source impedance was measured to be 25 Ω and the characteristic impedance of the transmission line was measured to be 50 Ω. Then the reflection coefficient at the source end is:
A. -0.33333
B. -0.1111
C. 0.678
D. 0.2345
Answer: A
Clarification: For a two port network, the reflection coefficient ГS seen looking towards the load is (ZS –Z0)/ (ZS –Z0). Substituting the given values in the above equation, reflection coefficient at the source end is -0.3333.

6. For a unilateral transistor, the S parameter that is zero is:
A. S11
B. S12
C. S21
D. S22
Answer: B
Clarification: In a unilateral transistor power flow occurs only in one direction and hence S12 is sufficiently small and can be ignored. Also for a unilateral transistor the reflection coefficients reduce to Гin=S11 and Гout=S22.

7. Gain of an amplifier is independent of the operating frequency.
A. True
B. False
Answer: B
Clarification: Gain of an amplifier depends on the operating frequency. Gain of a conjugate matched FET amplifier drops off as 1/f2 or 6dB per octave.

8. Gain of a conjugate matched FET amplifier is given by the relation:
A. Rds (fT)2/ 4Ri (f)2
B. 4Ri (f)2/Rds (fT)2
C. Rds/ Ri
D. None of the mentioned
Answer: A
Clarification: Gain of FET amplifier is given by the relation Rds (fT)2/ 4Ri (f)2. Gain depends on the drain to source resistance, input resistance and also on the frequency of operation of the amplifier.

9. When both input and output of an amplifier are matched to zero reflection (in contrast to conjugate matching), the transducer power gain is:
A. │S21│2
B. │S22│2
C. │S12│2
D. |S11│2
Answer: A
Clarification: When both input and output of an amplifier are matched to zero reflection, ГL=0 and ГS=0. This reduces the complex transducer gain equation to the s parameter of the amplifier S21. S21 signifies the power at port 2 due to input applied at port 1.

10. If the load impedance of a two port network is 40 Ω and the characteristic impedance is 50 Ω, then the reflection coefficient of the two port network at the load end is:
A. -0.111
B. -0.333
C. -0.987
D. None of the mentioned
Answer: A
Clarification: Reflection coefficient at the load end of a two port network is given by the ratio (ZL-Z0)/ (ZL+Z0). ZL is the load impedance and Z0 is the characteristic impedance. Substituting, reflection at load end is -0.1111.


all areas of Microwave Engineering,

250+ TOP MCQs on Noise Characteristics of Receivers and Answers

Microwave Engineering Multiple Choice Questions on “Noise Characteristics of Receivers”.

1. The noise power will determine the maximum detectable signal level for a receiver.
A. True
B. False
Answer: B
Clarification: The noise power will determine the minimum detectable signal level of the receiver for a given transmitter power, maximum range of a communication link. There is a limit on the maximum noise that can be associated with a signal in spite of which the signal can be recovered from the noise.

2. Equivalent noise temperature of a transmission line connecting the antenna to the receiver is:
A. TP (LP-1)
B. TP (LP + 1)
C. TP/ (LP-1)
D. TP / (LP + 1)
Answer: A
Clarification: The transmission line connecting the antenna to the receiver has a loss of LT and is at a physical temperature TP. its noise equivalent temperature is given by TP (LP-1).

3. In a receiver, if the noise figure of the mixer stage in the receiver is 7 dB, then the equivalent noise temperature is given that the receiver is operating at 290 K:
A. 1163 K
B. 1789 K
C. 1000 K
D. 1234 K
Answer: A
Clarification: Equivalent noise temperature for a given noise figure is given by To (FM-1). FM is the noise figure in dB. Substituting the given values for noise figure and temperature, noise equivalent temperature is 1163 K.

4. If a transmission line connecting the antennas to the receiver has a loss of 1.5 dB, given the physical temperature is 270C, noise equivalent temperature is:
A. 123 K
B. 145 K
C. 345 K
D. 234 K
Answer: A
Clarification: The noise equivalent temperature of the transmission line is given by TP(LP-1). Converting the value from dB scale and substituting, noise equivalent temperature is 123 K.

5. Given that the antenna efficiency is 0.9, equivalent brightness temperature is 200 K; physical temperature is 300 K, noise temperature of an antenna is:
A. 220 K
B. 210 K
C. 240 K
D. None of the mentioned
Answer: B
Clarification: Noise temperature of an antenna is given by rad Tb + (1- raD. TP. Tb is the equivalent brightness temperature and TP is the physical temperature. Substituting the given values, noise temperature of the antenna is 210 K.

6. If a receiver is operating at a bandwidth of 1 MHz and has antenna noise temperature of 210 K, then the input noise power is:
A. -90 dBm
B. -115 dBm
C. -56 dBm
D. -120 dBm
Answer: B
Clarification: Input noise power is given the expression kBTA. Here k is the Boltzmann’s constant, B is the operational bandwidth of the antenna and TA is the antenna noise temperature. Substituting in the above expression, input noise power is -115 dBm.

7. Antenna noise temperature of a system is 210 K, noise temperature of transmission line is 123 K, loss of a transmission line connecting the antenna to receiver is 1.41 and noise temperature of the receiver cascade is 304 K. then the total system noise temperature is:
A. 840 K
B. 762 K
C. 678 K
D. 1236 K
Answer: B
Clarification: The total system noise temperature is given by the expression TA+TTL+LTTREC. TA is the antenna noise temperature, TTL is the transmission line noise temperature, TREC is the noise temperature of receiver cascade. Substituting the given values, total system noise temperature is 762 K.

8. If the received power at antenna terminals is -80dBm, and if the input noise power is -115 dBm, then the input SNR is:
A. 45 dB
B. -195 dB
C. -35 dB
D. 35 dB
Answer: D
Clarification: Input SNR of a system is (Si-Ni) in dB. Substituting the given signal power and noise power in dB, input SNR of the system is 35 dB.

9. A receiver system is operating at a bandwidth of 1 MHz and has a total system noise temperature of 762 K. then the output noise power is:
A. -110 dBm
B. -234 dBm
C. -145 dBm
D. -124 dBm
Answer: A
Clarification: Output noise power of a receiver system is kBTsys. B is the operating bandwidth and Tsys is the total system noise temperature. Substituting the given values in the given equation, output noise power is -110 dBm.

10. If the received power at the antenna terminals is Si=-80 dBm and the output noise power is -110 dBm then the output signal to noise ratio is given by:
A. 30 dB
B. -30 dB
C. 35 dB
D. -35 dB
Answer: A
Clarification: Output signal to noise ratio in dB is given by (So-No). Substituting the given values in the above equation, the output SNR is 30 dB.


Microwave Engineering,

250+ TOP MCQs on Lumped Element Circuit Model of Transmission Line and Answers

Basic Microwave Engineering Questions on “Lumped Element Circuit Model of Transmission Line”.

1. The key difference between circuit theory and transmission line theory is:
A. circuit elements
B. voltage
C. current
D. electrical size
Answer: D
Clarification: Circuit theory assumes physical dimensions of the network smaller than electrical wavelength, while transmission lines may be considerable fraction of wavelength.

2. Transmission line is a _________ parameter network.
A. lumped
B. distributed
C. active
D. none of the mentioned
Answer: B
Clarification: Since no lumped elements like resistors, capacitors are used at microwave frequencies, only transmission lines are used. Hence they are called distributed parameter network.

3. For transverse electromagnetic wave propagation, we need a minimum of:
A. 1 conductor
B. 2 conductors
C. 3 conductors
D. bunch of conductors
Answer: B
Clarification: With a single conductor, transverse electromagnetic wave propagation is not possible. Hence we need a minimum of 2 conductors.

4. To model a transmission line of infinitesimal length Δz, the lumped element that is not used is:
A. resistor
B. inductor
C. capacitor
D. transistor
Answer: D
Clarification: In the lumped element circuit model of a transmission line, we use only resistor, capacitor and inductor. Hence no transistor is used.

5. _________ and __________ contribute to the impedance of a transmission line in the lumped element representation.
A. resistor, inductor
B. resistor, capacitor
C. capacitor, inductor
D. transistor, capacitor
Answer: A
Clarification: Z=R+jωL. Hence, both resistor and inductor contribute to the impedance of the transmission line.

6. _________ and __________ contribute to the admittance of a transmission line in the lumped element representation.
A. conductance G, capacitor
B. conductance, inductor
C. resistor, capacitor
D. resistor, inductor
Answer: A
Clarification:Y=G+jωC. Hence, both conductance and capacitance contribute to the admittance of the transmission line.

7. Characteristic impedance of a transmission line is:
A. impedance Z of a transmission line
B. impedance which is a constant at any point on the transmission line
C. reciprocal of admittance of a transmission line
D. none of the mentioned
Answer: B
Clarification: Characteristic impedance is defined as that impedance of a line which is a constant when measured at any point on the line, Hence B.

8. Propagation constant γ is a :
A. real value
B. none of the mentioned
C. imaginary value
D. complex value
Answer: C
Clarification: Since propagation constant is a complex value, containing attenuation constant α, phase constant β respectively as their real and imaginary parts.

9. Attenuation constant α signifies:
A. real part of propagation constant
B. loss that the transmission line causes
C. none of the mentioned
D. all of the mentioned
Answer: D
Clarification: α is the real value of propagation constant, also signifies the loss that the transmission line causes and hence the total amount of energy transmitted. Hence all the mentioned.

10. Propagation constant γ is given by:
A. α+jβ
B. α-jβ
C. α/jβ
D. α.jβ
Answer: A
Clarification: Propagation constant is a complex sum of α and β, α being the real value and β being the complex part.

11. Characteristic impedance Zₒ is given by:
A. √Z/Y
B. √ZY
C. √Z+√Y
D. √Z-√Y
Answer: A
Clarification: Characteristic impedance Zₒ is the square root of ratio of impedance and admittance of the transmission line.

12. Propagation constant γ in terms of admittance and impedance of the transmission line is:
A. √Z/Y
B. √ZY
C. ZY
D. ZY*
Answer: B
Clarification: Propagation constant is the root of product of impedance and admittance of the transmission line.