Category: Microwave Engineering Objective Questions

250+ TOP MCQs on Lange Coupler and Answers

Microwave Engineering Multiple Choice Questions on “Lange Coupler”.

1. The major disadvantage of coupled line coupler is:
A. complex construction
B. low power gain
C. higher loss
D. none of the mentioned
Answer: D
Clarification: Coupling in a coupled line coupler is too loose to achieve coupling factors of 3 or 6 dB. Since higher efficiency is not achievable, this a major disadvantage. One method of improving coupling is designing in such a way that the fringing field contributes to coupling.

2. Lange couplers consist of four parallel lines that are ______ coupled.
A. tightly
B. loosely
C. partially
D. none of the mentioned
Answer: A
Clarification: Lange couplers consist of four parallel lines. These parallel lines are tightly coupled to achieve 3 dB coupling ratio. Achieving 3dB coupling implies that 70% of the applied power is coupled.

3. Lange coupler is a type of Quadrature hybrid.
A. true
B. false
Answer: A
Clarification: Lange coupler ha two output ports. The output voltage measured at these ports is 900 out of phase with one another. Hence, they are called a type of Quadrature hybrid.

4. A major disadvantage of Lange coupler is:
A. high power loss
B. complex construction
C. low gain
D. none of the mentioned
Answer: B
Clarification: The main disadvantage of Lange coupler is probably practical, as the lines are very narrow and close together, and the required bonding wires across the lines increases complexity.

5. For even mode excitation of a Lange coupler, all the four lines of the Lange coupler are at:
A. equal potential
B. zero potential
C. unequal potential
D. negative potential
Answer: A
Clarification: during the analysis of Lange coupler, when they are excited in the even mode, all the four conductors of the coupler are at equal potential. Hence the capacitance Cm has no effect and need not be considered for further analysis.

6. Even mode characteristic impedance of a Lange coupler:
A. (Vp.Cep)-1
B. Vp.Cep
C. (Vp.Lep)-1
D. Vp.Lep
Answer: A
Clarification: The even mode characteristic impedance of a Lange coupler is given by (Vp.Cep)-1. Here Vpp is the phase velocity and Cep is the capacitance of any of the four conductors of the Lange coupler measured during even mode of analysis.

7. In a Lange coupler having 4 lines, the adjacent lines are connected so as to act as two conductors in parallel.
A. true
B. false
Answer: B
Clarification: In a Lange coupler consisting of 4 lines, the alternating lines are connected together. This simplifies the computation effort for finding equivalent capacitance and the construction is simpler. Also ports are easily distinguished and helps in flow of power in required direction.

8. For a 4 wire Lange coupler, if the odd mode and even mode capacitances are 12pF and 9pF respectively, then the equivalent capacitance in terms of two wired coupled line in even mode is:
A. 15.1 pF
B. 16.7 pF
C. 12 pF
D. 9 pF
Answer: B
Clarification: Equivalent capacitance in terms of two wired coupled model is Ce (3Ce+C0)/(Ce+C0), Where Ce and C0 are the equivalent capacitances in even mode and odd mode for 4 wire model. Substituting the given values in the above expression, the equivalent capacitance is 16.7 pF.

9. For a 4 wire Lange coupler, if the odd mode and even mode capacitances are 12pF and 9pF respectively, then the equivalent capacitance in terms of two wired coupled line in odd mode is:
A. 18.56 pF
B. 25.71 pF
C. 42 pF
D. 31.6 pF
Answer: B
Clarification: Equivalent capacitance in terms of two wired coupled model is C0 (3C0+Ce)/ (Ce+C0), Where Ce and C0 are the equivalent capacitances in even mode and odd mode for 4 wire model. Substituting the given values in the above expression, the equivalent capacitance is 25.71 pF.

10. The phase velocity of the waves in both even and odd mode of a Lange coupler is equal.
A. true
B. false
Answer: B
Clarification: The phase velocity of the waves in the even mode and odd mode are not practically equal and there are lot of variations. But only for theoretical analysis, they are assumed to be equal.


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250+ TOP MCQs on Heterojunction BJT – 2 and Answers

Microwave Engineering Multiple Choice Questions & Answers on “Heterojunction BJT – 2”.

1. The hybrid-π model of a BJT is useful for analysis at all frequency ranges and variation of other transistor parameters.
A. true
B. false
Answer: A
Clarification: The element values of the hybrid-π model are fairly constant over a wide range of operating points, bias conditions, load conditions and frequency. Otherwise, the element values become frequency, bias or load dependent in which case the hybrid –π model becomes less useful.

2. If the S11 and S22 parameters of a common emitter operated BJT is high:
A. then the output and input ports are matched well
B. there is mismatch in the ports
C. the gain of the amplifier is high
D. none of the mentioned
Answer: B
Clarification: S11 and S22 parameters of a two port network signify the amount of signal to the same port that is excited by the source, a high value of these values imply that these ports are not impedance matched properly.

3. If a common emitter configuration BJT is treated as a two port network, the gain of the amplifier is roughly given by the S parameter:
A. S11
B. S12
C. S21
D. S22
Answer: C
Clarification: When a BJT is represented as a two port network, where the base is port 1 and collector is port 2. Gain of the amplifier is given by the parameter S21. This parameter drops quickly with an increase in the operating frequency.

4. Short circuit current gain of BJT is given by the expression:
A. gm/ωC
B. ωC/ gm
C. gm/C
D. none of the mentioned
Answer: A
Clarification: Short circuit current gain of BJT is defined as the ratio of output collector current to the input base current assuming the base resistance to be zero. The frequency at which the short circuit current gain of the amplifier is unity is called upper frequency limit.

5. The output collector to emitter current of a BJT amplifier is independent of the input base current of the amplifier.
A. true
B. false
Answer: B
Clarification: BJT is a current controlled device. Output collector current is controlled by the input base current. If the input base current is increased, the collector current also increases.

6. The current gain of a BJT ________ with frequency.
A. increases
B. decreases
C. remains constant
D. none of the mentioned
Answer: A
Clarification: The short circuit current gain of a BJT amplifier is given by the expression gm/ωC. From the equation, it is seen that gain is inversely proportional to frequency. As the frequency of operation of BJT increases, current gain of the transistor reduces.

7. If a transistor has a short circuit current gain of 25 and the capacitance measured in the hybrid-π model of the transistor was 60 pF. Then the threshold frequency of operation of the transistor is:
A. 60 MHz
B. 45.6 GHz
C. 66.3 GHz
D. 34.8 GHz
Answer: C
Clarification: The threshold frequency for a BJT is given by gm/2πC. substituting the given values; the threshold operating frequency of the transistor is 66.3 GHz.

8. Hetero junction bipolar transistors have the same working principle and operation as that of a BJT.
A. true
B. false
Answer: B
Clarification: The operation of hetero junction BJT is same as that of BJT, but an HBT has a base emitter junction made from a compound semiconductor material such as GaAs in junction with thin layers of other materials.

9. Advantage of HJT over BJT is that it has:
A. higher gain
B. high frequency of operation
C. sophisticated construction
D. none of the mentioned
Answer: B
Clarification: The gain of BJT reduces with the increase in the operating frequency. This problem is overcome in HJT which gives sustained gain even at frequencies of about 100 GHz. They are also suitable for application in low power circuits.

10. The S21 parameter of a HJT increases with increase in the operating frequency of the transistor.
A. true
B. false
Answer: B
Clarification: S21 parameter of a transistor signifies the gain of the transistor. With the increase in the operating frequency of the transistor, the gain of the transistor reduces, but it is to be noted that the fall in gain of a HJT is not as rapid as fall in gain in a BJT.


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250+ TOP MCQs on Frequency Multipliers and Answers

Microwave Engineering Multiple Choice Questions on “Frequency Multipliers”.

1. Oscillators operating at millimeter wavelength are difficult to realize and are also less efficient.
A. true
B. false
Answer: A
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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 %.


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250+ TOP MCQs on General Solutions-TE,TEM,TM Waves and Answers

Microwave Engineering Problems on “General Solutions-TE,TEM,TM Waves”.

1. Maxwell’s equation for electromagnetic waves in a waveguide is:
A. ∇×E = -jωµ(vector H)
B. ∇×E =-jωμ(vector E)
C. ∇×H=-jωμ(vector H)
D. ∇×H=jωμ(vector H)
Answer: A
Clarification: Maxwell’s equation governs the propagation of waves in a waveguide. This equation describes the relation between electric field and magnetic field inside the waveguide.

2. If the wavelength of a signal is 10 mm, then the wavenumber of the material when a waveguide is filled with that material is:
A. 628
B. 345
C. 123
D. None of the mentioned
Answer: A
Clarification: The wave number is given by the expression 2π/λ. Substituting the given wavelength, wave number is 628.

3. If a waveguide is filled with a lossy material then the expression for ∈ for that material is:
A. ∈=∈∈r(1-jtan δ)
B. ∈= ∈ₒ∈r (1/ j tanδ)
C. ∈=∈ₒ∈r/(1+ j tan δ )
D. ∈=∈ₒ∈r/(1-j tan δ )
Answer: A
Clarification: When a waveguide is filled with a dielectric of or any lossy material, the relative permittivity is given by ∈=∈∈r(1-jtan δ).

4. If a waveguide is filled with a lossless material of relative permeability 2, then the wave impedance in the TEM mode is:
A. 188.5 Ω
B. 170 Ω
C. 123 Ω
D. 345 Ω
Answer: A
Clarification: The wave impedance is given by the expression /√∈. is the intrinsic impedance of the medium substituting the given values in the above expression, wave impedance is 188.5 Ω.

5. If the wave impedance of a medium is 200 Ω, then what is the relative permittivity of that medium?
A. 1.885
B. 2
C. 2.2
D. 2.5
Answer: A
Clarification: The wave impedance is given by the expression /√∈. is the intrinsic impedance of the medium substituting the given values in the above expression, the relative permittivity is 1.885.

6. If p=0.3 and the wave number of air in TM mode is 16, then the intrinsic impedance of air in TM mode given wave number is 125 is:
A. 1 Ω
B. 0.9 Ω
C. 0.8 Ω
D. 2 Ω
Answer: B
Clarification: Intrinsic impedance for TM mode of propagation is given by β/k. substituting the given values in the above equation; the intrinsic impedance is 0.9 Ω.

7. If the intrinsic impedance of a medium is 0.8 Ω, with wave number 125 and β being 0.2, then the relative permeability of the medium is:
A. 1.326
B. 2.34
C. 4.5
D. 6.7
Answer: A
Clarification: Intrinsic impedance for TM mode of propagation is given by β/k. substituting the given values in the above equation; the permeability of the medium is 1.326.

8. The losses that occur in a transmission line is:
A. Conduction losses
B. Di-electric loss
C. Both of the mentioned
D. None of the mentioned
Answer: C
Clarification: Both conductor loss and dielectric loss occur in a transmission medium. Conductor loss is due to the property of the transmission line while dielectric loss is due to the medium inside the transmission line.

9. Which of the following is true regarding attenuation?
A. Conductor loss
B. Di-electric loss
C. Sum of both conductor loss and di electric loss
D. Attenuation is different from the losses
Answer: C
Clarification: Attenuation of a propagating wave is due to both the irregularities in the waveguide and as the property of the dielectric material. Hence attenuation of a propagating wave is due to both conductor loss and dielectric loss.

10. If the wave number of a medium is 20 and loss tangent is 0.4 , then the dielectric loss caused by the medium is:
A. 4
B. 2
C. 3
D. 6
Answer: A
Clarification: Dielectric loss in a medium is given by the expression k tan δ/2. Substituting the given value of loss tangent and wave number dielectric loss is 4 Np/m.

11. If the dielectric loss of a medium is 0.2 Np/m with a wave number of 12, then the value of loss tangent is:
A. 0.0334
B. 0.05
C. 0.08
D. 0.09
Answer: A
Clarification: Dielectric loss in a medium is given by the expression k tan δ/2. Substituting the given values, the loss tangent of the medium is 0.0334.


Problems,

250+ TOP MCQs on Reflections and Answers

Microwave Engineering Multiple Choice Questions on “Reflections”.

1. Discontinuities in the matching quarter wave transformer are not of considerable amount and are negligible.
A. True
B. False
Answer: B
Clarification: Discontinuities in the matching network cause reflections which result in considerable attenuation of the transmitted signal. Hence, discontinuities in transformers are not negligible.

2. The overall reflection coefficient of a matching quarter wave transformer cannot be calculated because of physical constraints.
A. True
B. False
Answer: B
Clarification: Though the computation of total reflection is complex, the total reflection can be computed in two ways. They are the impedance method and the multiple reflection method.

3. In the multiple reflections analysis method, the total reflection is:
A. An infinite sum of partial reflections
B. An infinite sum of partial reflection and transmissions
C. Constant value
D. Finite sum of partial reflections
Answer: B
Clarification: The number of discontinuities in the matching circuit (quarter wave transmission line) is theoretically infinite since the exact number cannot be practically determined. Hence, the total reflection is an infinite sum of partial reflections and transmission.

4. The expression for total reflection in the simplified form is given by:
A. Г=Г1+ Г3e-2jθ
B. Г=Г11+Г3
C. Г=Г12+ Г3e-2jθ
D. Г= Г1+ Г2e-2jθ
Answer: A
Clarification: This expression dictates that the total reflection is dominated by the reflection from the initial discontinuity between Z1 and Z2 (Г1), and the first reflection from the discontinuity between Z2 and ZL (Г3e-2jθ).

5. The e-2jθ term in the expression for total reflection in a single section quarter wave transformer impedance matching network Г=Г1+ Г3e-2jθ signifies:
A. Phase delay
B. Frequency change
C. Narrowing bandwidth
D. None of the mentioned
Answer: A
Clarification: The term e-2jθ in Г=Г1+ Г3e-2jθ accounts for phase delay when the incident wave travels up and down the line. This factor is a result of multiple reflections.

6. If the first and the third reflection coefficients of a matched line is 0.2 and 0.01, then the total reflection coefficient if quarter wave transformer is used for impedance matching is:
A. 0.2
B. 0.01
C. 0.21
D. 0.19
Answer: D
Clarification: The total reflection co-efficient of a matched line due to discontinuities is given by Г=Г1+ Г3e-2jθ. Given that Г1=0.2 and Г3=0.01, β=2π/λ, l=λ/4. θ=βl, Substituting the given values in the above 2 given equations, the total reflection coefficient is 0.19.

7. If a λ/4 transmission line is used for impedance matching, then always Г1> Г3.
A. True
B. False
Answer: A
Clarification: Since the load is matched to the transmission line the reflection from the load towards the source will be very less (Г3). Г1 is the reflection from the junction of the transmission line and the λ/4 matching section. Since this end will have some improper matching and discontinuities, Г1 is always greater than Г3.

8. To compute the total reflection of a multi-section transmission line, the lengths of the transmission lines considered are all unequal.
A. True
B. False
Answer: B
Clarification: The computation of total reflection of a matched line due to discontinuities is theoretically complex. In order to obtain an approximated simple expression, the lengths of the multi section matching transformers is a constant or all of them are equal.

9. If ZL< Z0, then the reflection coefficient at that junction is:
A. ГN<0
B. ГN>0
C. ГN>1
D. None of the mentioned
Answer: A
Clarification: When there is no proper matching between load impedance and the characteristic impedance of a transmission line and given the condition that ZL< Z0, then the reflection coefficient at that junction is always negative. That is, ГN<0.

10. The total approximate reflection coefficient is a finite sum of reflection co-efficient of individual matching section of the matching network.
A. True
B. False
Answer: A
Clarification: In a multi section transformer there are N sections, if the reflection from each section is ГN, then the total reflection is the sum of reflections that occur due to individual sections. There is an exponential component associated with each reflection coefficient that decays exponentially.

11. Using the relation for total reflection co-efficient certain designs of matching networks can be made as per practical requirements.
A. True
B. False
Answer: A
Clarification: We can synthesize any desired reflection coefficient response as a function of frequency by properly choosing the ГN and using enough sections (N).


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250+ TOP MCQs on 180 Degree Hybrids and Answers

Microwave Engineering test on “180 Degree Hybrids”.

1. 1800 hybrid is a network in which there is a phase shift of 1800 between the input signal applied and the output taken.
A. true
B. false
Answer: B
Clarification: 1800 hybrid is a four port network that has one input port and two output ports. The phase difference between the 2 output ports is 1800.

2. Port 1 and port 4 of 1800 hybrid are called sum and difference ports respectively because of their behavior and action mechanism.
A. true
B. false
Answer: A
Clarification: When a 1800 hybrid is used as a combiner, with input signals applied at port 2 and port 3, the sum of the inputs will be formed at port 1, while the difference will be formed at port 4. Hence they are referred to as sum and difference ports.

3. S matrix of 1800 hybrid consists of all diagonal elements zero.
A. true
B. false
Answer: A
Clarification: If all the ports of 1800 hybrid are properly matched, no power is reflected back to the same port. Hence all the diagonal elements of the S matrix, Sii=0.

4. In 1800 hybrid, different power levels can be received at the two output ports of the hybrid.
A. true
B. false
Answer: B
Clarification: 1800 hybrid is a symmetrical coupler. Hence, the input power applied at the input port can be divided equally and obtained at the 2 output ports. Unequal division of power is not possible in 1800 hybrid.

5. 18000 ring hybrid with system impedance of 50 Ω has to be designed. Then the characteristic impedance of the arms of the 1800 hybrid is:
A. 50 Ω
B. 70.70 Ω
C. 100 Ω
D. none of the mentioned
Answer: B
Clarification: Given that the system impedance is 50 Ω, the characteristic impedance of the arms is 50√2. Hence the characteristic impedance is 70.70 Ω.

6. In a waveguide magic-T there is no coupling of power between port 1 and port 4.
A. true
B. false
Answer: A
Clarification: Consider TE10 mode incident at port 1. There is odd symmetry about guide 4. Because the field lines of a TE10 mode in guide 4 would have even symmetry, hence there is no coupling between port 1 and port 4.

7. When a TE10 wave is incident on port 4 of a magic-T, all the power is coupled to port 1.
A. true
B. false
Answer: B
Clarification: When port 4 of a Magic-T is excited, port 1 and port 4 are decoupled, due to symmetry. Port 2 and port 3 are excited equally by the incident wave with a phase difference of 1800.

8. The tapered coupled line 1800 hybrid can provide an arbitrary power division at the 2 output ports of the coupler.
A. true
B. false
Answer: A
Clarification: When the arms of the coupled line is tapered, they result in division of power at the two output ports unequally and they offer a bandwidth of one decade or more.

9. The plot of frequency V/s S11 parameter for a tapered line coupler has a dip at the frequency at which it is designed.
A. true
B. false
Answer: A
Clarification: When port 1 of a tapered coupler is excited, no power flows back to port 1 since the ports are matched. Hence, S11 value is almost zero or negligibly small and hence has a dip at the designed frequency.

10. For a tapered line coupler, the curves of S12 and S13 are identical and have the same magnitude at all frequencies.
A. true
B. false
Answer: B
Clarification: Tapered line couplers result in unequal power division at the output ports. These are useful for various applications. Hence the S12 and S13 curves are not identical since the power outputs are not equal.


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