## [MCQ’s]Microwave Engineering

Exit Intent

#### Module 01

1. Micro strip can be fabricated using:
a) Photo lithographic process
b) Electrochemical process
c) Mechanical methods
d) None of the mentioned
Explanation: Microstrip lines are planar transmission lines primarily because it can be fabricated by photolithographic processes and is easily miniaturized and integrated with both passive and active microwave devices.

2. The mode of propagation in a microstrip line is:
a) Quasi TEM mode
b) TEM mode
c) TM mode
d) TE mode
Explanation: The exact fields of a microstrip line constitute a hybrid TM-TE wave. In most practical applications, the dielectric substrate is very thin and so the fields are generally quasi-TEM in nature.

3. Microstrip line can support a pure TEM wave.
a) True
b) False
c) Microstrip supports only TM mode
d) Microstrip supports only TE mode
Explanation: The modeling of electric and magnetic fields of a microstrip line constitute a hybrid TM-TE model. Because of the presence of the very thin dielectric substrate, fields are quasi-TEM in nature. They do not support a pure TEM wave.

4. The effective di electric constant of a microstrip line is:
a) Equal to one
b) Equal to the permittivity of the material
c) Cannot be predicted
d) Lies between 1 and the relative permittivity of the micro strip line
Explanation: The effective dielectric constant of a microstrip line is given by (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). Along with the relative permittivity, the effective permittivity also depends on the effective width and thickness of the microstrip line.

5. Effective dielectric constant of a microstrip is given by:
a) (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w)
b) (∈r+1)/2 + (∈r-1)/2
c) (∈r+1)/2 (1/√1+12d/w)
d) (∈r + 1)/2-(∈r-1)/2
Explanation: The effective dielectric constant of a microstrip line is (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). This relation clearly shows that the effective permittivity is a function of various parameters of a microstrip line, the relative permittivity, effective width and the thickness of the substrate.

6. The effective dielectric constant of a micro strip line is 2.4, then the phase velocity in the micro strip line is given by:
a) 1.5*108 m/s
b) 1.936*108 m/s
c) 3*108 m/s
d) None of the mentioned
Explanation: The phase velocity in a microstrip line is given by C/√∈r. substituting the value of relative permittivity and the speed of light in vacuum, the phase velocity is 1.936*108 m/s.

7. The effective di electric constant of a micro strip line with relative permittivity being equal to 2.6, with a width of 5mm and thickness equal to 8mm is given by:
a) 2.6
b) 1.97
c) 1
d) 2.43
Explanation: The effective dielectric constant of a microstrip line is given by (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). Substituting the given values of relative permittivity, effective width, and thickness, the effective dielectric constant is 1.97.

8. If the wave number of an EM wave is 301/m in air , then the propagation constant β on a micro strip line with effective di electric constant 2.8 is:
a) 602
b) 503.669
c) 150
d) 200
Explanation: The propagation constant β of a microstrip line is given by k0√∈e. ∈e is the effective dielectric constant. Substituting the relevant values, the effective dielectric constant is 503.669.

9. For most of the micro strip substrates:
a) Conductor loss is more significant than di electric loss
b) Di electric loss is more significant than conductor loss
c) Conductor loss is not significant
d) Di-electric loss is less significant
Explanation: Surface resistivity of the conductor (microstrip line) contributes to the conductor loss of a microstrip line. Hence, conductor loss is more significant in a microstrip line than dielectric loss.

10. The wave number in air for EM wave propagating on a micro strip line operating at 10GHz is given by:
a) 200
b) 211
c) 312
d) 209
Explanation: The wave number in air is given by the relation 2πf/C. Substituting the given value of frequency and ‘C’, the wave number obtained is 209.

11. The effective dielectric constant ∈r for a microstrip line:
a) Varies with frequency
b) Independent of frequency
c) It is a constant for a certain material
d) Depends on the material used to make microstrip
Explanation: The effective dielectric constant of a microstrip line is given by (∈r + 1)/2 + (∈r-1)/2 * 1/ (√1+12d/w). The equation clearly indicates that the effective dielectric constant is independent of the frequency of operation, but depends only on the design parameters of a microstrip line.

12. With an increase in the operating frequency of a micro strip line, the effective di electric constant of a micro strip line:
a) Increases
b) Decreases
c) Independent of frequency
d) Depends on the material of the substrate used as the microstrip line
Explanation: As the relation between effective permittivity and the other parameters of a microstrip line indicate, effective dielectric constant is not a frequency dependent parameter and hence remains constant irrespective of the operation of frequency.

13. Which mode of propagation is supported by a strip line?
a) TEM mode
b) TM mode
c) TE mode
d) None of the mentioned
Explanation: Since a stripline has 2 conductors and a homogeneous dielectric, it supports a TEM wave, and this is the usual mode of operation.

14. The higher order wave guide modes of propagation can be avoided in a strip line by:
a) Restricting both the ground plate spacing and the sidewall width to less than λd/2
b) Restricting both the ground facing plate spacing and the sidewall width to less than λd
c) Filling the region between 2 plates with di electric
d) Restricting both the ground plate spacing and the sidewall width between λg and λg/2
Explanation: When stripline is used as a media for propagation, it is always preferred that only certain modes of propagation are allowed. Hence, in order to avoid the higher order modes, it is achieved by restricting both the ground plate spacing and the sidewall width to less than λd/2.

15. Stripline can be compared to a:
a) Flattened rectangular waveguide
b) Flattened circular waveguide
c) Flattened co axial cable
d) None of the mentioned
Explanation: A stripline has an enter conductor enclosed by an outer conductor and are uniformly filled with a dielectric medium, these are similar to a coaxial cable. Hence it can be compared to a flattened coaxial cable.

16. If the dielectric material filled between the round plates of a microstrip line has a relative permittivity of 2.4, then the phase velocity is:
a) 1.3*108 m/s
b) 1.9*108 m/s
c) 3*108 m/s
d) 2*108 m/s
Explanation: Phase velocity is given by the expression C/√∈ for a stripline. Substituting the given values, the phase velocity for the above case is 1.9*108 m/s.

17. Expression for propagation constant β of a strip line is:
a) ω(√µ∈∈r)
b) ω(√µₒ/√∈r)
c) ω/(√µₒ∈ₒ∈r)
d)c/(√µₒ∈ₒ∈r)
Explanation: Propagation constant is associated with the propagating wave in the strip line. This propagation constant for a wave is defined by the expression ω(√µ∈∈r).

18. If the phase velocity in a stripline is 2.4*108m/s, and the capacitance per unit length of a micro stripline is 10pF/m, then the characteristic impedance of the line:
a) 50 Ω
b) 41.6 Ω
c) 100 Ω
d) None of the mentioned
Explanation: Characteristic impedance of a stripline is given by 1/ (vPc). Substituting the given values of phase velocity and capacitance, the characteristic impedance of the line is 41.6 Ω.

19. The expression for characteristic impedance Zₒ of a stripline is:
a) (30πb/√∈r)(1/We+0.441b)
b) (30πb) (1/We+0.441b)
c) 30π/√∈r
d) (1/We+0.441b)
Explanation: Characteristic impedance of a stripline is a function of the various parameters of the stripline. They are effective width, thickness and relative permittivity of the dielectric material. Changing any one of these parameters results in changing the characteristic impedance of the line The derived expression is hence (30πb/√∈r)(1/We+0.441b).

20. If the effective width of the center conductor is 3 mm and the distance between the two ground plates is 0.32 cm with the material of the dielectric used having a relative permittivity of 2.5, then what is the characteristic impedance of the strip line?
a) 50Ω
b) 71.071Ω
c) 43.24Ω
d) 121Ω
Explanation: The characteristic impedance of a stripline is given by the expression (30πb/√∈r)(1/We+0.441b). Substituting the given values in the given expression and hence solving, the characteristic impedance of the line is 43.24 Ω.

21. The wave number of a stripline operating at a frequency of 10 GHz is:
a) 401
b) 155
c) 206
d) 310
Explanation: The wave number of a microstrip line is given by the expression 2πf√∈r/c, c is the speed of light in space, ∈r is the relative permittivity of the dielectric medium. Substituting the given values in the equation, the wave number is 310.

22. If the loss tangent is 0.001 for a stripline operating at 12 GHz with the relative permittivity of the dielectric material being used equal to 2.6, then the conductor loss is:
a) 0.102
b) 0.202
c) 0.001
d) 0.002
Explanation: Conductor loss in a stripline is given by the expression k*tanδ/2. K is given by the expression 2πf√∈r/C which is the wave number. Substituting the values in the above two equations, conductor loss is 0.202.

23. If the dielectric material used between the grounded plates of a stripline is 2.2, when the strip line operating at 8 GHz, the wavelength on stripline is:
a) 1.2 cm
b) 2.52 cm
c) 0.15 cm
d) 3.2 cm
Explanation: The propagating wavelength on the stripline is defined by the relation C/f√∈r. substituting in the above relation, the propagating wavelength on the microstrip line is 2.52 cm.

24. Fields of TEM mode on strip line must satisfy:
a) Laplace’s equation
b) Ampere’s circuital law
c) Gaussian law
d) None of the mentioned
Explanation: If φ(x,y) is the function of potential in the stripline varying along the width and thickness, this potential function must satisfy the Laplace’s equation.

25. __________ is a region of Electromagnetic spectrum having frequency ranging from 1GHz to 100 GHz
A. Micro wave
B. UV
C. IR
D. None of these

26. Which of the following is the main advantage of microwave
A. Highly directive
B. Moves at the speed of light
C. Greater S/N ratio
D. High penetration power

27. Slotted line is a transmission line configuration that allows the sampling of:
a) electric field amplitude of a standing wave on a terminated line
b) magnetic field amplitude of a standing wave on a terminated line
c) voltage used for excitation
d) current that is generated by the source
Explanation: Slotted line allows the sampling of the electric field amplitude of a standing wave on a terminated line. With this device, SWR and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.

28. A slotted line can be used to measure _____ and the distance of _____________ from the load.
a) SWR, first voltage minimum
b) SWR, first voltage maximum
c) characteristic impedance, first voltage minimum
d) characteristic impedance, first voltage maximum
Explanation: With a slotted line, SWR and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.

29. A modern device that replaces a slotted line is:
a) Digital CRO
b) generators
c) network analyzers
d) computers
Explanation: Although slotted lines used to be the principal way of measuring unknown impedance at microwave frequencies, they have largely been superseded by the modern network analyzer in terms of accuracy, versatility and convenience.

30. If the standing wave ratio for a transmission line is 1.4, then the reflection coefficient for the line is:
a) 0.16667
b) 1.6667
c) 0.01667
d) 0.96
Explanation: ┌= (SWR-1)/ (SWR+1). Substituting for SWR in the above equation for reflection co-efficient, given SWR is 1.4, reflection co-efficient is 0.16667.

31. If the reflection coefficient of a transmission line is 0.4, then the standing wave ratio is:
a) 1.3333
b) 2.3333
c) 0.4
d) 0.6
Explanation: SWR= (1+┌)/ (1-┌). Where ┌ is the reflection co-efficient. Substituting for the reflection co-efficient in the equation, SWR is 2.3333.

32. Expression for ϴ means phase angle of the reflection co efficient r=|r|-e^jθ, the phase of the reflection co-efficient is:
a) θ=2π+2βLmin
b) θ=π+2βLmin
c) θ=π/2+2βLmin
d) θ=π+βLmin
Explanation: here, θ is the phase of the reflection co-efficient. Lmin is the distance from the load to the first minimum. Since voltage minima repeat every λ/2, any multiple of λ/2 can be added to Lmin .

33. In the expression for phase of the reflection coefficient, Lmin stands for :
a) distance between load and first voltage minimum
b) distance between load and first voltage maximum
c) distance between consecutive minimas
d) distance between a minima and immediate maxima
Explanation: Lmin is defined as the distance between the terminating load of a transmission line and the first voltage minimum that occurs in the transmission line due to reflection of waves from the load end due to mismatched termination.

34. If SWR=1.5 with a wavelength of 4 cm and the distance between load and first minima is 1.48cm, then the reflection coefficient is:
a) 0.0126+j0.1996
b) 0.0128
c) 0.26+j0.16
d) none of the mentioned
Explanation: ┌= (SWR-1)/ (SWR+1). Substituting for SWR in the above equation for reflection co-efficient, magnitude of the reflection co-efficient is 0.2. To find θ, θ=π+2βLmin, substituting Lmin as 1.48cm, θ=86.4⁰. Hence converting the polar form of the reflection co-efficient into rectangular co-ordinates, reflection co-efficient is 0.0126+j0.1996.

35. If the characteristic impedance of a transmission line 50 Ω and reflection coefficient is 0.0126+j0.1996, then load impedance is:
a) 47.3+j19.7Ω
b) 4.7+j1.97Ω
c) 0.26+j0.16
d) data insufficient
Explanation: ZL=Z0 (1+┌)/ (1-┌). Substituting the given values of reflection co-efficient and characteristic impedance, ZL is 47.3+j19.7Ω .

36. If the normalized load impedance of a transmission line is 2, then the reflection co-efficient is:
a) 0.33334
b) 1.33334
c) 0
d) 1
Explanation: ZL=Z0 (1+┌)/ (1-┌), this is the expression for load impedance. Normalized load impedance is the ratio of load impedance to the characteristic impedance, taking ZLL/Z0 as 2, the reflection co-efficient is equal to 0.33334.

37. The major advantage of single stub tuning over other impedance matching techniques is:
a) Lumped elements are avoided
b) It can be fabricated as a part of transmission line media
c) It involves two adjustable parameters
d) All of the mentioned
Explanation: Single stub matching does not involve any lumped elements, it can be fabricated as a part of transmission media and it also involves to adjustable parameters namely length and distance from load giving more flexibility.

38. Shunt stubs are preferred for:
a) Strip and microstrip lines
b) Coplanar waveguides
c) Circular waveguide
d) Circulators
Explanation: Since microstrip and strip lines are simple structures, impedance matching using shunt stubs do not increase the complexity and structure of the transmission line. Hence, shunt stubs are preferred for strip and microstrip lines.

39. The two adjustable parameters in single stub matching are distance‘d’ from the load to the stub position, and _________
a) Susceptance or reactance provided by the stub
b) Length of the stub
c) Distance of the stub from the generator
d) None of the mentioned
Explanation: Reactance or susceptance of the matching stub must be known before it used for matching, since it is the most important parameter for impedance matching between the load and the source.

40. In shunt stub matching, the key parameter used for matching is:
a) Admittance of the line at a point
c) Impedance of the stub
Explanation: In shunt stub tuning, the idea is to select d so that the admittance Y, seen looking into the line at distance d from the load is of the form Yₒ+jb) Then the stub susceptance is chosen as –jB, resulting in a matched condition.

41. For series stub matching, the parameter used for matching is:
a) Impedance of the transmission line at a point
b) Voltage at a point on the transmission line
c) Admittance at a point on the transmission line
Explanation: In series sub matching, the distance‘d’ is selected so that the impedance, Z seen looking into the line at a distance‘d’ from the load is of the form Zₒ+jX. Then the stub reactance is chosen as –jX resulting in a matched condition.

42. For co-axial lines and waveguides, ________ is more preferred.
a) Open circuited stub
b) Short circuited stub
c) Slotted section
d) Co-axial lines cannot be impedance matched
Explanation: For co-axial cables and waveguides, short-circuited stub is usually preferred because the cross-sectional area of such an open-circuited line may be large enough to radiate, in which case the stub is no longer purely reactive.

43. For a load impedance of ZL=60-j80. Design of 2 single-stub shunt tuning networks to match this load to a 50Ω line is to be done. What is the normalized admittance obtained so as to plot it on smith chart?
a) 1+j
b) 0.3+j0.4
c) 0.4+j0.3
d) 0.3-j0.4
Explanation: To impedance match a load to a characteristic impedance of the transmission line, first the load has to be normalized. That is, zL=ZL/Z0. For impedance matching using shunt stubs, admittance is used. Taking the reciprocal of impedance, normalized load admittance is 0.3+j0.4.

44. If the normalized admittance at a point on a transmission line to be matched is 1+j1.47. Then the normalized susceptance of the stub used for shunt stub matching is:
a) 1Ω
b) 1.47 Ω
c) -1.47 Ω
d) -1 Ω
Explanation: When shunt stubs are used for impedance matching between a load and transmission line, the susceptance of the shunt stub must be negative of the line’s susceptance at that point for impedance matching.

45. After impedance matching, if a graph is plot with frequency v/s reflection co-efficient of the transmission line is done, then at the frequency point for which the design is done, which of the following is true?
a) There is a peak at this point of the curve
b) There is a dip at this point of the curve
c) The curve is a straight line
d) Such a plot cannot be obtained
Explanation: Since the plot is frequency v/s reflection co-efficient, after impedance matching the reflection co-efficient will be zero or minimum. Hence, there is a dip at that point of the curve.

46. In series stub matching, if the normalized impedance at a point on the transmission line to be matched is 1+j1.33. Then the reactance of the series stub used for matching is:
a) 1 Ω
b) -1.33 Ω
c) -1 Ω
d) 1.33 Ω
Explanation: The reactance of the series stub is negative of the reactance of the line at the point at which it has to be matched. That is, if the line reactance is inductive, the series stub’s reactance is capacitive.

47. Discontinuities in the matching quarter wave transformer are not of considerable amount and are negligible.
a) True
b) False
Explanation: Discontinuities in the matching network cause reflections which result in considerable attenuation of the transmitted signal. Hence, discontinuities in transformers are not negligible.

48. The overall reflection coefficient of a matching quarter wave transformer cannot be calculated because of physical constraints.
a) True
b) False
Explanation: 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.

49. 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
Explanation: 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.

50. The expression for total reflection in the simplified form is given by:
a) Г=Г1+ Г3e-2jθ
b) Г=Г113
c) Г=Г12+ Г3e-2jθ
d) Г= Г1+ Г2e-2jθ
Explanation: 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).

51. 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
Explanation: 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.

52. 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
Explanation: 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.

53. If a λ/4 transmission line is used for impedance matching, then always Г1> Г3.
a) True
b) False
Explanation: 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.

54. 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
Explanation: 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.

55. 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
Explanation: 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.

56. 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
Explanation: 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.

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

58. S parameters are expressed as a ratio of:
a) Voltage and current
b) Impedance at different ports
c) Indecent and the reflected voltage waves
d) None of the mentioned
Explanation: S matrix can be used to represent any n port network. S parameters are defined for microwave networks. Hence instead of voltage and current measurement, the amplitude of the incident and reflected voltage waves is measured.

59. The relation between incident voltage matrix , reflected voltage matrix and S matrix for a microwave network:
a) [v-] = [s] [v+].
b) [v+] = [s] [v-].
c) [v-] [v] = [s].
d) [s] = [v] [v-].
Explanation: S parameter for a microwave network is defined as the ratio of reflected voltage wave to the incident voltage wave. When represented in the form of a matrix, reflected voltage matrix is the product of S parameter and the incident voltage wave at that port.

60. The specific element Sij of the scattering matrix can be determined as:
a) SIJ= Vi-/Vj+
b) SIJ= Vi+/Vj-
c) S= Vj+/Vi-
d) None of the mentioned
Explanation: The parameter Sij is found by driving port j with an incident wave of voltage Vj+ coming out of ports i. The incident waves on all ports except the jth port are set to zero.

61. The device used to get the measurement of S parameters of n- port micro wave network is:
a) CRO
b) Network analyzer
c) Circulator
d) Attenuator
Explanation: Network analyzer is a device to which any microwave network can be externally connected with the help of probes and the s parameters of the network can be obtained.

62. For a one port network , the scattering parameter S₁₁ in terms of impedance parameter Z₁₁ is:
a) (Z11-1)/ (Z11+1)
b) (Z11+1)/ (Z11-1)
c) (Z11+1) (Z11-1)
d) Z11
Explanation: If Z matrix of a one port network is computed, then the s matrix of the same can be computed using the Z11 coefficient. To compute the S11 parameter of the network, the relation used is (Z11-1)/ (Z11+1).

63. Scattering matrix for a reciprocal network is:
a) Symmetric
b) Unitary
c) Skew symmetric
d) Identity matrix
Explanation: For a reciprocal network, the input to port I and output at port j is the same as the input at port j and output measured at port i. Hence, the ports are interchangeable. As the ports are interchangeable, this is reflected in the matrix and the matrix becomes symmetric.

64. S₁₂=0.85-45⁰ and S₁₂=0.85 +45⁰ for a two port network. Then the two port network is:
a) Non-reciprocal
b) Lossless
c) Reciprocal
d) Lossy
Explanation: For a reciprocal network, the S matrix is symmetric. For the matrix to be symmetric, Sij=Sji. Since this condition is not satisfied in the above case, the matrix is non reciprocal.

65. Scattering matrix for a lossless matrix is:
a) Unitary
b) Symmetric
c) Identity matrix
d) Null matrix
Explanation: For a lossless network, the scattering matrix has to be unitary. That is, the law of conservation of energy is to be verified for this case. Using appropriate formula, this condition can be verified.

66. If the reflection co efficient of a 2 port network is 0.5 then the return network loss in the network is:
a) 6.5 dB
b) 0.15 dB
c) 6.020 dB
d) 10 dB
Explanation: Given the reflection coefficient of the network, return loss of the network is calculated using the formula -20 log │Г│. Substituting for reflection coefficient, the return loss of the network is 6.02 dB.

67. If the reflection co efficient of a 2 port network is 0.25 then the return network loss in the network is:
a) 12.05 dB
b) 0.15 dB
c) 20 dB
d) 10 dB
Explanation: Given the reflection coefficient of the network, return loss of the network is calculated using the formula -20 log │Г│. Substituting for reflection coefficient, the return loss of the network is 12.05 dB.

#### Module 02

1. Microwave resonators can be constructed from open sections of waveguide.
a) True
b) False
Explanation: For resonance to occur in waveguides, a closed structure is required. They resonate between the walls of the rectangular waveguide. Also radiation loss from an open ended waveguide can be significant.

2. There is no energy stored inside a rectangular waveguide cavity resonator.
a) True
b) False
Explanation: Energy is stored in a waveguide resonator in the form of electric field and magnetic field. Power is dissipated in the metallic walls of the cavity as well as in the dielectric material that may fill the cavity.

3. A rectangular cavity supports:
a) TEM mode of resonance
b) TM mode of resonance
c) TE mode of resonance
d) TE, TM modes of resonance
Explanation: A rectangular wave guide supports both TE and TEM mode of propagation. Likewise, when a rectangular waveguide is used as resonator, it supports both TE and TM modes of resonance.

4. A waveguide is open circuited at both the ends to use it as a waveguide resonator.
a) True
b) False
Explanation: A closed cavity structure is required in order to bring resonance in the rectangular cavity. Also open ended waveguides result in radiation losses. Hence the waveguide is short circuited to form a resonator.

5. In order to obtain the resonant frequency of a rectangular waveguide, the closed cavity has to satisfy:
a) Gaussian equation
b) Helmholtz equation
c) Ampere’s law
d) None of the mentioned
Explanation: Helmholtz wave equation is considered and solved using variable separable form. Then the boundary conditions are applied to the wave equation considering the walls of the cavity. Solving this gives the expression for resonant frequency.

6. Given the dimension of the waveguide as b<a<d, no resonant mode exists for this specification of dimensions.
a) True
b) False
Explanation: For the given dimensional specification b<a<d, the dominant resonant mode (lowest resonant frequency) will be the TE101 mode, corresponding to the TE10 dominant waveguide mode in a shorted guide of length λg/2.

7. Unloaded Q of a rectangular waveguide cavity resonator:
a) Does not exist
b) Defined as the ratio of length of the waveguide to breadth of the waveguide
c) Defined as the ratio of stored energy to the power dissipated in the walls
d) None of the mentioned
Explanation: Quality factor signifies the power loss in the circuit. It is defined as the ratio of stored energy to the power dissipated in the walls. Higher the power dissipation in the walls, lower is the quality factor of the waveguide resonator.

8. Find the wave number of a rectangular cavity resonator filled with a dielectric of 2.25 and designed to operate at a frequency of 5 GHz.
a) 157.08
b) 145.2
c) 345.1
d) 415.08
Explanation: The wave number of rectangular wave resonator is 2πf√∈r/C, substituting the given values in the above equation, the wave number of the rectangular cavity resonator is 157.08.

9. The required length of the cavity resonator for l=1 mode (m=1, n=0) given that the wave number of the cavity resonator is 157.01 and the broader dimension of the waveguide is 4.755 cm:
a) 1.10 cm
b) 2.20 cm
c) 2.8 cm
d) 1.8 cm
Explanation: The required length of the cavity resonator for the given mode is given by the expression d=lπ/√(k>sup>2-(π/a)2. Substituting the given values in the equation, the required length of the waveguide is 2.20 cm.

10. If the loss tangent of a rectangular waveguide is 0.0004, then Q due to dielectric loss is:
a) 1250
b) 2450
c) 2500
d) 1800
Explanation: Q of a rectangular waveguide due to dielectric loss is given by 1/tanδ. Substituting for tanδ in the above equation, Q due to dielectric loss is 2500.

11. A cylindrical cavity resonator can be constructed using a circular waveguide.
a) shorted at both the ends
b) open at both the ends
c) matched at both the ends
d) none of the mentioned
Explanation: A cylindrical cavity resonator is formed by shorting both the ends of the cylindrical cavity because open ends may result in radiation losses in the cavity.

12. The dominant mode in the cylindrical cavity resonator is TE101 mode.
a) true
b) false
Explanation: The dominant mode of propagation in a circular waveguide is TE111 mode. Hence, the dominant mode of resonance in a cylindrical cavity made of a circular waveguide is TE111 mode. In a cylindrical resonator, the mode of propagation depends on the length of the cavity.

13. Circular cavities are used for microwave frequency meters.
1a) true
b) false
Explanation: Circular cavities are used for microwave frequency meters. The cavity is constructed with a movable top wall to allow the mechanical tuning of the resonant frequency.

14. The mode of the circular cavity resonator used in frequency meters is:
a) TE011 mode
b) TE101 mode
c) TE111 mode
d) TM111 mode
Explanation: Frequency resolution of a frequency meter is determined from its quality factor. Q factor of TE011 mode is much greater than the quality factor of the dominant mode of propagation.

15. The propagation constant of TEmn mode of propagation for a cylindrical cavity resonator is:
a) √ (k2-(pnm/a)2)
b) √ pnm/a
c) √ (k2+(pnm/a)2)
d) none of the mentioned
Explanation: The propagation constant for a circular cavity depends on the radius of the cavity, and the wave number. If the mode of propagation is known and the dimension of the cavity is known then the propagation constant can be found out.

16. A circular cavity resonator is filled with a dielectric of 2.08 and is operating at 5GHz of frequency. Then the wave number is:
a) 181
b) 151
c) 161
d) 216
Explanation: Wave number for a circular cavity resonator is given by the expression 2πf011√∈r/C. substituting the given values in the above expression; the wave number of the cavity resonator is 151.

17. Given that the wave number of a circular cavity resonator is 151 (TE011 mode), and the length of the cavity is twice the radius of the cavity, the radius of the circular cavity operating at 5GHz frequency is:
a) 2.1 cm
b) 1.7 cm
c) 2.84 cm
d) insufficient data
Explanation: For a circular cavity resonator, wave number is given by √( (p01/a)2 +(π/d)2). P01 for the given mode of resonance is 3.832. Substituting the given values the radius of the cavity is 2.74 cm.

18. The loss tangent for a circular cavity resonator is 0.0004.Then the unloaded Q due to dielectric loss is:
a) 1350
b) 1560
c) 560
d) 2500
Answer: Unloaded Q due to the dielectric loss in a circular cavity resonator is the reciprocal of the loss tangent. Hence, taking the reciprocal of the loss tangent, unloaded Q due to dielectric loss is 2500.

19. A circular cavity resonator has a wave number of 151, radius of 2.74 cm, and surface resistance of 0.0184Ω. If the cavity is filled with a dielectric of 2.01, then unloaded Q due to conductor loss is:
a) 25490
b) 21460
c) 29390
d) none of the mentioned
Explanation: Unloaded Q of a circular resonator due to conductor loss is given by ka/2Rs. is the intrinsic impedance of the medium. Substituting the given values in the equation for loaded Q, value is 29390.

20. If unloaded Q due to conductor loss and unloaded Q due to dielectric loss is 29390 and 2500 respectively, then the total unloaded Q of the circular cavity is:
a) 2500
b) 29390
c) 2300
d) 31890
Explanation: The total unloaded Q of a circular cavity resonator is given by the expression (Qc-1+ Qd-1)-1. Substituting the given values in the above expression, the total unloaded Q for the resonator is 2300.

21. Lumped elements can be used to make resonators that rare to be operated at microwave frequencies.
a) True
b) False
Explanation: Lumped elements cannot be used at microwave frequencies since their behavior is not deterministic at these frequencies and the required response cannot be achieved.

22. Short circuited λ/2 transmission line has a quality factor of:
a) β/2α
b) 2β/α
c) β/α
d) Z0/ZL
Explanation: 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.

23. Quality factor of a coaxial cable transmission line is independent of the medium between the wires of the transmission line.
a) True
b) False
Explanation: 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.

24. 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
Explanation: 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.

25. 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
Explanation: 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.

26. 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
Explanation: 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.

27. 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
Explanation: 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.

28. 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
Explanation: 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.

29. 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
Explanation: 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.

30. Inductance of an open circuited λ/2 transmission line is dependent on the characteristic impedance of the transmission line.
a) True
b) False
Explanation: 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.

31. The major advantage of single stub tuning over other impedance matching techniques is:
a) Lumped elements are avoided
b) It can be fabricated as a part of transmission line media
c) It involves two adjustable parameters
d) All of the mentioned
Explanation: Single stub matching does not involve any lumped elements, it can be fabricated as a part of transmission media and it also involves to adjustable parameters namely length and distance from load giving more flexibility.

32. Shunt stubs are preferred for:
a) Strip and microstrip lines
b) Coplanar waveguides
c) Circular waveguide
d) Circulators
Explanation: Since microstrip and strip lines are simple structures, impedance matching using shunt stubs do not increase the complexity and structure of the transmission line. Hence, shunt stubs are preferred for strip and microstrip lines.

33. The two adjustable parameters in single stub matching are distance‘d’ from the load to the stub position, and _________
a) Susceptance or reactance provided by the stub
b) Length of the stub
c) Distance of the stub from the generator
d) None of the mentioned
Explanation: Reactance or susceptance of the matching stub must be known before it used for matching, since it is the most important parameter for impedance matching between the load and the source.

34. In shunt stub matching, the key parameter used for matching is:
a) Admittance of the line at a point
c) Impedance of the stub
Explanation: In shunt stub tuning, the idea is to select d so that the admittance Y, seen looking into the line at distance d from the load is of the form Yₒ+jb) Then the stub susceptance is chosen as –jB, resulting in a matched condition.

35. For series stub matching, the parameter used for matching is:
a) Impedance of the transmission line at a point
b) Voltage at a point on the transmission line
c) Admittance at a point on the transmission line
Explanation: In series sub matching, the distance‘d’ is selected so that the impedance, Z seen looking into the line at a distance‘d’ from the load is of the form Zₒ+jX. Then the stub reactance is chosen as –jX resulting in a matched condition.

36. For co-axial lines and waveguides, ________ is more preferred.
a) Open circuited stub
b) Short circuited stub
c) Slotted section
d) Co-axial lines cannot be impedance matched
Explanation: For co-axial cables and waveguides, short-circuited stub is usually preferred because the cross-sectional area of such an open-circuited line may be large enough to radiate, in which case the stub is no longer purely reactive.

37. For a load impedance of ZL=60-j80. Design of 2 single-stub shunt tuning networks to match this load to a 50Ω line is to be done. What is the normalized admittance obtained so as to plot it on smith chart?
a) 1+j
b) 0.3+j0.4
c) 0.4+j0.3
d) 0.3-j0.4
Explanation: To impedance match a load to a characteristic impedance of the transmission line, first the load has to be normalized. That is, zL=ZL/Z0. For impedance matching using shunt stubs, admittance is used. Taking the reciprocal of impedance, normalized load admittance is 0.3+j0.4.

38. If the normalized admittance at a point on a transmission line to be matched is 1+j1.47. Then the normalized susceptance of the stub used for shunt stub matching is:
a) 1Ω
b) 1.47 Ω
c) -1.47 Ω
d) -1 Ω
Explanation: When shunt stubs are used for impedance matching between a load and transmission line, the susceptance of the shunt stub must be negative of the line’s susceptance at that point for impedance matching.

39. After impedance matching, if a graph is plot with frequency v/s reflection co-efficient of the transmission line is done, then at the frequency point for which the design is done, which of the following is true?
a) There is a peak at this point of the curve
b) There is a dip at this point of the curve
c) The curve is a straight line
d) Such a plot cannot be obtained
Explanation: Since the plot is frequency v/s reflection co-efficient, after impedance matching the reflection co-efficient will be zero or minimum. Hence, there is a dip at that point of the curve.

40. In series stub matching, if the normalized impedance at a point on the transmission line to be matched is 1+j1.33. Then the reactance of the series stub used for matching is:
a) 1 Ω
b) -1.33 Ω
c) -1 Ω
d) 1.33 Ω
Explanation: The reactance of the series stub is negative of the reactance of the line at the point at which it has to be matched. That is, if the line reactance is inductive, the series stub’s reactance is capacitive.

41. Discontinuities in the matching quarter wave transformer are not of considerable amount and are negligible.
a) True
b) False
Explanation: Discontinuities in the matching network cause reflections which result in considerable attenuation of the transmitted signal. Hence, discontinuities in transformers are not negligible.

42. The overall reflection coefficient of a matching quarter wave transformer cannot be calculated because of physical constraints.
a) True
b) False
Explanation: 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.

43. 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
Explanation: 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.

44. The expression for total reflection in the simplified form is given by:
a) Г=Г1+ Г3e-2jθ
b) Г=Г113
c) Г=Г12+ Г3e-2jθ
d) Г= Г1+ Г2e-2jθ
Explanation: 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).

45. 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
Explanation: 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.

46. 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
Explanation: 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.

47. If a λ/4 transmission line is used for impedance matching, then always Г1> Г3.
a) True
b) False
Explanation: 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.

48. 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
Explanation: 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.

49. 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
Explanation: 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.

50. 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
Explanation: 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.

51. S parameters are expressed as a ratio of:
a) Voltage and current
b) Impedance at different ports
c) Indecent and the reflected voltage waves
d) None of the mentioned
Explanation: S matrix can be used to represent any n port network. S parameters are defined for microwave networks. Hence instead of voltage and current measurement, the amplitude of the incident and reflected voltage waves is measured.

52. The relation between incident voltage matrix , reflected voltage matrix and S matrix for a microwave network:
a) [v-] = [s] [v+].
b) [v+] = [s] [v-].
c) [v-] [v] = [s].
d) [s] = [v] [v-].
Explanation: S parameter for a microwave network is defined as the ratio of reflected voltage wave to the incident voltage wave. When represented in the form of a matrix, reflected voltage matrix is the product of S parameter and the incident voltage wave at that port.

53. The specific element Sij of the scattering matrix can be determined as:
a) SIJ= Vi-/Vj+
b) SIJ= Vi+/Vj-
c) S= Vj+/Vi-
d) None of the mentioned
Explanation: The parameter Sij is found by driving port j with an incident wave of voltage Vj+ coming out of ports i. The incident waves on all ports except the jth port are set to zero.

54. The device used to get the measurement of S parameters of n- port micro wave network is:
a) CRO
b) Network analyzer
c) Circulator
d) Attenuator
Explanation: Network analyzer is a device to which any microwave network can be externally connected with the help of probes and the s parameters of the network can be obtained.

55. For a one port network , the scattering parameter S₁₁ in terms of impedance parameter Z₁₁ is:
a) (Z11-1)/ (Z11+1)
b) (Z11+1)/ (Z11-1)
c) (Z11+1) (Z11-1)
d) Z11
Explanation: If Z matrix of a one port network is computed, then the s matrix of the same can be computed using the Z11 coefficient. To compute the S11 parameter of the network, the relation used is (Z11-1)/ (Z11+1).

56. Scattering matrix for a reciprocal network is:
a) Symmetric
b) Unitary
c) Skew symmetric
d) Identity matrix
Explanation: For a reciprocal network, the input to port I and output at port j is the same as the input at port j and output measured at port i. Hence, the ports are interchangeable. As the ports are interchangeable, this is reflected in the matrix and the matrix becomes symmetric.

57. S₁₂=0.85-45⁰ and S₁₂=0.85 +45⁰ for a two port network. Then the two port network is:
a) Non-reciprocal
b) Lossless
c) Reciprocal
d) Lossy
Explanation: For a reciprocal network, the S matrix is symmetric. For the matrix to be symmetric, Sij=Sji. Since this condition is not satisfied in the above case, the matrix is non reciprocal.

58. Scattering matrix for a lossless matrix is:
a) Unitary
b) Symmetric
c) Identity matrix
d) Null matrix
Explanation: For a lossless network, the scattering matrix has to be unitary. That is, the law of conservation of energy is to be verified for this case. Using appropriate formula, this condition can be verified.

59. If the reflection co efficient of a 2 port network is 0.5 then the return network loss in the network is:
a) 6.5 dB
b) 0.15 dB
c) 6.020 dB
d) 10 dB
Explanation: Given the reflection coefficient of the network, return loss of the network is calculated using the formula -20 log │Г│. Substituting for reflection coefficient, the return loss of the network is 6.02 dB.

60. If the reflection co efficient of a 2 port network is 0.25 then the return network loss in the network is:
a) 12.05 dB
b) 0.15 dB
c) 20 dB
d) 10 dB
Explanation: Given the reflection coefficient of the network, return loss of the network is calculated using the formula -20 log │Г│. Substituting for reflection coefficient, the return loss of the network is 12.05 dB.

61. The one below among others is not a type TEM line used in microwave networks:
a) Co-axial wire
b) Micro strip line
c) Strip lines
d) Surface guide
Explanation: Coaxial micro strip and strip lines all support TEM mode of propagation through them. But surface guides do not support TEM mode of propagation in them. Hence it cannot be called a TEM line.

62. The one below is the only micro wave network element that is a TEM line:
a) Co-axial cable
b) Rectangular wave guide
c) Circular wave guide
d) Surface wave guide
Explanation: Coaxial cables support TEM mode of propagation in them and rectangular waveguide, circular wave guide, surface waveguides do not support TEM mode of propagation in them.

63. The relation between voltage, current and impedance matrices of a microwave network is:
a) [V] = [Z][I].
b) [Z] = [V][I].
c) [I] = [Z][V].
d) [V] = [Z]-[I].
Explanation: In microwave networks, at any point in a network, the voltage at a point is the product of the impedance at that point and current measured. This can be represented in the form of a matrix.

64. The relation between voltage, current and admittance matrices of a microwave network is:
a) [I] = [Y] [V].
b) [Y] = [V] [I].
c) [I] = [Z] [V].
d) [V] = [Z]-1[I].
Explanation: The relation between voltage current and admittance matrices is [I] = [Y] [V]. here I represents the current matrix, Y is the admittance matrix and V is the voltage matrix.

65. Admittance and impedance matrices of a micro waves network are related as:
a) [Y] = [Z]-1.
b) [Y] = [Z].
c) [V] = [Z] [Z]-1.
d) [Z] = [V] [V]-1.
Explanation: Both admittance and impedance matrix can be defined for a microwave network. The relation between these admittance and impedance matrix is [Y] = [Z]-1. Admittance matrix is the inverse of the impedance matrix.

66. The element of a Z matrix, Zij can be given in terms of voltage and current of a microwave network as:
a) ZIJ = VI/IJ
b) ZIJ = VIIJ
c) 1//ZIJ = 1/JIVI
d) VIJ = IJ/JI
Explanation: The element Zij of a Z matrix is defined as the ratio of voltage at the ith port to the current at the jth port given that all other currents are set to zero.

67. In a two port network, if current at port 2 is 2A and voltage at port 1 is 4V, then the impedance Z₁₂ is:
a) 2 Ω
b) 8 Ω
c) 0.5 Ω
d) Insufficient data
Explanation: Z12 is defined as the ratio of the voltage at port 1 to the current at port 2. Substituting the given values in the above equation, Z12 parameter of the network is 2 Ω.

68. In a 2 port network, if current at port 2 is 2A and voltage at port 1 is 4 V, then the admittance Y₂₁ is:
a) 0.5 Ʊ
b) 8 Ʊ
c) 2 Ʊ
d) 4 Ʊ
Explanation: Admittance parameter Y12 is defined as the ratio of current at port 1 to the voltage at port 2. Taking the ratio, the admittance Y12 is 0.5 Ʊ.

69. For a reciprocal network, Z matrix is:
a) A unit matrix
b) Null matrix
c) Skew symmetric matrix
d) Symmetric matrix
Explanation: For a reciprocal matrix, the impedance measured at port Zij is equal to the impedance measured at port Zji. Since these parameters occupy symmetric positions in the Z matrix, the matrix becomes symmetric.

70. For a lossless network, the impedance and admittance matrices are:
a) Real
b) Purely imaginary
c) Complex
d) Rational
Explanation: For a network to be lossless, the network should be purely imaginary. Presence of any real component implies the presence of resistance in the network from which the network becomes lossy. So the matrices must be purely imaginary.

#### Module 03

1. The production of power at higher frequencies is much simpler than production of power at low frequencies.
a) True
b) False
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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
Explanation: 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.

13) HEMT used in the microwave circuit is a
a) source
b) high power amplifier
c) low noise amplifier
d) detector

14) Klystron operates on the principle of
a) Amplitude Modulation
b) Frequency Modulation
c) Pulse Modulation
d) Velocity Modulation

15) A cavity resonator can be represented by
a) an LC circuit
b) an LCR circuit
c) a lossy inductor
d) a lossy capacitor

16) Ionospheric preparation is not possible for microwaves because
a) Microwaves will be fully absorbed by the ionospheric layers
b) There will be an abrupt scattering in all directions
c) Microwave will penetrate through the ionospheric layers
d) There will be dispersion of microwave energy

17) If the peak power of pulsed microwave system is 104 W and the average power is 800 W, the the duty cycle will be
a) 80%
b) 8%
c) 0.8%
d) 0.08%

18) The noise produced in a microwave tube due to random nature of emission and electron flow is called
a) Partition noise
b) Shot noise
c) Johnson noise
d) Shannon noise

19) Which of the following is not possible in a circular wave guide ?
a) TE10
b) TE01
c) TE11
d) TE12

20) The pulse frequency is equal to
a) Duty cycle/pulse width
b) The reciprocal of the pulse repetition rate
c) Pulse width x peak power/average power
d) All of these

21) The maximum theoretical output circuit efficiency of a double resonator klystron amplifier is
a) 25%
b) 50%
c) 58%
d) 85%

22) Which of the following can be used for amplification of microwave energy?
a) Travelling wave tube
b) Magnetron
c) Reflex Klystron
d) Gunn diode

23. A microwave tube amplifier uses an axial magnetic field and a radial electric field. This is the
a. reflex klystron
b. coaxial magnetron
c. traveling-wave magnetron
d. CFA

24. One of the following is unlikely to be used as a pulsed device. It is the
a. multicavity klystron
b. BWO
c. CFA
d. TWT

25. One of the reasons why vacuum tubes eventually fail at microwave frequencies is that their
a. noise figure increases
b. transit time becomes too short
c. shunt capacitive reactances become too large
d. series inductive reactances become too small

26. Indicate the false statement. Transit time in microwave tubes will be reduced if
a. the electrodes are brought closer together
b. a higher anode current is used
c. multiple or coaxial leads are used
d. the anode voltage is made larger

27. The multicavity klystron
a. is not a good low-level amplifier because of noise
b. has a higher repeller voltage to ensure a rapid transit time
c. is not suitable for pulsed operation
d. needs a long transit time through the buncher cavity to ensure current modulation

28. Indicate the false statement. Klystron amplifiers may use intermediate cavities to
a. prevent the oscillations that occur in two-cavity klystrons
b. increase the bandwidth of the device
c. improve the power gain
d. increase the efficiency of the klystron

29. The TWT is sometimes preferred to the multicavity klystron amplifier, because it
a. is more efficient
b. has a greater bandwidth
c. has a higher number of modes
d. produces a higher output power

30. The transit time in the repeller space of a reflex klystron must be n + ¾ cycles to ensure that
a. electrons are accelerated by the gap voltage on their return
b. returning electrons give energy to the gap oscillations
c. it is equal to the period of the cavity oscillations
d. the repeller is not damaged by striking electrons

31. The cavity magnetron uses strapping to
a. prevent mode jumping
b. prevent cathode back-heating
c. ensure bunching
d. improve the phase-focusing effect

32. A magnetic field is used in the cavity magnetron to
a. prevent anode current in the absence of oscillation
b. ensure that the oscillations are pulsed
c. help in focusing the electron beam, thus preventing spreading
d. ensure that the electrons will orbit around the cathode

33. To avoid difficulties with strapping at high frequencies, the type of cavity structure used in the magnetron is the
a. hole-and-slot
b. slot
c. vane
d. rising sun

34. The primary purpose of the helix in a traveling-wave tube is to
a. prevent the electron beam from spreading in the long tube
b. reduce the axial velocity of the RF field
d. reduce the noise figure

35. The attenuator is used in the traveling-wave tube to
a. help bunching
b. prevent oscillations
c. prevent saturation
d. increase gain

36. Periodic permanent-magnet focusing is used with TWTs to
a. allow pulsed operation
b. improve electron bunching
c. avoid the bulk of an electromagnet
d. allow coupled-cavity operation at the highest frequencies

37. The TWT is sometimes preferred to the magnetron as a radar transmitter output tube because it is
a. capable of a longer duty cycle
b. a more efficient bandwidth
d. less noisy

38. A magnetron whose oscillating frequency is electronically adjustable over a wide range is called a
a. coaxial magnetron
b. dither-tuned magnetron
c. frequency-agile magnetron
d. VTM

39. Indicate which of the following is nota TWT slow-wave structure:
a. Periodic-permanent magnet
b. Coupled cavity
c. Helix
d. Ring-bar

40. The glass tube of a TWT may be coated with aquadag to
a. help focusing
b. provide attenuation
c. improve bunching
d. increase gain

41. A backward-wave oscillator is based on the
a. rising-sun magnetron
b. crossed-field amplifier
c. coaxial magnetron
d. traveling-wave tube

42. ______is not a microwave tube.
A. cathode_ray tube
B. magnetron
C. travelling_wave tube
D. Both A and B
E. None of these

43. one of the bands that come under Microwave Band.
A. D
B. C
C. E
D. All the above
E. None of these

44. ______is depend on the velocity factor of a transmission line.
A. relative permittivity of dielectric
B. impedance
C. temperature
D. Both A and B
E. None of these

45. One of the following is used for amplification of microwave energy.
A. magnetron
B. travelling wave tube
C. gunn diode
D. Both A and B
E. None of these

46. Reflex klystron is used in_____.
A. Oscillator
B. mixer
C. frequency multiplier
D. None of these

47. ______ is the main advantage of a microwave.
A. frequency multiplier
B. Moves at the speed of light
C. Highly directive
D. High penetration power
E. None of these

48. _____ devices use a helix?
A. TWT
B. Klystron oscillator
C. Klystron amplifier
D. Both Aand B
E. None of these

49. ____is the most common antenna to the predetermined radiation pattern.
A. corner reflector
B. sectrol horn
C. helical antenna
D. array antenna
E. None of these

50. ______ principle does Klystron operates
A. Velocity Modulation
B. Pulse Modulation
C. Frequency Modulation
D. Amplitude Modulation

51. fabrication of microstrip line is_____.
A. photo etching
B. oxidation
C. printed circuit technique
E. None of these

52.HEMT used in the microwave circuit is a
A Low Noise Amplifier
B Detector
C Source
D High Power Amplifier

53.Klystron operates on the principle of
A Velocity Modulation
B Pulse Modulation
C Amplitude Modulation
D Frequency Modulation

54.A cavity resonator can be represented by
A an LC circuit
B a lossy capacitor
C a lossy inductor
D an LCR circuit

55.______ principle does Klystron operates
A Amplitude Modulation
B Pulse Modulation
C Frequency Modulation
D Velocity Modulation

56._____ devices use a helix?
A Klystron amplifier
B Klystron oscillator
C TWT
D None of these

57.Reflex klystron is used in_____.
A Oscillator
B mixer
C frequency multiplier
D None of these

58.One of the following is used for amplification of microwave energy.
A magnetron
B travelling wave tube
C gunn diode
D None of these

59.Reflex klystron is a ______
A. Amplifier
B. Oscillator
C. Attenuator
D. Filter

60.On which of the following principle does Klystron operates
A. Amplitude Modulation
B. Frequency Modulation
C. Pulse Modulation
D. Velocity Modulation

61.In multicavity klystron additional cavities are inserted between buncher & catcher cavities to achieve
A. Higher Gain
B. Higher Efficiency
C. Higher Frequency
D. Higher Bandwidth

62.Which of the following is one of the mode in Reflex Klystron
A. Give same frequency but different transit time
B. Are caused by spurious frequency modulation
C. Are just for theoretical consideration
D. Result from excessive transit time across resonator gap

63.A space between two cavities in two cavity klystron is _______
A. Drift space
B. Free space
C. Running space
D. Normal space

64.Magnetron is an _______
A. Amplifier
B. Oscillator
C. Phase shifter
D. Both phase shifter & amplifier

65.Traveling Wave Tube is __________
A. Oscillator
B. Tuned Amplifier
C. Wide Band Amplifier
D. Both Amplifier & Oscillator

#### Module 04

1. A PIN diode consists of ______number of semiconductor layers.
a) Three
b) Two
c) Four
d) One
Explanation: PIN diode is a p-type, intrinsic, n-type diode consisting of narrow layer of p-type semiconductor and a narrow layer of n-type semiconductor material, with a thicker region of intrinsic or very lightly n doped semiconductor sandwiched between them.

2. The material out of which PIN diode is made is:
a) Silicon
b) Germanium
c) GaAs
d) None of the mentioned
Explanation: Silicon is the semiconductor normally used because of its power handling capability and it offers high resistivity for the intrinsic region. But depending on the application, these days GaAs is also used in fabricating PIN diodes.

3. The behavior of a PIN diode is entirely different from normal diodes at all frequency of operation.
a) True
b) False
Explanation: PIN diode acts as a ordinary diode at frequencies up to about 100MHz. at high frequencies it stops to rectify and then acts as a variable resistance.

4. The junction resistance and capacitance of the intrinsic region in a PIN diode are connected______ in the equivalent circuit of PIN diode.
a) Series
b) Parallel
c) Connected across package capacitance
d) None of the mentioned
Explanation: The junction capacitance Cj and junction resistance Rj of a PIN diode are connected in parallel in the equivalent circuit of a PIN diode. The package resistance and package capacitance are connected in series to these junction parameters.

5. The resistance of the PIN diode with positive bias voltage:
a) Increases
b) Decreases
c) Remains constant
d) Insufficient data
Explanation: When the bias is varied on the PIN diode, its microwave resistance RJ changes from a typical value of 6 KΩ under negative bias to perhaps 5 Ω under forward bias. Thus if the diode is mounted on a 50Ω coaxial line, it will not significantly load this line.

6. A PIN diode can be used in either a series or a shunt configuration to form a __________
a) Single pole single throw switch
b) Single pole double throw switch
c) Amplifier
d) Oscillator
Explanation: A PIN diode can be used in either a series or a shunt configuration to form a single pole single throw switch. In the series configuration, the switch is on when the diode is forward biased and off when the diode is reverse biased.

7. The working principle of series and shunt configuration single pole single throw switch is the same.
a) True
b) False
Explanation: In the series configuration, the switch is on when the diode is forward biased and off when the diode is reverse biased. In the shunt configuration, forward biasing the diode cuts-off the transmission while reverse biasing the diode ensures transmission from input to output.

8. Under ideal conditions, when a PIN diode is used as a switch, the switch must have _______ insertion loss in the ON state.
a) Maximum
b) Zero
c) Average
d) Insertion loss cannot be defined for a switch
Explanation: Ideally, when PIN diode is used as switch, the switch should have zero insertion loss in the ON state and infinite attenuation in the OFF state. These are ideal conditions. But practically a good operating switch must have low insertion loss.

9. When PIN diode is used as a switch, the expression for insertion loss of the switch is given by:
a) 10 log (Po/PL)
b) 10 log (PL/P0)
c) 10 log (PL. Pₒ)
d) None of the mentioned
Explanation: Insertion loss of a switch is defined as the ratio of incident power applied to the load when switch is absent to the actual power delivered to the load.

10. For a shunt configuration switch, the diode impedance is 40 Ω and the terminated line characteristic impedance is 50 Ω. Then the insertion loss of the switch is:
a) 2.2 dB
b) 4.2 dB
c) 8.4 dB
d) 3.6 dB
Explanation: Insertion loss of a shunt configuration switch is given by 20 log (2ZD+Z0/2ZD). Substituting the given values in the above expression, the insertion loss of the shunt configuration switch is 4.2 dB.

11. In the series configuration of a PIN diode switch, the terminated load impedance was found to be 50 Ω and the diode impedance was 60 Ω. Then the insertion loss of the switch is:
a) 4 dB
b) 2 dB
c) 3.6 dB
d) 4.8 dB
Explanation: Insertion loss of a shunt configuration switch is given by 20 log (2ZD+Z0/2ZD). Substituting the given values in the above equation, the insertion loss is 4 dB.

12. The number of PIN diodes used in SPST switch and SPDT switch are the same.
a) True
b) False

Explanation: The number of PIN diodes in SPST switch is one, while the number of PIN diodes used in single pole double throw switch is two.

13. Varactor diode is a semiconductor diode in which the _________ can be varied as a function of reverse voltage of the diode.
a) Junction resistance
b) Junction capacitance
c) Junction impedance
d) None of the mentioned
Explanation: Varactors (variable-capacitor) have non-linearity of capacitance which is fast enough to follow microwaves. Varactor diode is a semiconductor diode in which the junction capacitance can be varied as a function of reverse voltage of the diode.

14. Any semiconductor diode has a junction capacitance varying with reverse bias. If such a diode has microwave characteristics, it is called:
a) IMPATT diode
b) TRAPITT diode
c) SKOTTKY diode
d) None of the mentioned
Explanation: Any semiconductor diode has a junction capacitance varying with reverse bias. If such a diode has microwave characteristics, it is called varactor diode. With the reverse bias, the junction is depleted of mobile carriers resulting in a capacitance that is the diode behaves as a capacitance with the junction acting as dielectric between two conducting plates.

15. The width of depletion region of a varactor diode ________with increase in reverse bias voltage.
a) Increases
b) Decreases
c) Remains constant
d) None of the mentioned
Explanation: The width of the depletion region goes on increasing with increase in reverse bias voltage of the varactor diode. As the width of depletion region is an inverse function for capacitance, as the width increases, capacitance decreases.

16. Diffused junction mesa silicon diodes are widely used at microwave frequencies.
a) True
b) Fals
Explanation: Diffused junction mesa silicon diodes are widely used at microwave frequencies. They are capable handling large powers and large break down voltages. They have relative independence of ambient temperature and low noise.

17. Varactors made of ______ have higher frequency range of operation compared to silicon fabricated varactor diodes.
a) Germanium
b) GaAs
c) GaN
d) None of the mentioned
Explanation: Varactor diodes made of silicon have frequency range of operation of 25 GHz. Varactor diodes made of gallium arsenide operate in the frequency range of 90 GHz. Varactors of gallium arsenide also have better at low temperature.

18. Varactor diodes are operated in _________ region to achieve maximum efficiency possible.
a) Cutoff region
b) Saturation region
c) Reverse saturation region
d) Active region
Explanation: Varactors are used between the reverse saturation point and a point just above the avalanche region. The capacitance variation and the reverse voltage swing are limited to between the operating regions mentioned above.

19. The cutoff frequency for operation of a varactor diode at a specific bias is given by:
a) 1/2πRSCjv
b) 1/2πCSRjv
c) 1/2π√LC
d) None of the mentioned
Explanation: Cutoff frequency for a certain bias voltage applied is given by 1/2πRSCjv. Here, Rs is the wafer resistance and Cjv is the junction capacitance measured in the varactor diode for a given specific bias voltage V.

20. ___________ is an amplifier constructed using a device whose reactance is varied to produce amplification.
a) Travelling wave tube
b) Parametric amplifier
c) Common emitter
d) Klystron amplifier
Explanation: Parametric amplifier is an amplifier constructed using a device whose reactance is varied to produce amplification. Varactor diode is the most widely used element in a parametric amplifier.

21. Parametric amplifier is a ________ amplifier.
a) Low noise
b) High gain
c) Low gain
d) High noise
Explanation: Parametric amplifiers are constructed using varactor diodes. Since they do not involve any resistance, they result in low noise levels. There will be no thermal noise, as the active element used involved is reactive and not resistive.

22. Parametric amplifiers find their application in long range RADAR and satellite ground stations.
a) True
b) False
Explanation: Due to the advantage of low noise amplification, parametric amplifiers are used in applications when noise levels are high at the receiving end but the amplification of noise must not occur. Such applications include long range RADARS satellite ground stations, radio telescopes to name a few.

23. Gain of a parametric amplifier in terms of the frequencies involved in their operation is:
a) (fP – fS)/fS
b) fS/ (fP – fS)
c) fP/fS
d) None of the mentioned
Explanation: Gain of a parametric amplifier in terms of the frequencies involved in their operation is (fP – fS)/fS. Here fP is the pump frequency, fS is the signal frequency and fi is the idler frequency.

24. Silicon and germanium are called ___________ semiconductors.

a) direct gap
b) indirect gap
c) band gap
d) indirect band gap
Explanation: The forbidden energy gap for silicon and germanium are respectively 1.21 eV in Si and 0.79 eV in germanium. Silicon and germanium are called indirect gap semiconductors because the bottom of the conduction band does not lie directly above the top of the valence band.

25. GaAs is used in the fabrication of GUNN diodes because:
a) GaAs is cost effective
b) It less temperature sensitive
c) it has low conduction band electrons
d) less forbidden energy gap
Explanation: In GaAs, the conduction band lies directly above the top of the valence band. The lowest energy conduction band in GaAs is called as primary valley. GaAs consists of six secondary valleys. The bottom of one of the secondary valley is at an energy difference of 0.35 eV with the bottom of the primary valley in conduction band.

26. In a GaAs n-type specimen, the current generated is constant irrespective of the electric filed applied to the specimen.
a) true
b) false
Explanation: In a GaAs n-type specimen, when the electric field applied reaches a threshold value of Eth, the current in the specimen becomes suddenly oscillatory and with respect to time and these oscillations are in the microwave frequency range. This effect is called Gunn Effect.

27. When the electric field applied to GaAs specimen is less than the threshold electric field, the current in the material:
a) increases linearly
b) decreases linearly
c) increases exponentially
d) decreases exponentially
Explanation: When the electric field applied is less than the threshold value of electric field, the electrons jump from the valence band to the primary valley of the conduction band and current increases linearly with electric field.

28. When the applied electric field exceeds the threshold value, electrons absorb more energy from the field and become:
a) hot electrons
b) cold electrons
c) emission electrons
d) none of the mentioned
Explanation: When the applied electric field exceeds the threshold value, electrons absorb more energy from the field and become hot electrons. These electrons jump into the lowest secondary valley in the conduction band. When the electrons become hot, their mobility reduces.

29. GaAs is used in fabricating Gunn diode. Gunn diode is:
a) bulk device
b) sliced device
c) made of different type of semiconductor layers
d) none of the mentioned
Explanation: A GUNN diode is a bulk device, that is, it does not contain any junction but it is a slice of n-type GaAs. P-type GaAs does not exhibit Gunn Effect. Hence it is a reversible and can be operated in both directions.

30. The electrodes of a Gunn diode are made of:
a) molybdenum
b) GaAs
c) gold
d) copper
Explanation: Gunn diode is grown epitaxially onto a gold or copper plated molybdenum electrode, out of gallium arsenide doped with silicon, tellurium or selenium to make it n-type.

31. When either a voltage or current is applied to the terminals of bulk solid state compound GaAs, a differential ______ is developed in that bulk device.
a) negative resistance
b) positive resistance
c) negative voltage
d) none of the mentioned
Explanation: When either a voltage or current is applied to the terminals of a sample of bulk solid state compound formed by group 5 and 3 elements of periodic table, a differential resistance is developed in the bulk device. This fundamental concept is called RWH theory.

32. The number of modes of operation for n type GaAs is:
a) two
b) three
c) four
d) five
Explanation: n-type GaAs used for fabricating Gunn diode has four modes of operation. They are Gunn oscillation mode, limited space charge accumulation mode, and stable amplification mode bias circuit oscillation mode.

33. The free electron concentration in N-type GaAs is controlled by:
a) effective doping
b) bias voltage
c) drive current
d) none of the mentioned
Explanation: The free electron concentration in n-type GaAs is controlled through effective doping so that they range from 1014 to 1017 per cc at room temperature. The typical specimen of n-type GaAs has the dimensions 150 µm by 150 µm.

34. The modes of operation of a Gunn diode are illustrated in a plot of voltage applied to the Gunn diode v/s frequency of operation of Gunn diode.
a) true
b) false

Explanation: A graph of plot of product of frequency and the length of the device plotted along y-axis versus the product of doping concentration and length along X- axis. These are the parameters on which the four modes of operation of Gunn diode are explained.

35. The mode of operation in which the Gunn diode is not stable is:
a) Gunn oscillation mode
b) limited space charge accumulation mode
c) stable amplification mode
d) bias circuit oscillation mode
Explanation: In Gunn oscillation mode, the device is unstable due to the formation of accumulation layer and field domain. This high field domain moves from cathode to anode.

36. The frequency of oscillation in Gunn diode is given by:
a) vdom/ Leff
b) Leff/ Vdom
c) Leff/ WVdom
d) none of the mentioned
Explanation: In Gunn oscillation mode, the frequency of oscillation is given by vdom/ Leff, where vdom is the domain velocity, Leff is effective length that the domain moves from the time it is formed until the time a new domain is formed.

37. In Gunn diode oscillator, the Gunn diode is inserted into a waveguide cavity formed by a short circuit termination at one end
a) true
b) false
Explanation: The Gunn diode is mounted at the centre of the broad wall of a shorted waveguide since for the dominant TE10 mode; the electric field is maximum at the centre.

38. In a Gunn diode oscillator, the electron drift velocity was found to be 107 cm/second and the effective length is 20 microns, then the intrinsic frequency is:
a) 5 GHz
b) 6 GHz
c) 4 GHz
d) 2 GHz
Explanation: The intrinsic frequency for a Gunn oscillator is given by Vd/L. Here VD is the drift velocity and L is the effective length. Substituting the given values in the above equation, intrinsic frequency is 5 GHz.

39. The material used to fabricate IMPATT diodes is GaAs since they have the highest efficiency in all aspects.
a) true
b) false
Explanation: IMPATT diodes can be fabricated using silicon, germanium, GaAs or indium phosphide. Out of these materials, GaAs have highest efficiency, low noise and high operating frequencies. But GaAs has a major disadvantage of complex fabrication process and higher cost. So, GaAs are not preferred over silicon and germanium.

40. When a reverse bias voltage exceeding the breakdown voltage is applied to an IMPATT diode, it results in:
a) avalanche multiplication
b) break down of depletion region
c) high reverse saturation current
d) none of the mentioned
Explanation: A reverse bias voltage exceeding the breakdown voltage is applied to an IMPATT diode, a high electric field appears across the n+ p junction. This high field imparts sufficient energy to the holes and also to valence electrons to raise themselves to the conduction band. This results in avalanche multiplication of electron hole pair.

41. To prevent an IMPATT diode from burning, a constant bias source is used to maintain _______ at safe limit.
a) average current
b) average voltage
c) average bias voltage
d) average resistance
Explanation: Avalanche multiplication is a cumulative process resulting in rapid increase of carrier density. To prevent the diode from burning due to this increased carrier density, a constant bias source is used to maintain average current at safe limit.

42. The number of semiconductor layers in IMPATT diode is:
a) two
b) three
c) four
d) none of the mentioned
Explanation: IMPATT diode consists of 4 layers according to the construction. It consists of a p+ region and n+ layers at the two ends. In between these layers, a p type layer and an intrinsic region is sandwiched.

43. The resonant frequency of an IMPATT diode is given by:
a) Vd/2l
b) Vd/l
c) Vd/2πl
d) Vdd/4πl
Explanation: The resonant frequency of an IMPATT diode is given by the expression Vd/2l. Here VD is the carrier drift velocity; L is the length of the intrinsic region in the IMPATT diode.

44. If the length of the intrinsic region in IMPATT diode is 2 µm and the carrier drift velocity are 107 cm/s, then the drift time of the carrier is:
a) 10-11 seconds
b) 2×10-11 seconds
c) 2.5×10-11 seconds
d) none of the mentioned
Explanation: The drift time of the carrier is defined as the ratio of length of the intrinsic region to the carrier drift velocity. Substituting the given values in this relation, the drift time of the carrier is 2×10-11 seconds.

45. If the length of the intrinsic region in IMPATT diode is 2 µm and the carrier drift velocity are 107 cm/s, then the nominal frequency of the diode is:
a) 12 GHz
b) 25 GHz
c) 30 GHz
d) 24 GHz
Explanation: Nominal frequency is defined as the ratio of the carrier drift velocity to twice the length of the intrinsic region. Substituting the given values in the above equation, the nominal frequency is 25 GHz.

46. IMPATT diodes employ impact ionization technique which is a noisy mechanism of generating charge carriers.
a) true
b) false
Explanation: IMPATT devices employ impact ionization techniques which is too noisy. Hence in order to achieve low noise figure, impact ionization is avoided in BARITT diodes. The minority injection is provided by punch through of the intermediate region.

47. An essential requirement for the BARITT diode is that the intermediate drift region be completely filled to cause the punch through to occur.
a) true
b) false
Explanation: An essential requirement for the BARITT diode is that the intermediate drift region be completely filled to cause the punch through to the emitter-base junction without causing avalanche breakdown of the base collector junction.

48. If the RMS peak current in an IMPATT diode is 700 mA and if DC input power is 6 watt, with the load resistance being equal to 2.5 Ω, the efficiency of the diode is:
a) 10.1 %
b) 10.21 %
c) 12 %
d) 15.2 %
Explanation: Efficiency of IMPATT diode is defined as the ratio of output RMS power to the input DC power. Calculating the RMS output power from the given RMS current and substituting in the equation of efficiency, the efficiency is 10.21%.

50. If the critical field in a Gunn diode oscillator is 3.2 KV/cm and effective length is 20 microns, then the critical voltage is:
a) 3.2 V
b) 6.4 V
c) 2.4 V
d) 6.5 V
Explanation: Critical voltage of a Gunn diode oscillator is given by the expression lEc where l is the effective length and Ec is the critical field. Substituting the given values in the above equation, critical voltage is 6.4 volts.

51. BJTs are bipolar junction transistors. The name bipolar is given because:
a) they are made of n type and p type semiconductor
b) they have holes as charge carriers
c) they have electrons as charge carriers
d) none of the mentioned

Explanation: In bipolar junction transistors, both electrons and holes are charge carriers and both of them together constitute current flow in transistors. Since both carriers result in current, they are called bipolar devices.

52. BJTs are suitable for RF applications because:
a) good performance in terms of frequency
b) power capacity
c) noise characteristics
d) all of the mentioned
Explanation: BJTs designed to operate at certain frequency can be operated over a wide range of frequencies hence offering higher bandwidth. Also they have high power handling capacity and very good noise characteristics.

53. Bipolar junction transistors have _______ 1/f characteristics hence making them suitable for oscillators.
a) high
b) low
c) constant
d) decreasing exponential
Explanation: Bipolar junction transistors have very low 1/f noise. 1/f noise is nothing but thermal noise. Hence BJTs are not very temperature and can be used at high temperature applications as well.

54. Silicon junction transistors are used as amplifiers at frequency range of about:
a) 5-10 MHz
b) 2-10 GHz
c) 40-50 MHz
d) 12-45 GHz
Explanation: Silicon junction transistors have unconditional stability as a two port device at a wide range of frequencies. They are more suitable as amplifiers in the frequency range of about 2-10 GHz. Junction transistors when used as oscillators are used in the frequency range of about 20 GHz.

55. At frequency range of about 2-4 GHz, BJTs are preferred over FETs.
a) true
b) false
Explanation: At about 2-4 GHz frequency range, BJTs have higher gain as compared to FETs, power capacity is high and biasing can be done using a single power supply. Because of these advantages, BJTs are preferred over FETs.

56. One major disadvantage of BJTs over FETs is that:
a) they have low gain
b) they do not have a good noise figure
c) low bandwidth
d) none of the mentioned
Explanation: Bipolar junction transistors are subject to shot noise as well as thermal noise effects, so their noise figure is not as good as that of FET. Noise figure can pose serious problems at high operating frequencies.

57. Bipolar junction transistor is a ________ driven device.
a) current
b) voltage
c) power
d) none of the mentioned
Explanation: Bipolar junction transistor is a current driven device where the collector output current directly depends on the input base current. Base current modulates the collector current of the device.

58. The upper frequency limit of BJT depends on the:
a) collector length in the transistor
b) base length
c) emitter length
d) driving voltage
Explanation: The upper operating frequency limit of a BJT depends on the base length of the transistor. Typical base length of a transistor is in the range of a 0.1 µm. the operating frequency is a few GHz for this base length.

59. In the hybrid –π model of a BJT, the capacitance Cc between the base and collector in the hybrid –π model is ignored.
a) true
b) false
Explanation: The capacitance Cc in the hybrid –π model is small and can be neglected. This has the effect of making the S12 parameter of the BJT equal to zero, implying that the power flows only in one direction through the device.

60. with the increase in the operating frequency of a BJT, the S22 parameter of the transistor:
a) increases
b) decreases
c) remains constant
d) none of the mentioned
Explanation: With increase in the operating frequency of the transistor, S22 parameter of the transistor decreases. S22 parameter signifies the voltage reflected back to port 2. S22 parameter has a value of about 0.93 at 0.1 GHz frequency and 0.33 at 4 GHz frequency.

61. There exists no difference between the construction of GaAs MESFET and silicon MOSFET except for the material used in their construction.
a) True
b) False
Explanation: There exists a difference between the construction of MESFET and MOSFET. There is a thin insulating layer of silicon dioxide between the gate contact and the channel region. Because the gate is insulated, it does not conduct DC bias current.

62. MOSFETs can provide a power of several hundred watts when the devices are packaged in:
a) Series
b) Parallel
c) Diagonal
d) None of the mentioned
Explanation: MOSFETs can be used at frequencies into the UHF range and can provide powers of several hundred watts when devices are packaged in parallel. Laterally diffused MOSFETs have direct grounding of the source and can operate at low microwave frequencies with high power.

63. High electron mobility transistors can be constructed with the use of single semiconductor material like GaAs that have high electron mobility.
a) True
b) False
Explanation: High electron mobility transistor is a hetero junction FET, meaning that it does not use a single semiconductor material, but instead is constructed with several layers of compound semiconductor materials.

64. The curve of IDS v/s VDS of an FET does not vary with the gate to source voltage applied.
a) True
b) False
Explanation: 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.

65. High-power circuits generally use higher values of:
a) Gate to source current
b) Drain to source current
c) Drain current
c) Gate to source voltage
Explanation: 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.

66. High drain current at RF levels is achieved with the biasing and decoupling circuitry for a dual polarity supply.
a) True
b) False
Explanation: 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 very high impedance at RF frequencies to isolate the signal from the bias supply.

67. Since multiple layers of semiconductor materials is used in high electron mobility transistors, this results in:
a) High gain
b) Power loss
c) Temperature sensitivity
d) Thermal stress
Explanation: The multiple layers in the high electron mobility transistor result in the thermal and mechanical stress in the layers. To avoid this, the layers usually have matched crystal lattice.

68. A major disadvantage of high electron mobility transistor is that:
a) They have low gain
b) High manufacturing cost
c) Temperature sensitive
d) High driving voltage is required
Explanation: High electron mobility transistors are devices containing multiple layers of different semiconductor materials. This complicated structure of HEMT requires sophisticated fabrication techniques leading to relatively high cost.

69. HEMT fabricated using GaN and aluminum gallium nitride on a silicon substrate can be used in :
a) High power transmitters
d) Smart antennas
Explanation: GaN HEMT operate with drain voltages in the range of 20-40 V and can deliver power up to 100 W at frequencies in the low microwave range, making these devices popular for high power transmitters.

70. The scattering parameter S11 for GaN HELMT increases with increase in frequency of operation
a) True
b) False
Explanation: For GaN, the S11 parameter of the amplifier decreases with increase in frequency of operation. Experimental results have shown that S11 parameter was 0.96 at 0.5 GHz of frequency and 0.88 at 4 GHz of frequency.

#### Module 05

1. Slotted line is a transmission line configuration that allows the sampling of:
a) electric field amplitude of a standing wave on a terminated line
b) magnetic field amplitude of a standing wave on a terminated line
c) voltage used for excitation
d) current that is generated by the source
Explanation: Slotted line allows the sampling of the electric field amplitude of a standing wave on a terminated line. With this device, SWR and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.

2. A slotted line can be used to measure _____ and the distance of _____________ from the load.
a) SWR, first voltage minimum
b) SWR, first voltage maximum
c) characteristic impedance, first voltage minimum
d) characteristic impedance, first voltage maximum
Explanation: With a slotted line, SWR and the distance of the first voltage minimum from the load can be measured, from this data, load impedance can be found.

3. A modern device that replaces a slotted line is:
a) Digital CRO
b) generators
c) network analyzers
d) computers
Explanation: Although slotted lines used to be the principal way of measuring unknown impedance at microwave frequencies, they have largely been superseded by the modern network analyzer in terms of accuracy, versatility and convenience.

4. If the standing wave ratio for a transmission line is 1.4, then the reflection coefficient for the line is:
a) 0.16667
b) 1.6667
c) 0.01667
d) 0.96
Explanation: ┌= (SWR-1)/ (SWR+1). Substituting for SWR in the above equation for reflection co-efficient, given SWR is 1.4, reflection co-efficient is 0.16667.

5. If the reflection coefficient of a transmission line is 0.4, then the standing wave ratio is:
a) 1.3333
b) 2.3333
c) 0.4
d) 0.6
Explanation: SWR= (1+┌)/ (1-┌). Where ┌ is the reflection co-efficient. Substituting for the reflection co-efficient in the equation, SWR is 2.3333.

6. Expression for ϴ means phase angle of the reflection co efficient r=|r|-e^jθ, the phase of the reflection co-efficient is:
a) θ=2π+2βLmin
b) θ=π+2βLmin
c) θ=π/2+2βLmin
d) θ=π+βLmin
Explanation: here, θ is the phase of the reflection co-efficient. Lmin is the distance from the load to the first minimum. Since voltage minima repeat every λ/2, any multiple of λ/2 can be added to Lmin .

7. In the expression for phase of the reflection coefficient, Lmin stands for :
a) distance between load and first voltage minimum
b) distance between load and first voltage maximum
c) distance between consecutive minimas
d) distance between a minima and immediate maxima
Explanation: Lmin is defined as the distance between the terminating load of a transmission line and the first voltage minimum that occurs in the transmission line due to reflection of waves from the load end due to mismatched termination.

8. If SWR=1.5 with a wavelength of 4 cm and the distance between load and first minima is 1.48cm, then the reflection coefficient is:
a) 0.0126+j0.1996
b) 0.0128
c) 0.26+j0.16
d) none of the mentioned
Explanation: ┌= (SWR-1)/ (SWR+1). Substituting for SWR in the above equation for reflection co-efficient, magnitude of the reflection co-efficient is 0.2. To find θ, θ=π+2βLmin, substituting Lmin as 1.48cm, θ=86.4⁰. Hence converting the polar form of the reflection co-efficient into rectangular co-ordinates, reflection co-efficient is 0.0126+j0.1996.

9. If the characteristic impedance of a transmission line 50 Ω and reflection coefficient is 0.0126+j0.1996, then load impedance is:
a) 47.3+j19.7Ω
b) 4.7+j1.97Ω
c) 0.26+j0.16
d) data insufficient
Explanation: ZL=Z0 (1+┌)/ (1-┌). Substituting the given values of reflection co-efficient and characteristic impedance, ZL is 47.3+j19.7Ω .

10. If the normalized load impedance of a transmission line is 2, then the reflection co-efficient is:
a) 0.33334
b) 1.33334
c) 0
d) 1
Explanation: ZL=Z0 (1+┌)/ (1-┌), this is the expression for load impedance. Normalized load impedance is the ratio of load impedance to the characteristic impedance, taking ZLL/Z0 as 2, the reflection co-efficient is equal to 0.33334.

11. Standing waves occurs due to
a) Impedance match
b) Impedance mismatch
c) Reflection
d) Transmission
Explanation: Impedance mismatches result in standing waves along the transmission line. It shows the variation of the wave amplitudes due to mismatching.

12. Standing wave ratio is defined as the
a) Ratio of voltage maxima to voltage minima
b) Ratio of current maxima to current minima
c) Product of voltage maxima and voltage minima
d) Product of current maxima and current minima
Explanation: SWR is defined as the ratio of the partial standing wave’s amplitude at an antinode (maximum) to the amplitude at a node (minimum) along the line. It is given by S = VMAX/VMIN.

13. Given that the reflection coefficient is 0.6. Find the SWR.
a) 2
b) 4
c) 6
d) 8
Explanation: The relation between reflection coefficient and SWR is given by S = 1 + R/1 – R. On substituting for R = 0.6, we get S = 1 + 0.6/1 – 0.6 = 1.6/0.4 = 4.

14. The maxima and minima voltage of the standing wave are 6 and 2 respectively. The standing wave ratio is
a) 2
b) 3
c) 1/2
d) 4
Explanation: The ratio of voltage maxima to voltage minima is given by the standing wave ratio SWR. Thus S = VMAX/VMIN. On substituting the given data, we get S = 6/2 = 3.

15. Find the standing wave ratio, when a load impedance of 250 ohm is connected to a 75 ohm line.
a) 0.3
b) 75
c) 250
d) 3.33
Explanation: The standing wave ratio is the ratio of the load impedance to the characteristic impedance. Thus S = ZL/Zo. On substituting for ZL = 250 and Zo = 75, we get S = 250/75 = 3.33.

16. Find the reflection coefficient of the wave with SWR of 3.5.
a) 0.55
b) 0.23
c) 0.48
d) 0.68
Explanation: The reflection coefficient in terms of the SWR is given by R = S – 1/S + 1. On substituting for S = 3.5, we get 3.5 – 1/3.5 + 1 = 0.55.

17. The range of the standing wave ratio is
a) 0 < S < 1
b) -1 < S < 1
c) 1 < S < ∞
d) 0 < S < ∞
Explanation: The standing wave ratio is given by S = 1 – R/1 + R. Thus the minimum value of S is 1. It can extend upto infinity for long lines. Thus the range is 1 < S < ∞.

18. For matched line, the standing wave ratio will be
a) 0
b) ∞
c) -1
d) 1
Explanation: In a matched line, maximum transmission occurs. The reflection will be zero. The standing wave ratio S = 1 – R/1 + R. For R = 0, the SWR is unity for matched line.

19. The maximum impedance of a 50 ohm transmission line with SWR of 3 is
a) 50/3
b) 3/50
c) 150
d) 450
Explanation: The maximum impedance is given by the product of the characteristic impedance and the SWR. Thus Zmax = S Zo. On substituting for S = 3 and Zo = 50, we get ZMAX = 3 X 50 = 150 units.

20. The minimum impedance of a 75 ohm transmission line with a SWR of 2.5 is
a) 100
b) 50
c) 25
d) 30
Explanation: The minimum impedance in terms of SWR is given by ZMIN = Zo/S. Substituting the given data for S = 2.5 and Zo = 75, we get Zmin = 75/2.5 = 30.

21. The standing wave ratio of short circuited and open circuited lines will be
a) 0
b) 1
c) -1
d) ∞
Explanation: The transmission line will reflect high power when it is short or circuited. This will lead to high reflection coefficient. Thus the standing wave ratio will be infinity for these extreme cases.

22. The current reflection coefficient of a line with voltage reflection coefficient of 0.65 is given by
a) 0
b) 0.65
c) -0.65
d) 0.35
Explanation: The current reflection coefficient at any point on the line is the negative of the voltage reflection coefficient at that point, i.e, -R. Given that the voltage reflection coefficient is 0.65, thus the current reflection coefficient is -0.65.

23. The power of the electromagnetic wave with electric and magnetic field intensities given by 12 and 15 respectively is
a) 180
b) 90
c) 45
d) 120
Explanation: The Poynting vector gives the power of an EM wave. Thus P = EH/2. On substituting for E = 12 and H = 15, we get P = 12 x 15/2 = 90 units.

24. The power of a wave of with voltage of 140V and a characteristic impedance of 50 ohm is
a) 1.96
b) 19.6
c) 196
d) 19600
Explanation: The power of a wave is given by P = V2/2Zo, where V is the generator voltage and Zo is the characteristic impedance. on substituting the given data, we get P = 1402/(2×50) = 196 units.

25. The power reflected by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)
a) 2
b) 8
c) 6
d) 4
Explanation: The fraction of the reflected to the incident power is given by the reflection coefficient. Thus Pref = R2xPinc. On substituting the given data, we get Pref = 0.52 x 16 = 4 units.

26. The power transmitted by a wave with incident power of 16 units is(Given that the reflection coefficient is 0.5)
a) 12
b) 8
c) 16
d) 4
Explanation: The fraction of the transmitted to the incident power is given by the reflection coefficient. Thus Pref = (1-R2) Pinc. On substituting the given data, we get Pref = (1- 0.52) x 16 = 12 units. In other words, it is the remaining power after reflection.

27. The incident and the reflected voltage are given by 15 and 5 respectively. The transmission coefficient is
a) 1/3
b) 2/3
c) 1
d) 3
Explanation: The ratio of the reflected to the incident voltage is the reflection coefficient. It is given by R = 5/15 = 1/3. To get the transmission coefficient, T = 1 – R = 1 – 1/3 = 2/3.

28. The current reflection coefficient is given by -0.75. Find the voltage reflection coefficient.
a) -0.75
b) 0.25
c) -0.25
d) 0.75
Explanation: The voltage reflection coefficient is the negative of the current reflection coefficient. For a current reflection coefficient of -0.75, the voltage reflection coefficient will be 0.75.

29. The attenuation is given by 20 units. Find the power loss in decibels.
a) 13.01
b) 26.02
c) 52.04
d) 104.08
Explanation: The attenuation refers to the power loss. Thus the power loss is given by 20 units. The power loss in dB will be 10 log 20 = 13.01 decibel.

30. The reflection coefficient is 0.5. Find the return loss.
a) 12.12
b) -12.12
c) 6.02
d) -6.02
Explanation: The return loss is given by RL = -20log R, where is the reflection coefficient. It is given as 0.5. Thus the return loss will be RL = -20 log 0.5 = 6.02 decibel.

31. The radiation resistance of an antenna having a power of 120 units and antenna current of 5A is
a) 4.8
b) 9.6
c) 3.6
d) 1.8
Explanation: The power of an antenna is given by Prad = Ia2 Rrad, where Ia is the antenna current and Rrad is the radiation resistance. On substituting the given data, we get Rrad = Prad/Ia2 = 120/52 = 4.8 ohm.

32. The transmission coefficient is given by 0.65. Find the return loss of the wave.
a) 9.11
b) 1.99
c) 1.19
d) 9.91
Explanation: The transmission coefficient is the reverse of the reflection coefficient, i.e, T + R = 1. When T = 0.65, we get R = 0.35. Thus the return loss RL = -20log R = -20log 0.35 = 9.11 decibel.

33. The return loss is given as 12 decibel. Calculate the reflection coefficient.
a) 0.35
b) 0.55
c) 0.25
d) 0.75
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 12, we get R = 10(-12/20) = 0.25.

34. Find the transmission coefficient of a wave, when the return loss is 6 decibel.
a) 0.498
b) 0.501
c) 0.35
d) 0.65
Explanation: The return loss is given by RL = -20log R. The reflection coefficient can be calculated as R = 10(-RL/20), by anti logarithm property. For the given return loss RL = 6, we get R = 10(-6/20) = 0.501. The transmission coefficient will be T = 1 –
R = 1-0.501 = 0.498.

35. The characteristic impedance of a quarter wave transformer with load and input impedances given by 30 and 75 respectively is
a) 47.43
b) 37.34
c) 73.23
d) 67.45
Explanation: In quarter wave transformer, the characteristic impedance will be the geometric mean of the input impedance and the load impedance. Thus Zo2 = ZIN ZL. On substituting for ZIN = 75 and ZL = 30, we get the characteristic impedance as 47.43 units.

36. The input impedance of a quarter wave line 50 ohm and load impedance of 20 ohm is
a) 50
b) 20
c) 1000
d) 125
Explanation: The characteristic impedance will be the geometric mean of the input impedance and the load impedance. Thus Zo2 = Zin ZL. On substituting for Zo = 50 and ZL = 20, we get the input impedance as 502/20 = 125 ohm.

37. For a matched line, the input impedance will be equal to
b) Characteristic impedance
c) Output impedance
d) Zero
Explanation: A matched line refers to the input and characteristic impedance being the same. In such condition, maximum transmission will occur with minimal losses. The reflection will be very low.

38. The reflection coefficient lies in the range of
a) 0 < τ < 1
b) -1 < τ < 1
c) 1 < τ < ∞
d) 0 < τ < ∞
Explanation: The reflection coefficient lies in the range of 0 < τ < 1. For full transmission, the reflection will be zero. For no transmission, the reflection will be unity.

39. When the ratio of load voltage to input voltage is 5, the ratio of the characteristic impedance to the input impedance is
a) 1/5
b) 5
c) 10
d) 25
Explanation: From the transmission line equation, the ratio of the load voltage to the input voltage is same as the ratio of the characteristic impedance to the input impedance. Thus the required ratio is 5.

40. The power of the transmitter with a radiation resistance of 12 ohm and an antenna current of 3.5A is
a) 147
b) 741
c) 174
d) 471
Explanation: The power in a transmitter is given by Prad = Iant2 Rrad. On substituting Irad = 3.5 and Rrad =12, we get Prad = 3.52 x 12 = 147 units.

41. The group delay of the wave with phase constant of 62.5 units and frequency of 4.5 radian/sec is
a) 13.88
b) 31.88
c) 88.13
d) 88.31
Explanation: The group delay is given by td = β/ω. Given that β = 62.5 and ω = 4.5, we get the group delay as td = 62.5/4.5 = 13.88 units.

42. The maximum impedance of a transmission line 50 ohm and the standing wave ratio of 2.5 is
a) 20
b) 125
c) 200
d) 75
Explanation: The maximum impedance of a line is given by Zmax = SZo. On substituting for S = 2.5 and Zo = 50, we get Zmax = 2.5 x 50 = 125 ohm.

43. The minimum impedance of a transmission line 75 ohm with a standing wave ratio of 4 is
a) 75
b) 300
c) 18.75
d) 150
Explanation: The minimum impedance of a line is given by Zmin = Zo/S. On substituting for Zo = 75 and S = 4, we get Zmin = 75/4 = 18.75 units.

44. The average power in an electromagnetic wave is given by
a) propagation constant
b) poynting vector
c) phase constant
d) attenuation constant
Explanation: The Poynting vector is the cross product of the electric field and magnetic field intensities. It gives the total power of an electromagnetic wave.

45. The characteristic impedance of a transmission line is normally chosen to be
a) 50
b) 75
c) 50 or 75
d) 100
Explanation: The characteristic impedance is always 50 ohm or 75 ohm for a transmission line. This is because of the GHz range of operation and the load impedences employed.

46. Identify the material which is not present in a transmission line setup.
a) waveguides
b) cavity resonator
c) antenna
d) oscillator
Explanation: The transmission line setup consists of antennae for transmitting and receiving power. It consists of waveguides and cavity resonator for guided transmission of electromagnetic waves. Thus oscillator is the odd one out.

47. A __________ is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space.
a) Transmitting antenna
b) Receiving antenna
d) Mixer
Explanation: A transmitting antenna is a device that converts a guided electromagnetic wave on a transmission line into a plane wave propagating in free space. It appears as an electrical circuit on one side, provides an interface with a propagating plane wave.

48. Antennas are bidirectional devices.
a) True
b) False
Explanation: Antennas can be used both as transmitters and receivers. As transmitters they radiate energy to free space and as receivers they receive signal from free space. Hence, they are called bidirectional devices as they are used at both transmitting end and receiving end.

49. Dipole antennas are an example for:
a) Wire antennas
b) Aperture antennas
c) Array antennas
d) None of the mentioned
Explanation: Dipoles, monopoles, oops, Yagi-Uda arrays are all examples for wire antennas. These antennas have low gains, and are mostly used at lower frequencies.

50. _________ antennas consist of a regular arrangement of antenna elements with a feed network
a) Aperture antennas
b) Array antennas
c) Printed antennas
d) Wire antennas
Explanation: Array antennas consist of a regular arrangement of antenna elements with a feed network. Pattern characteristics such as beam pointing angle and side lobe levels can be controlled by adjusting the amplitude and phase excitation of array elements.

51. A parabolic reflector used for reception with the direct broadcast system is 18 inches in diameter and operates at 12.4 GHz. The far-field distance for this antenna is:
a) 18 m
b) 13 m
c) 16.4 m
d) 17.3 m
Explanation: Far field distance for a reflector antenna is given by 2D2/λ. D is the diameter and λ is the operating signal wavelength. Substituting in the above expression, far field distance is 17.3 m.

52._________ of an antenna is a plot of the magnitude of the far field strength versus position around the antenna.
b) Directivity
c) Beam width
d) None of the mentioned
Explanation: Radiation pattern of an antenna is a plot of the magnitude of the far field strength versus position around the antenna. This plot gives the detail regarding the region where most of the energy of antenna is radiated, side lobes and beam width of an antenna.

53. Antennas having a constant pattern in the azimuthal plane are called _____________
a) High gain antenna
b) Omni directional antenna
c) Unidirectional antenna
d) Low gain antenna
Explanation: Omni directional antennas radiate EM waves in all direction. If the radiation pattern for this type of antenna is plotted, the pattern is a constant signifying that the radiated power is constant measured at any point around the antenna.

54. Beamwidth and directivity are both measures of the focusing ability of an antenna.
a) True
b) False
Explanation: Beamwidth and directivity are both measures of the focusing ability of an antenna. An antenna with a narrow main beam will have high directivity, while a pattern with low beam will have low directivity.

55. If the beam width of an antenna in two orthogonal planes are 300 and 600. Then the directivity of the antenna is:
a) 24
b) 18
c) 36
d) 12
Explanation: Given the beam width of the antenna in 2 planes, the directivity is given by 32400/θ*∅, where θ,∅ are the beam widths in the two orthogonal planes. Substituting in the equation, directivity of the antenna is 18.

56. If the power input to an antenna is 100 mW and if the radiated power is measured to be 90 mW, then the efficiency of the antenna is:
a) 75 %
b) 80 %
c) 90 %
d) Insufficient data
Explanation: Antenna efficiency is defined as the ratio of radiated power to the input power to the antenna. Substituting the given data in the efficiency equation, the efficiency of the antenna is 90%.

#### Module 06

1. Progress in ________ and other related semiconductors material processing led to the feasibility of monolithic microwave integrated circuits.
a) GaAs
b) Silicon
c) Germanium
d) GaAlAs
Explanation: Progress in GaAs and other related semiconductor material processing led to the feasibility of MMIC, where all the passive and active components required for a given circuit can be grown or implanted in a substrate.

2. MMICs are high cost devices that involve complex fabrication methods and contain multiple layers to contain even small circuits.
a) True
b) False
Explanation: Min MIC can be made at low cost because the labor involved with fabricating hybrid MIC’s is reduced. In addition, a single wafer can contain a large number of circuits, all of which can be processed and fabricated simultaneously.

3. The substrate of an MMIC must be a _____________ to accommodate the fabrication of all the type of devices.
a) Semiconductor
b) Insulator
c) Partial conductors
d) Metals operable at high frequencies
Explanation: Substrate of MMIC must be a semiconductor material to accommodate the fabrication of active devices. The type of devices and the frequency range dictate the type of substrate material. One such material is GaAs MESFET.

4. GaAs MESFETs are versatile device because it finds application in:
a) Low-noise amplifiers
b) High gain amplifiers
c) Mixers
d) All of the mentioned
Explanation: GaAs MESFET find application in low noise amplifiers, high gain amplifiers, broadband amplifiers, mixers, oscillators, phase shifters, and switches. These are the mostly used and cost effective substrates.

5. Transmission lines and other conductors in microwave devices are usually made with ___________
a) Gold metallization
b) Silver metallization
c) Copper metallization
d) Zinc metallization
Explanation: Transmission lines and conductors at microwave operation are usually made with gold metallization. To improve the adhesion of gold to the substrate, a thin layer of chromium or titanium may be deposited first since these metals are relatively lossy.

6. For the capacitors used in MMICs, the insulating dielectric films used are:
a) Air
b) SiO
c) Titanium
d) GaAs
Explanation: Capacitor and overlaying lines require insulating dielectric films, such as SiO, SiO2, SiN4 and Ta2O5. These materials have high dielectric constants and low loss and are compatible with integrated circuit processing.

7. Resistors used at normal operating frequencies can be directly used at microwave frequencies in MMIc.
a) True
b) False
Explanation: Resistors used at normal operating frequencies cannot be used directly in MMICs. Resistors require the deposition of the lossy films like NiCr, Ta, Ti, and doped GaAs commonly used.

8. Processing in MMICs is done by __________
a) Ion implantation
b) Net list generation
c) Floor planning
d) None of the mentioned
Explanation: Processing begins by forming an active layer in the semiconductor substrate for the necessary active devices. This is done by ion implantation or epitaxial techniques.

9. MMICs are the best microwave integrated circuit fabrication methodologies without any drawbacks in it.
a) True
b) False
Explanation: Major drawback of MMIC is that they tend to waste large area of relatively expensive semiconductor substrate for components such as transmission lines and hybrids.

10. MMICs have higher circuit flexibility as compared to other microwave integrated fabrication methods.
a) True
b) False
Explanation: Since the fabrication of additional FETs in an MMIC design is much easy, the circuit flexibility and performance can be enhanced with only little additional cost and the requirement for the fabrication of the entire device is prevented.

11. A multichip circuit :
Option-1 : consists of a number of interconnected thin film circuits
Option-2 : consists of several interconnected monolithic wafers
Option-3 : consists of a number of interconnected thin film and thick film circuits
Option-4 : none of these
Ans-2
12. Capacitors for integrated circuits ;
Option-1 : can not be integrated and are therefore connected externally as discreteelements
Option-2 : cannot be fabricated using diffusion techniques
Option-3 : can be fabricated using SiO2 of dielectric
Option-4 : none of these
Ans:3
13 : Advantages of integrated circuits include the possibility of
Option-1 : repairing individual circuits elements
Option-2 : obtaining low tolerance resistors
Option-3 : using high values of capacitors
Option-4 : obtaining high stability at low cost
Ans:4
14 : The overcall cost of an integrated circuit :
Option-1 : is always lower than the corresponding discrete component assembly
Option-2 : depends on the total number manufactured
Option-3 : is tending to increase
Option-4 : is always dominated by the design cost
Ans-2
15 : Reliability of IC’s may be increase by:
Option-1 : reducing the number of interconnections
Option-3 : reducing component size
Option-4 : operation at high temperature
Ans-1
16: Microwave integrated circuits :
Option-1 : can be made using thin film on ceramics
Option-2 : can be made using thick film on ceramics
Option-3 : are very difficult to fabricate
Ans-2
17: Passive elements in thin film circuits ;
Option-1 : can be made entirely from tantalum
Option-2 : are made of ceramic
Option-3 : require special isolation process
Option-4 : none of these
Ans-1
18: A diffused resistor in an IC ;Option-1 : is formed along with the fabrication of transistors
Option-2 : can be fabricated with ease for any resistance value
Option-3 : is fabricated before transistor diffusion
Option-4 : is fabricated after transistor diffusion
Ans-1
19 : In a monolithic IC :
Option-1 : all circuit elements are formed on a single wafer of silicon
Option-2 : active elements are formed on separate wafers
Option-3 : circuit elements are assembled on intrinsic silicon substrate
Option-4 : none of these
Ans-1
20 : An IC comprises of 40 logic gates, each of which consists of 4 components.
This constitutes case of :
Option-1 : small scale integration
Option-2 : medium scale integration
Option-3 : large scale integration
Option-4 : none of these
Ans-2