Category: Microwave Engineering Objective Questions

250+ TOP MCQs on Terminated Lossless Transmission Lines – 2 and Answers

Microwave Engineering Interview Questions on “Terminated Lossless Transmission Lines – 2”.

1. Expression for input impedance of a transmission line in terms of load impedance and characteristic impedance is:
A. Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl)
B. (Z0+j ZLtan βl)/ (ZL+j Z0tan βl)
C. Z0 (ZL-j Z0tan βl)/ (Z0-j ZLtan βl)
D. (Z0-j ZLtan βl)/ (ZL-j Z0tan βl)
Answer: A
Clarification: Representing the input voltage as the ratio of voltage at current, representing voltage and currents in hyperbolic function form and simplifying, we get Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl).

2. Input impedance of a short circuited transmission line is :
A. -jZ0tanβl
B. jZ0tanβl
C. jZ0cotβl
D. – jZ0cotβl
Answer: B
Clarification: Since the load impedance of a short circuited transmission line is zero, substituting ZL=0 in the expression for input impedance of a transmission line Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl), input impedance of the transmission line comes out to be jZ0tanβl.

3. Input impedance of a transmission line can be represented in terms of this simple trigonometry function.
A. sine function
B. cosine function
C. cotangent function
D. tangent function
Answer: D
Clarification: The input impedance of a transmission line is expressed in the standard form as Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl) which is represented in terms of a tangent function.

4. If a ƛ/3 transmission line is short circuited that has a characteristic impedance of 50 Ω, then its input impedance is:
A. -j100Ω
B. 50Ω
C. 86.60Ω
D. –j86.60Ω
Answer: D
Clarification: For a short circuited transmission line, the input impedance is given by jZ0tanβl.substituting for characteristic impedance and ‘l’ in the above equation, input impedance is –j86.60Ω.

5. Expression for input impedance of an Open circuited transmission line is:
A. -jZ0tanβl
B. jZ0tanβl
C. jZ0cotβl
D. -jZ0cotβl
Answer: D
Clarification: Since the load impedance of a open circuited transmission line is infinity, substituting ZL=infinity (1/ ZL =0) in the expression for input impedance of a transmission line Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl), input impedance of the open circuited transmission line comes out to be- jZ0cotβl.

6. Input impedance of a open circuited transmission line is represented using this trigonometric function:
A. sine function
B. cosine function
C. cotangent function
D. tangent function
Answer: C
Clarification: The input impedance of a transmission line is expressed in the standard form as Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl). With ZL equal to infinity for open circuit termination, 1/ ZL equal to 0, substituting this, we get input impedance in terms of a cotangent function.

7. For a λ/2 transmission line, if the characteristic impedance of the line is 50 Ω and the terminated with a load of 100 Ω, then its input impedance is:
A. 100Ω
B. 50Ω
C. 88.86Ω
D. none of the mentioned
Answer: A
Clarification: Input impedance of a transmission line is given by Z0 (ZL+j Z0tan βl)/ (Z0+j ZLtan βl). Substituting β=2π/λ, and l=λ/2, we get input impedance of the transmission line equal to the load impedance or the terminated load.

8. If a λ/3 transmission line is open circuited and has characteristic impedance of 50 Ω then the input impedance is:
A. j28.86Ω
B. 50Ω
C. j50Ω
D. 28.86Ω
Answer: A
Clarification: Input impedance of an open circuited transmission line is given by – jZ0cotβl. Substituting l=λ/3 and β=2π/λ in the above equation, input impedance is j28.86Ω.

9. Expression for a transmission co-efficient of a transmission line is :
A. 2ZL/ ( ZL+Z0)
B. (ZL-Z0)/ (ZL+Z0)
C. 2Z0/( ZL+Z0)
D. (ZL+Z0)/ (ZL-Z0)
Answer: A
Clarification: T=┌+1, where T is the transmission co-efficient and ┌ is the reflection co-efficient substituting ┌= (ZL-Z0)/ (ZL+Z0) in the equation for transmission co-efficient, we get 2ZL/ ( ZL+Z0).

10. For a transmission line, if the reflection coefficient is 0.4, then the transmission coefficient is:
A. 0.4
B. 1.4
C. 0.8
D. 2.8
Answer: B
Clarification: T=┌+1, where T is the transmission co-efficient and ┌ is the reflection co-efficient substituting ┌=0.4 in the above equation, transmission co-efficient is equal to 1.4.

11. If the transmission coefficient of a transmission line is 1.6, then the reflection co efficient is:
A. 0.8
B. 0.6
C. 0.4
D. 0.3
Answer: B
Clarification: T=┌+1, where T is the transmission co-efficient and ┌ is the reflection co-efficient substituting T=1.6, we get ┌=0.6.

12. For a transmission line, if the transmission coefficient is 1.4, then the insertion loss in dB is:
A. -2.922dB
B. 29dB
C. 1.46dB
D. -29dB
Answer: A
Clarification: Insertion loss for a transmission line is given by the expression -20loglTl in dB. Substituting T=1.4 and taking logarithm to base 10, insertion loss is -2.922dB.

13. The relation between nepers and decibels is:
A. 1 Np= 8.686 dB
B. 1 dB=8.868 dB
C. Np≥dB
D. dB≥Np
Answer: A
Clarification: 1 Np=10log e2 dB. Substituting e=2.718 in the above equation , 1Np=8.686 dB.


250+ TOP MCQs on Aperture Coupling and Answers

Microwave Engineering Multiple Choice Questions on “Aperture Coupling”.

1. The matched network is placed between:
A. load and transmission line
B. source and the transmission line
C. source and the load
D. none of the mentioned
Answer: A
Clarification: At microwave frequencies, for maximum power transmission, the characteristic impedance of the transmission line must be matched to the load impedance with which the line is terminated. Hence to match these impedances, the matched network is laced between load and transmission line.

2. When a transmission line is matched to a load using a matched network, reflected waves are present:
A. between the load and the matched network
B. between the matched network and the transmission line
C. between the source and the transmission line
D. between the matched network and source
Answer: A
Clarification: The matching circuit is used to match the transmission line and the load. This circuit prevents the reflection of the waves reaching the source. Hence, reflected waves are present between the load and the matched network.

3. Impedance matching sensitive receiver components may improve the _____ of the system.
A. noise
B. SNR
C. amplification factor
D. thermal noise
Answer: B
Clarification: SNR (signal to noise ratio) of the system defines the ratio of signal power to noise power. An increase in this value results in increase of the signal strength. Impedance matching certain sensitive receiver components helps in delivering maximum power to the load and increased signal strength.

4. One of the most important factors to be considered in the selection of a particular matching network is:
A. noise component
B. amplification factor
C. bandwidth
D. none of the mentioned
Answer: C
Clarification: Any type of matching network can ideally give a perfect match at a single frequency. But it is desirable to match a load over a band of frequencies. Hence, bandwidth plays an important role in the selection of the matching network.

5. The simplest type of matching network, L section consists of _______ reactive elements.
A. one
B. two
C. four
D. six
Answer: B
Clarification: As the name of the matching circuit indicates, ‘L’ section consists of 2 reactive elements, one element vertical and another horizontal. 2 types of ‘L’ sections exist. The best one is chose based on the normalized value of the load impedance.

6. The major limitation of a lumped elements matching ‘L’ network is:
A. they are not equally efficient at higher frequencies as they are at lower frequencies.
B. size of the network
C. they restrict flow of current
D. none of the mentioned
Answer: A
Clarification: Since we use lumped elements like inductors and capacitors as the components of the matching network, they behave differently at frequencies higher than 1GHz, because of the frequency dependent factor of inductive and capacitive reactance. This is one of the major limitations.

7. An ‘L’ network is required to match a load impedance of 40Ω to a transmission line of characteristic impedance 60Ω. The components of the L network are:
A. 28.28+j0 Ω
B. 28.28+j1 Ω
C. 50Ω
D. 48.9Ω
Answer: A
Clarification: Since both load impedance and characteristic impedance are resistive (real), the imaginary part of the matching network is 0. Real part of the matching network is given by the expression ±√(RL(Z0– RL))-XL. Substituting the values given, the matching network impedance is 28.28Ω.

8. The imaginary part of the matching network is given by the relation:
A. ±(√(Z0– RL)/RL)Z00
B. ±(√(Z0– RL)/RL)
C. ±(√(Z0– RL)/ Z0
D. None of the mentioned
Answer: A
Clarification: By theoretical analysis, the expressions for real and imaginary parts of the impedance of the matching network are derived in terms of the load impedance and the characteristic impedance of the transmission line. This expression derived is ±(√(Z0– RL)/RL)Z0 .

9. Which of the following material is not used in the fabrication of resistors of thin films?
A. nichrome
B. tantalum nitride
C. doped semiconductor
D. pure silicon
Answer: D
Clarification: Certain physical properties are to be met in order to use a material to make thin film resistors. These properties are not found in pure silicon which is an intrinsic semiconductor.

10. Large values of inductance can be realized by:
A. loop of transmission line
B. spiral inductor
C. coils of wires
D. none of the mentioned
Answer: B
Clarification: Loop of transmission lines are used to make inductors to realize lower values of inductance. Coils of wires cannot be used to realize inductors at higher frequencies. Spiral conductors can be used to realize inductors of higher values at higher frequencies.

11. A short transmission line stub can be used to provide a shunt capacitance of:
A. 0-0.1µF
B. 0-0.1pF
C. 0-0.1nF
D. 1-10pF
Answer: B
Clarification: Since a transmission line consists of two wires, which can act plates of a capacitor, they can be used as a capacitor of very low values of the range 0-0.1pF.


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250+ TOP MCQs on Power Dividers( T Junction) and Answers

Microwave Engineering Interview Questions and Answers for Experienced people on “Power Dividers( T Junction)”.

1. A T junction power divider can be used only for division of power.
A. True
B. False
Answer: B
Clarification: A T junction power divider is a 3 port network that can be used either for power dividing or power combining. For power division, one of the ports is excited with the source and the other two ports are used to receive power. For power combining, 2 ports are excited with the source and output is taken at the third port.

2. The lossless T junction dividers can be can all be modeled as a junction of three transmission lines.
A. True
B. False
Answer: A
Clarification: A T junction consists of three ports; hence they can be modeled with 3 transmission lines. However, they cannot be both lossless and matched simultaneously.

3. For the realization of lossless T-junction power divider using transmission lines, the characteristic impedance of the transmission line has to be real.
A. True
B. False
Answer: A
Clarification: From various analyses, it is found that for a transmission line must have real characteristic impedance. If they are not real, they capacitive and inductive passive elements result in retaining some energy in the junction making them lossy.

4. The output power measured at the 2 ports of the T junction:
A. Is a constant
B. Variable
C. Is not real power
D. None of the mentioned
Answer: B
Clarification: The output line impedance determines the power delivered to the 2 output ports in a T junction coupler. Depending on the value of the power desired, the impedance can be changed and the corresponding power is obtained.

5. Hybrid couplers are also a type of directional couplers.
A. True
B. False
Answer: A
Clarification: Hybrid couplers are also a type of directional couplers that give a coupling factor of 3dB. A coupling factor of 3dB means that 70.7% of the total input power is received at the output port.

6. If the input power is divided in the ratio of 2:1 in a T- junction coupler and the characteristic impedance of the 2 output lines is 150Ω and 75Ω, then the impedance of the input line is:
A. 100Ω
B. 50Ω
C. 150Ω
D. None of the mentioned
Answer: B
Clarification: The input impedance of the T junction is the equivalent of the 2 output impedances in parallel. That is 150││75. Solving this, (150*75/(150+75)), the input impedance is 50 Ω.

7. A lossy T junction can be matched at all the three ports.
A. True
B. False
Answer: A
Clarification: If a T junction is constructed using resistors, the T junction becomes lossy, but it can be simultaneously matched at all the three ports.

8. The diagonal elements of the s matrix of a resistive T junction are:
A. 0
B. 1
C. 0.5
D. 1.5
Answer: A
Clarification: A resistive junction can be matched at all the three ports of the junction. Hence no power is reflected back. As a result, the diagonal elements are all 0 for a resistive T junction.

9. The power delivered to the input port of a resistive power divider is equal to the source voltage applied.
A. True
B. False
Answer: B
Clarification: Since power is applied to a resistive power divider, there is loss and hence not all the supply power is delivered to the input port of the power divider.

10. The power input at the port 1 of resistive T junction is equally divided among the 2 output ports of the T junction.
A. True
B. False
Answer: B
Clarification: The power division ratio of a resistive T junction depends on the resistance of the resistors used in forming those junctions. Depending on the resistors used, the power gets divided accordingly.


250+ TOP MCQs on GUNN Diodes and Answers

Microwave Engineering Multiple Choice Questions on “GUNN Diodes”.

1. Silicon and germanium are called ___________ semiconductors.
A. direct gap
B. indirect gap
C. band gap
D. indirect band gap
Answer: B
Clarification: 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.

2. 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
Answer: D
Clarification: 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.

3. In a GaAs n-type specimen, the current generated is constant irrespective of the electric filed applied to the specimen.
A. true
B. false
Answer: B
Clarification: 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.

4. 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
Answer: A
Clarification: 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.

5. 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
Answer: A
Clarification: 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.

6. 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
Answer: A
Clarification: 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.

7. The electrodes of a Gunn diode are made of:
A. molybdenum
B. GaAs
C. gold
D. copper
Answer: A
Clarification: 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.

8. 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
Answer: A
Clarification: 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.

9. The number of modes of operation for n type GaAs is:
A. two
B. three
C. four
D. five
Answer: C
Clarification: 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.

10. 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
Answer: A
Clarification: 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.

11. 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
Answer: B
Clarification: 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.

12. 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
Answer: A
Clarification: 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.

13. The frequency of oscillation in Gunn diode is given by:
A. vdom/ Leff
B. Leff/ Vdom
C. Leff/ WVdom
D. none of the mentioned
Answer: A
Clarification: 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.

14. 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
Answer: A
Clarification: 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.

15. 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
Answer: A
Clarification: 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.


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250+ TOP MCQs on Crystal Oscillators and Answers

Microwave Engineering Multiple Choice Questions on “Crystal Oscillators”.

1. One condition to be satisfied in an oscillator circuit so that stable oscillations are produced is:
A. positive feedback is to be achieved
B. negative feedback is to be achieved
C. 1800 phase shift is required between the transistor input and output.
D. none of the mentioned
Answer: C
Clarification: In an oscillator a total of 3600 of phase shift is to be achieved in the entire circuit to produce oscillations. The transistor used in the oscillator circuit must produce a phase shift of 1800 to achieve stable oscillations. Hence this condition has to be satisfied by the oscillator.

2. In an oscillator, the resonant feedback circuit must have must have a low Q in order to achieve stable oscillation.
A. true
B. false
Answer: B
Clarification: If the resonant feedback circuit has a high Q, so that there is random phase shift with frequency, the oscillator will have good frequency stability.

3. Quartz crystals are more efficient as a feedback network because:
A. less circuit complexity
B. cost effective
C. crystals operate at high voltage levels
D. LC circuits have unloaded Q of a few hundreds
Answer: D
Clarification: At frequencies below a few hundred MHz, where LC resonators seldom have unloaded Qs greater than a few hundred. Quartz crystals have unloaded Q of about 10000 and have a temperature drift of 0.001%/C0.

4. Quartz crystal and tourmaline used in oscillators work on the principle of:
A. photo electric effect
B. piezo electric effect
C. Raman effect
D. black body radiation
Answer: B
Clarification: Quartz crystals work on the principle of piezo electric effect. When electrical energy is applied to these crystals, they vibrate in a direction perpendicular to the application of energy producing oscillations.

5. A quartz crystals equivalent circuit is a series LCR circuit and has a series resonant frequency.
A. true
B. false
Answer: B
Clarification: A quartz crystal has an equivalent circuit such that a series LCR network is in parallel with a capacitor. A quartz crystal thus has both series and parallel resonant frequencies.

6. Quartz crystal is used in the _______region, where the operating point of the crystal is fixed.
A. resistive
B. inductive reactance
C. capacitive reactance
D. none of the mentioned
Answer: B
Clarification: Quartz crystal is always operated in the inductive reactance region so that the crystal is used in place of an inductor in a Colpitts or pierce oscillator.

7. In the plot of reactance v/s frequency of a crystal oscillator, the reactance between series resonant frequency and parallel resonant frequency is:
A. capacitive
B. inductive
C. both capacitive and inductive
D. none of the mentioned
Answer: B
Clarification: In the plot of reactance v/s frequency of a crystal oscillator, the reactance between series resonant frequency and parallel resonant frequency is inductive. In this region between the series and parallel and series resonant frequencies, the operating point of the crystal is fixed and hence can be used as part of other circuits.

8. In the equivalent circuit of a quartz crystal, LCR arm has an inductance of 4 mH and capacitor has a value of 4nF, then the series resonant frequency of the oscillator is:
A. 0.25 MHz
B. 2.5 MHz
C. 25 MHz
D. 5 MHz
Answer: A
Clarification: The series resonant frequency of a crystal oscillator is given by 1/√LC. Substituting the given values of L and C in the expression, the series resonant frequency is 0.25 MHz.

9. Parallel resonant frequency of quartz crystal is given by:
A. 1/ √(LCₒC/(Cₒ+C.)
B. 1/√LC
C. 1/√LCₒ
D. 1/ √(L(Cₒ+C. )
Answer: A
Clarification: Parallel resonant frequency of an oscillator is given by√(LCₒC/(Cₒ+C.). Here L and C are the inductance and capacitance in the LCR arm of the equivalent circuit of the crystal. Co is the capacitance existing in parallel to this LCR arm.

10. The equivalent circuit of a quartz crystal has LCR arm capacitance of 12nF and inductance of 3mH and parallel arm capacitance of 4nF. Parallel resonant frequency for the circuit is:
A. 3 MHz
B. 0.3 MHz
C. 6 MHz
D. 9 MHz
Answer: A
Clarification: The parallel resonant frequency of a crystal oscillator is given by 1/√(LCₒC/(Cₒ+C.). Substituting the given values in the equation, the parallel resonant frequency is found to be 3 MHz.


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250+ TOP MCQs on Antenna Radiation and Answers

Microwave Engineering Multiple Choice Questions on “Antenna Radiation”.

1. An antenna source that radiates energy uniformly in all the directions is called:
A. Isotropic source
B. Anisotropic source
C. Point source
D. None of the mentioned
Answer: A
Clarification: Isotropic source radiates energy in all the direction uniformly. For such a source, the radial component Sr of the pointing vector is independent of θ and φ. The three dimensional power pattern of n isotropic source is a sphere.

2. Antennas that radiate energy only in a specified are called anisotropic antennas.
A. True
B. False
Answer: A
Clarification: All physically realizable, simplest antennas also have directional properties. That is, they radiate energy in one direction than in any other direction. Such sources are called anisotropic point sources.

3. The expression for pointing vector of an isotropic point source at a distance ‘r’ from the source is given by:
A. P/ 4πR2
B. P/4π
C. P/ 4πR
D. P×4πR2
Answer: A
Clarification: The pointing field vector for an isotropic source is given by the expression P/ 4πR2.P is the total power radiated y the source. As the distance of the point from the source increases, the magnitude of pointing vector decreases.

4. A source has a cosine radiation-intensity pattern given by U=UM cos (θ). The directivity of this source is:
A. 2
B. 4
C. 6
D. 8
Answer: B
Clarification: To find the directivity of the given source, the power radiated by the given source is found out by the method of integration. Taking the ratio of the power radiated by the given source to the power radiated by an isotropic source gives the directivity. Following the above steps, the directivity of the given source is 4.

5. A source has a cosine power pattern that is bidirectional. Given that the directivity of a unidirectional source with cosine power pattern has a directivity of 4, then the directivity of the unidirectional source is:
A. 1
B. 2
C. 4
D. 8
Answer: B
Clarification: Given the directivity of unidirectional power pattern, the directivity of bidirectional power pattern is half of it. Hence the directivity of the source is 2.

6. A source has a radiation intensity pattern given by U=UM sin θ. The directivity of the source with this power pattern is:
A. 1
B. 1.27
C. 2.4
D. 3.4
Answer: B
Clarification: To find the directivity of the given source, the power radiated by the given source is found out by the method of integration. Taking the ratio of the power radiated by the given source to the power radiated by an isotropic source gives the directivity. Following the above steps, the directivity of the given source is 1.27.

7. A source has a sine squared radiation intensity power pattern. The directivity of the given source is:
A. 1.5
B. 3
C. 2.5
D. 3.5
Answer: A
Clarification: To find the directivity of the given source, the power radiated by the given source is found out by the method of integration. Taking the ratio of the power radiated by the given source to the power radiated by an isotropic source gives the directivity. Following the above steps, the directivity of the given source is 1.5.

8. A source with a unidirectional cosine squared radiation intensity pattern is given by UMcos2 (θ). The directivity of the given source is:
A. 6
B. 8
C. 2
D. 7
Answer: A
Clarification: To find the directivity of the given source, the power radiated by the given source is found out by the method of integration. Taking the ratio of the power radiated by the given source to the power radiated by anisotropic source gives the directivity. Following the above steps, the directivity of the given source is 6.

9. Considering distance as a parameter, two types of field zones can be defined around an antenna.
A. True
B. False
Answer: A
Clarification: Considering distance as a parameter, two types of field zones can be defined around an antennA. .The field near the antenna is called near field or Fresnel region and the other region is the far field that is also called as Fraunhofer region.

10. If the field strength at receiving antenna is 1 µV/m, and the effective aperture area is 0.4 m2 and the intrinsic impedance of the medium is 377 Ω, then the power received by the antenna is:
A. 1.06 pW
B. 1.06 fW
C. 2 µW
D. None of the mentioned
Answer: B
Clarification: The received power by the antenna is given by E2Ae/Zₒ. Substituting the known values in the above equation, the power received is 1.06×10-15 watts.


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