Category: Microwave Engineering Objective Questions

250+ TOP MCQs on Stability Circles and Answers

Microwave Engineering Multiple Choice Questions on “Stability Circles”.

1. For a transistor amplifier to be stable, either the input or the output impedance must have a real negative part.
A. True
B. False
Answer: A
Clarification: For a transistor amplifier to be stable, either the input or the output impedance must have a real negative part. This would imply that │Гin│>1 or │Гout│>1, because these reflection coefficients depend on the source and load matching network.

2. ____________ condition, if met then the transistor can be impedance matched for any load.
A. Conditional stability
B. Unconditional stability
C. Infinite gain
D. Infinite input impedance
Answer: B
Clarification: A network is said to be unconditionally stable if │Гin│<1 and │Гout│<1 for all passive source and load impedance. Transistors that are unconditionally stable can be easily matched.

3. A network is said to be conditionally stable if:
A. │Гin│<1, │Гout│<1.
B. │Гin│>1, │Гout│>1
C. │Гin│>1, │Гout│<1
D. │Гin│<1, │Гout│>1
Answer: A
Clarification: For conditional stability, the condition to be satisfied is │Гin│<1, │Гout│<1. But this condition will be valid only for a certain range of passive source and load Impedance. His condition is also called potentially unstable.

4. Stability condition of an amplifier is frequency independent and hence can be operated at any frequency.
A. True
B. False
Answer: A
Clarification: Stability condition of an amplifier is frequency dependent since the input and output matching networks generally depend on frequency. Hence it is possible for an amplifier to be stable at the designed frequency and unstable at other frequencies.

5. For a unilateral device condition for unconditional stability in terms of S parameters is:
A. │S11│<1, │S22│<1
B. │S11│>1, │S22│>1
C. │S11│>1, │S22│<1
D. │S11│<1, │S22│>1
Answer: A
Clarification: For a unilateral device, the condition for unconditional stability is │S11│<1, │S22│<1. S11 parameter signifies the amount of power reflected back to port 1, which is the input port of the transistor. If this S parameter is greater is than 1, more amount of power is reflected back implying the amplifier is unstable.

6. If │S11│>1 or │S22│>1, the amplifier cannot be unconditionally stable.
A. True
B. False
Answer: A
Clarification: If │S11│>1 or │S22│>1, the amplifier cannot be unconditionally stable because we can have a source or load impedance of Zₒ leading to Гs=0 or ГL=0, thus causing output and input reflection coefficients greater than 1.

7. For any passive source termination ГS, Unconditional stability implies that:
A. │Гout│<1
B. │Гout│>1
C. │Гin│<1
D. │Гin│>1
Answer: A
Clarification: Unconditional stability implies that │Гout│<1 for any passive source termination, Гs. The reflection coefficient for passive source impedance must lie within the unit circle of the smith chart, nd the other boundary of the circle is written as Гs=ejφ.

8. The condition for unconditional stability of a transistor as per the K-∆ test is │∆│> 1 and K<1.
A. True
B. False
Answer: B
Clarification: The condition for unconditional stability of a transistor is │∆│< 1 and K>1. Here, │∆│ and K are defined in terms of the s parameters of the transistor by defining the S matrix. To determine the unconditional stability of a transistor in K-∆ method, the S matrix of the transistor must be known.

9. If the S parameters of a transistor given are
S11=-0.811-j0.311
S12= 0.0306+j0.0048
S21=2.06+j3.717
S22=-0.230-j0.4517
Then ∆ for the given transistor is:
A. 0.336
B. 0.383
C. 0.456
D. None of the mentioned
Answer: A
Clarification: Given the S parameters of a transistor, the ∆ value of the transistor is given by │S11S22-S12S21│. Substituting the given values in the above equation, the ∆ of the transistor is 0.336.

10. By performing the K-∆ test for a given transistor the values of K and ∆ were found to be equal to 0.383 and 0.334 respectively. The transistor with these parameters has unconditional stability.
A. True
B. False
Answer: B
Clarification: The condition for unconditional stability of a transistor is │∆│< 1 and K>1. Here, │∆│ and K are defined in terms of the s parameters of the transistor by defining the S matrix. Here │∆│< 1 but the second condition is not satisfied. Hence they are not unconditionally stable.


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250+ TOP MCQs on Radar Systems and Answers

Microwave Engineering Multiple Choice Questions on “Radar Systems”.

1. In a RADAR system the transmitter of the radar is more sensitive than the receiver.
A. True
B. False
Answer: B
Clarification: The basic operation of RADAR is that the transmitter sends out a signal, which is partially reflected by the distant target, and then detected by a sensitive receiver. Because of the presence of noise in the received signal, the receiver has to be more sensitive.

2. For radar system, antennas with a large beam width are preferred over narrow beam antennas.
A. True
B. False
Answer: B
Clarification: If a narrow beam width antenna is used in radar, the target’s direction can be accurately given by the angular position of the antenna. Hence narrow beam antennas give more accurate position of the objects.

3. The radar in which both transmission and reception is done using the same antenna are called:
A. Monostatic radar
B. Bistatic radar
C. Monopole radar
D. Dipole radar
Answer: A
Clarification: Radar transmits electromagnetic waves and receives the waves that are reflected by objects. If a single antenna is used both for transmission and reception of the signals, they are called monostatic radar.

4. For applications like missile fire control, bistatic radars are used.
A. True
B. False
Answer: A
Clarification: In missile fire control, the target is illuminated with one antenna and the reflected wave is received from another antenna in the radar. In situations where continuous transmission and reception of signals is required, bistatic radars are used.

5. When a power Pt is transmitted by an antenna, amount of energy incident on the target is given by the expression:
A. Pt×G/4πR2
B. Pt/4πR2
C. Pt×4 πR2/G
D. None of the mentioned
Answer: A
Clarification: The amount of energy incident on the target is proportional to the energy radiated; gain of the antenna G, and R is the distance of the target from the radar system. As the distance from the radar system, the energy incident on the target reduces.

6. The term radar cross section defines the:
A. Scattering ability of the target
B. Power radiating ability of the radar
C. Amount of energy scattered by unwanted objects
D. Cross section of radar area through which energy is emitted
Answer: A
Clarification: Radar cross section is defined as the ratio of scattered power in a given direction to the power incident on it. The power incident is the energy radiated by the transmitting antenna of the radar.

7. A ________ determines the target range by measuring the round trip time of a pulsed microwave signal.
A. Pulse radar
B. Doppler radar
C. Cross section radar
D. None of the mentioned
Answer: A
Clarification: The working principle of pulse radar is that continuous pulses are transmitted and time is recorded until the pulse is received back by the radar. Based on this delay recorded, the range of target is estimated.

8. Construction of pulse radar is much simpler than a Doppler radar.
A. True
B. False
Answer: B
Clarification: In Doppler radar the power / signal is continuously radiated by the transmitting antenna. In pulse radar, pulses are transmitted to the target. Generation and transmission of pulses is more complex as compared to continuous signal.

9. In military applications the radar cross sections of vehicles is minimized.
A. True
B. False
Answer: A
Clarification: In military applications the radar cross sections of vehicles is minimized so that the military vehicles remain undetected. Lower the radar cross section, lower is the power scattered, and hence the object remains undetected.

10. Pulse radar operating at 10GHz frequency has an antenna with a gain of 28 dB and a transmitted power of 2kW. If it is desired to detect a target of cross section 12m2, and the minimum detectable signal is -90 dBm, the maximum range of the radar is:
A. 8114 m
B. 2348 m
C. 1256 m
D. 4563 m
Answer: A
Clarification: The maximum range of a radar system is given the expression, [PtG2σλ22/ (4π) Pmin] 0.25. Pt is the transmitted power, σ is the radar cross section, G is the antenna gain. Substituting the given values in the above equation, the maximum range of the radar is 8114 m.


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250+ TOP MCQs on Lossless Lines and Answers

Microwave Engineering Multiple Choice Questions on “Lossless Lines”.

1. The value of ‘α’ for a lossless line is:
A. 0
B. 1
C. Infinity
D. Data insufficient
Answer: A
Clarification: α-for a transmission line signifies the attenuation constant. For a lossless transmission line attenuation constant is zero and the propagation occurs without losses.

2. If propagation constant is 12:60°, then the value of phase constant and attenuation constant is:
A. α=6, β=10.39
B. α=61, β=78
C. α=12, β=20.6
D. none of the mentioned
Answer: A
Clarification: The given propagation constant is in polar form .converting from polar form to rectangular form and equating the real and imaginary parts, we get α=6 and β=10.39.

3. If a transmission line with inductive reactance of 41.97 Ω and capacitive reactance of 1132.5Ω is operated at 1 GHz , then its phase constant is:
A. 0.0305
B. 0.3
C. 30.3
D. 0.6
Answer: A
Clarification: From the given inductive reactance and capacitive reactance, L and C are calculated using XL =2πfL and Xc = 1/2πfC. β=ω√LC, substituting the calculated L and C, we get β=0.0305.

4. The expression for a phase velocity of a transmission line is:
A. √LC
B. 1/√LC
C. XL+Xc
D. XL/Xc
Answer: B
Clarification: The expression for phase velocity is derived from known basic transmission line equations and the derived equation comes out to be 1/√LC .

5. If the admittance and the impedance of a transmission line are 100 Ω and 50 Ω of a respectively, then value of phase constant β is:
A. 0
B. 20
C. 132
D. 50
Answer: A
Clarification: β=ω√LC. Since both the line impedance and line admittance are both real, there is no phase difference caused and hence substituting in the above equation, we get β=0.

6. For a lossless line, which of the following is true?
A. γ=jβ
B. γ=α
C. γ=α+jβ
D. γ=α*jβ
Answer: A
Clarification: For a lossless line, attenuation constant α is 0. Hence substituting α=0 in γ=α+jβ, we get γ= jβ.

7. Expression for phase constant β is:
A. √LC
B. ω √LC
C.1/ (ω √LC.
D. None of the mentioned
Answer: B
Clarification: From the equation of γ in terms of Z and Y(impedance and admittance of the transmission line respectively), expanding the equation and making certain approximations, β= ω √LC.

8. A microwave generator at 1.2 GHz supplies power to a microwave transmission line having the parameters R=0.8Ω/m, G=O.8millisiemen/m, L=0.01µH/m and C=0.4PF/m. Propagation constant of the transmission line is:
A. 0.0654 +j0.48
B. 0.064+j4.8
C. 6.4+j4.8
D. none of the mentioned
Answer: A
Clarification: Z=R+jωL and Y=G+jωC, hence finding out Z and Y from these equations, substituting in γ=√ZY, value of γ is found out to be 0.0654+j0.48.

9. In a certain microwave transmission line, the characteristic impedance was found to be 210 10°Ω and propagation constant 0.2 78°.What is the impedance Z of the line, if the frequency of operation is 1 GHz?
A. 0.035+j41.97
B. 0.35+j4.97
C. 35.6+j4.28
D. 9.254+j4.6
Answer: A
Clarification: Impedance Z of a transmission line is given by the product of propagation constant γ and characteristic Zₒ, Z= γZₒ , we get Z=0.035+j41.97.

10. For a transmission line, L=1.8mh/m C=0.01pF/m, then the phase constant of the line when operated at a frequency of 1 GHz is:
A. 4.2426
B. 2.2
C. 0.3
D. 1
Answer: A
Clarification: Formula to calculate the phase constant β is β=ω√LC.substituting the given values of L,C and f, the value of β is 4.2426.


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250+ TOP MCQs on Micro-Strip Lines and Answers

Microwave Engineering Multiple Choice Questions on “Micro-Strip Lines”.

1. Micro strip can be fabricated using:
A. Photo lithographic process
B. Electrochemical process
C. Mechanical methods
D. None of the mentioned
Answer: A
Clarification: 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
Answer: A
Clarification: 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
Answer: B
Clarification: 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
Answer: D
Clarification: 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
Answer: A
Clarification: 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
Answer: B
Clarification: 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
Answer: B
Clarification: 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
Answer: B
Clarification: 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
Answer: A
Clarification: 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
Answer: D
Clarification: 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
Answer: B
Clarification: 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
Answer: C
Clarification: 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.


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250+ TOP MCQs on Circular Waveguide Cavity Resonators and Answers

Microwave Engineering Multiple Choice Questions on “Circular Waveguide Cavity Resonators”.

1. 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
Answer: A
Clarification: 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.

2. The dominant mode in the cylindrical cavity resonator is TE101 mode.
A. true
B. false
Answer: B
Clarification: 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.

3. Circular cavities are used for microwave frequency meters.
A. true
B. false
Answer: A
Clarification: 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.

4. The mode of the circular cavity resonator used in frequency meters is:
A. TE011 mode
B. TE101 mode
C. TE111 mode
D. TM111 mode
Answer: A
Clarification: 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.

5. 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
Answer: A
Clarification: 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.

6. 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
Answer: B
Clarification: 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.

7. 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
Answer: D
Clarification: 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.

8. 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: D
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.

9. 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
Answer: C
Clarification: 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.

10. 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
Answer: C
Clarification: 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.


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250+ TOP MCQs on Noise Figure and Answers

Microwave Engineering Multiple Choice Questions on “Noise Figure”.

1. ___________ is defined as the ratio of desired signal power to undesired noise power.
A. Signal to noise ratio
B. Noise to signal ratio
C. Noise figure
D. Noise temperature

Answer: A
Clarification: SNR is defined as the ratio of desired signal power to undesired noise power, and so is dependent on the signal power. When noise and a desired signal are applied to the input of a noise less network, both noise and signal will be attenuated or amplified by the same factor, so that the signal to noise ratio will be unchanged.

2. __________ is defined as the ratio of input signal to noise ratio to the output signal to noise ratio.
A. Noise figure
B. Noise temperature
C. SNRo
D. None of the mentioned

Answer: A
Clarification: Noise figure is defined as the ratio of input signal to noise ratio to the output signal to noise ratio of a system or a receiver. SNRi is the signal to noise ratio measured at the input terminals of the device. SNR0 is the output signal to noise ratio measured at the output terminals of the device.

3. The equivalent noise temperature of a network given the noise figure of the network or system is:
A. T0(F-1)
B. T0(F+1)
C. T0(F)
D. T0/F

Answer: A
Clarification: The equivalent noise temperature of a network given the noise figure of the network or system is given by T0(F-1). In this expression, F is the noise figure of the system. T0 has the value 290 K. T0 is the standard temperature considered.

4. Noise figure can be defined for any microwave network irrespective of any other constraints.
A. True
B. False

Answer: B
Clarification: Noise figure is defined only for a matched input source and for a noise source equivalent to a matched load at a temperature T0= 290 K. noise figure and noise temperature are interchangeable noise properties.

5. Expression for noise of a two port network considering the noise due to transmission line and other lossy components is:
A. GkTB + GNadded
B. GkTB
C. GNadded
D. None of the mentioned

Answer: A
Clarification: Expression for noise of a two port network considering the noise due to transmission line and other lossy components is GkTB + GNadded. Here, G is the gain of the system. Nadded is the noise generated by the transmission line, as if it appeared at the input terminals of the line.

6. Noise equivalent temperature of a transmission line that adds noise to the noise of a device is:
A. T (L-1)
B. T (L+1)
C. T (L)
D. T/L

Answer: A
Clarification: Noise equivalent temperature of a transmission line that adds noise to the noise of a device is given by T (L-1). Here L is the loss factor of the line and T is the temperature at which the system is thermal equilibrium.

7. If the noise figures of the first stage of a two stage cascade network is 8 dB and the noise figure of the second stage is 7 dB and the gain of the first stage is 10, then the noise figure of the cascade is:
A. 8. 6 dB
B. 7.6 dB
C. 5.6 dB
D. 8.9 dB

Answer: A
Clarification: Noise figure of a two stage cascade network is given by F1+ (F2-1)/G1. Here F1, F2 are the noise figure of the first and the second stage respectively. G1 is the gain of the first stage. Substituting the given values in the above equation, noise figure of the cascade is 8.6 dB.

8. Noise equivalent temperature of a 2 stage cascade network is given by:
A. Te1 + Te2/ G1
B. Te1 + Te1
C. Te1 / Te1
D. None of the mentioned

Answer: A
Clarification: Noise equivalent temperature of a 2 stage cascade network is given by Te1 + Te1/ G1. Here, Te1 is the noise equivalent temperature of stage 1 and Te1 is the noise equivalent temperature of stage 2. G1 is the gain of the first stage of the amplifier.

9. When a network is matched to its external circuitry, the gain of the two port network is given by:
A. │S21│2
B. │S22│2
C. │S12│2
D. │S11│2

Answer: A
Clarification: The gain of a two port network is given by the product of SS21 of the network and reflection co-efficient at the source end. But when the two port network is matched to the external circuitry, reflection coefficient becomes zero and gain reduces to │S21│2.

10. For a Wilkinson power divider of insertion loss L and the coupler is matched to the external circuitry, and then the gain of the coupler in terms of insertion loss is:
A. 2L
B. 1/2L
C. L
D. 1/L

Answer: B
Clarification: To evaluate the noise figure of the coupler, third port is terminated with known impedance. Then the coupler becomes a two port device. Since the coupler is matched, ГS=0 and Гout=S22=0. So the available gain is │S21│2. This is equal to 1/2L from the available data.

11. Noise equivalent temperature of Wilkinson coupler having a gain of 1/2L is given as:
A. T (2L-1)
B. T (2L+1)
C. T (2L*1)
D. T / (2L-1)

Answer: A
Clarification: Noise equivalent temperature of the Wilkinson coupler is found using the relation
T (1-G21)/G21. Substituting for G21 in the above expression, equivalent noise temperature is T (2L-1).

12. Expression for over all noise figure of a mismatched amplifier is:
A. 1+ (F-1)/ (1 -│Г│2)
B. 1
C. 1+ (F-1)
D. (F-1)/ (1 -│Г│2)

Answer: A
Clarification: The overall noise figure of a mismatched amplifier is given by 1+ (F-1)/ (1 -│Г│2). Here F is the noise figure of the amplifier, when there is an impedance mismatch at the input of the amplifier; this impedance mismatch is given by Г.