Multiple choice question for engineering
Set 1
1. High gain is not achievable at microwave frequencies using BJT amplifiers because:
a) device construction
b) complex architecture
c) ports are not matched at high frequencies
d) none of the mentioned
Answer
Answer: c [Reason:] At higher frequencies, if higher bandwidth is desired, a compromise on maximum achievable gain is made. But at these higher frequencies, the ports of the amplifier are not matched to 50 Ω.
2. To flatten the gain response of a transistor:
a) biasing current has to be increased
b) input signal level has to increased
c) increase the operational bandwidth
d) give negative feedback to the amplifier
Answer
Answer: d [Reason:] Negative feedback can be used to increase the gain response of the transistor, improve the input and output match, and increase the stability of the device.
3. In conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. But this drawback can be overcome by using:
a) balanced amplifiers
b) distributed amplifiers
c) differential amplifiers
d) none of the mentioned
Answer
Answer: a [Reason:] In conventional amplifiers, a flat gain response is achieved at the cost of reduced gain. But this drawback can be overcome by using balanced amplifiers. This is overcome by using two 900 couplers to cancel input and output reflections from two identical amplifiers.
4. Bandwidth of balanced amplifier can be an octave or more, but is limited by the bandwidth of the coupler.
a) true
b) false
Answer
Answer: a [Reason:] In order to achieve flat gain response, balanced amplifiers use couplers to minimize reflections. But this in turn reduces the bandwidth of the amplifier to the coupler bandwidth.
5. Coupler that is mostly used in balanced amplifiers to achieve the required performance is:
a) branch line coupler
b) wilkinson coupler
c) lange coupler
d) waveguide coupler
Answer
Answer: c [Reason:] Lange couplers are broadband couplers and are compact in size. Since the bandwidth of a balanced amplifiers depends on the bandwidth of the coupler used. Lange coupler is thus preferred over couplers.
6. Distributed amplifiers offer very high _________
a) gain
b) bandwidth
c) attenuation
d) none of the mentioned
Answer
Answer: b [Reason:] Distributed amplifiers offer very high bandwidth of about 10 decade. But higher gain cannot be achieved using distributed amplifiers and matching at the ports is very important to achieve higher bandwidth.
7. In distributed amplifiers, all the FET stages in the amplifier are connected in series to one another.
a) true
b) false
Answer
Answer: b [Reason:] In distributed amplifiers, cascade of N identical FETs have their gates connected to a transmission line having a characteristic impedance of Zg with a spacing of lg while the drains are connected to a transmission line of characteristic impedance Zd, with a spacing ld.
8. ____________ uses balanced input and output, meaning that there are 2 signal lines, with opposite polarity at each port.
a) differential amplifier
b) distributed amplifier
c) balanced amplifier
d) none of the mentioned
Answer
Answer: a [Reason:] Differential amplifier uses balanced input and outputs, meaning that there are 2 signal lines, with opposite polarity at each port. It has two input ports and one output port. The difference of the 2 input signals is amplified.
9. A major advantage of differential amplifiers is:
a) high gain
b) low input impedance
c) higher output voltage swing
d) none of the mentioned
Answer
Answer: c [Reason:] Differential amplifiers can provide higher voltage swings that are approximately double that obtained with single ended amplifier.
10. Along with a differential amplifier, 1800 hybrid is used both at the input and output.
a) true
b) false
Answer
Answer: a [Reason:] A differential amplifier can be constructed using two single-ended amplifiers and 1800 hybrids at the input and output to split and then recombine the signals.
Set 2
1. In the course of development of microwave circuits, two distinct types of microwave integrated circuits have been developed according to the application requirements.
a) true
b) false
Answer
Answer: a [Reason:] There are two distinct types of microwave integrated circuits fabricated. They are hybrid microwave integrated circuits and monolithic microwave integrated circuits. They differ in the method of fabrication in the layers of metallization done.
2. __________ is an important consideration for a hybrid integrated circuit.
a) material selection
b) processing units
c) design complexity
d) active sources
Answer
Answer: a [Reason:] Material selection is an important consideration for a hybrid integrated circuit. Characteristics such as electrical conductivity, dielectric constant, loss tangent, thermal transfer and manufacturing compatibility of the material to be used for hybrid microwave circuits are evaluated first.
3. To fabricate a low frequency circuit using the hybrid microwave IC methodology, the material with _______ is preferred.
a) high dielectric constant
b) low dielectric constant
c) high resistivity
d) low resistivity
Answer
Answer: a [Reason:] At low frequency applications, a high dielectric constant is desirable because it results in smaller circuit size. At higher frequencies, however the substrate thickness must be decreased to prevent radiation loss and other spurious effects.
4. The mask in a hybrid microwave circuit is made of:
a) rubylith
b) silicon
c) quartz
d) arsenic
Answer
Answer: a [Reason:] The mask in hybrid microwave integrated circuits is made of Rubylith, a soft mylar film usually at a magnified scale for high accuracy. Then an actual size mask is made on a thin sheet of glass or quartz.
5. The metalized substrate is coated with __________ covered with the mast and exposed to light source.
a) photoresist
b) GaAs
c) germanium liquid
d) none of the mentioned
Answer
Answer: a [Reason:] The metalized substrate is coated with photoresist, covered with the mast and exposed to light source. The substrate can be etched to remove the unwanted areas of the metal.
6. Commonly used software packages for CAD of hybrid microwave integrated circuits are:
a) CADENCE
b) ADS
c) DESIGNER
d) all of the mentioned
Answer
Answer: d [Reason:] Before any microwave circuit design is implemented on the hardware, it is economical to simulate the same designs in software and check for the expected theoretical results. A few such software that provide such an environment is CADENCE, ADS, DESIGNER to name a few.
7. In hybrid microwave integrated circuits, the various components of the circuit are etched in the substrate.
a) true
b) false
Answer
Answer: b [Reason:] In hybrid integrated circuit design, after all initial design steps are completed, the discrete components are soldered or wire bonded to the conductors This can be done manually or through automated computer-controlled pick and place machines.
8. Once the circuit components are designed and fabricated for certain specific values, they cannot be changed as per the requirement later.
a) true
b) false
Answer
Answer: b [Reason:] IN HIC, provision is made for variations in component values and other circuit tolerances by providing tuning or trimming stubs that can be manually trimmed for each circuit. This increase circuit yield but also increases the cost of manufacture.
9.________ is a micromachining technique where suspended structures are formed on silicon substrates.
a) MMIC
b) HIC
c) RF MEMS
d) none of the mentioned
Answer
Answer: c [Reason:] RF MEMS switch technology is a micro machining technique where suspended structures are formed in silicon substrates. These can be used in microwave resonators, antennas and switches.
10. Depending on the single path (capacitive or direct contact) and the attenuation mechanism MEMS switch can be used for various configurations for various devices.
a) true
b) false
Answer
Answer: a [Reason:] A MEMS switch can be used in several different configurations depending on the single path, actuation mechanism, pull-back mechanism and the type of structure. One such example is switching the capacitance of a single path between high and low values by moving a flexible conductive membrane through the application of DC controlled voltage.
Set 3
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
Answer: a [Reason:] 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
Answer: b [Reason:] 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
Answer: a [Reason:] 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
Answer: a [Reason:] 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
Answer: a [Reason:] 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
Answer: b [Reason:] 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
Answer: d [Reason:] 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
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
Answer: c [Reason:] 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
Answer: c [Reason:] 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.
Set 4
1. In TE mode of a circular waveguide, EZ=0. The wave equation is:
a) ∇2HZ+k2HZ=0
b) ∇2HZ-k2HZ=0
c) ∇2HZ-HZ=0
d) ∇2HZ+HZ=0
Answer
Answer: a [Reason:] In TE mode, EZ=0. Hence, when we substitute it in the wave equation, we get ∇2HZ+k2HZ=0.
2. Bessel’s differential equation for a circular waveguide is:
a) ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0
b) n2(d2R/ dρ2) + n(dR/dρ) + (ρ2kC2– n2) R=0
c) d2R/ dρ2 + dR/dρ + (ρ2kC2– n2) R=0
d) None of the mentioned
Answer
Answer: a [Reason:] After solving the wave equation ∇2HZ+k2HZ=0 in TE mode by making suitable assumptions and making appropriate substitutions, the final equation obtained is ρ2(d2R/ dρ2) + ρ(dR/dρ) + (ρ2kC2– n2) R=0.
3. The lowest mode of TE mode propagation in a circular waveguide is:
a) TE10 mode
b) TE00 mode
c) TE01 mode
d) TE11 mode
Answer
Answer: c [Reason:] A circular waveguide can support various modes of propagation. Among these, the lowest mode of propagation supported by the waveguide is TE10 mode of propagation.
4. What is the cutoff frequency for TE₁₁ mode in a circular waveguide of radius 2 cm with P’₁₁= 1.841?
a) 5.5 GHz
b) 4.3 GHz
c) 7.7 GHz
d) 8.1 GHz
Answer
Answer: b [Reason:] The cutoff frequency for TE11 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.3 GHz.
5. In a circular waveguide, if the propagation is in TE21 mode with P21=3.054, with a diameter of 60 mm, then the cutoff frequency for the mode is:
a) 5.6 GHz
b) 6.4 GHz
c) 3.5 GHz
d) 4.8 GHz
Answer
Answer: d [Reason:] The cutoff frequency for TE21 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 4.8 GHz.
6. For a circular waveguide in TM11 mode of propagation with inner radius of 30mm, and the phase constant being equal to 0.3, then the wave impedance is equal to:
a) 0.16 Ω
b) 0.15 Ω
c) 0.5 Ω
d) 0.4 Ω
Answer
Answer: a [Reason:] For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.16 Ω.
7. For TM mode. The wave equation in cylindrical co ordinates is:
a) (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0
b) ∂2E2/∂ρ2 + 1/ρ ( ∂E/∂ρ)=0
c) ∂2E2/∂ρ2 + 1/ρ2 (∂2E2/∂∅2 ) = 0
d) None of the mentioned
Answer
Answer: a [Reason:] The wave propagation in a cylindrical waveguide in TM mode of propagation is governed by the equation (∂2/∂ρ2+1/ρ ∂/∂ρ + 1/ρ2 (∂2/∂∅2 + kc2) =0. This is a second order differential equation.
8. In TM mode, what is the first propagating mode?
a) TM01 mode
b) TM11 mode
c) TM12 mode
d) TM10 mode
Answer
Answer: a [Reason:] TM mode in a circular waveguide supports various modes of propagation. Among these modes of propagation, the first or the lowest mode of propagation is TM01 mode.
9. For TM01 mode of propagation in a circular waveguide with P01=2.405, with the inner diameter of the circular waveguide being equal to 25 mm. What is the cut off frequency for this mode of propagation?
a) 2.8 GHz
b) 6 GHz
c) 3.06 GHz
d) 4 GHz.
Answer
Answer: c [Reason:] The cutoff frequency for TM01 mode of propagation in a circular waveguide is given by Pnm/ 2πa√μϵ. Substituting the given values in the above expression, the cutoff frequency is 3.06 GHz.
10. If β is 0.3 for a circular wave guide operating in TM12 mode with P21=5.315, with the radius of the circular waveguide being equal to 25 mm, then the intrinsic impedance of the wave is:
a) 0.55 Ω
b) 0.4 Ω
c) 0.3 Ω
d) 1.2 Ω
Answer
Answer: a [Reason:] For a given mode of propagation in a circular waveguide, wave impedance is given by the expression k/β, where is the intrinsic impedance of the air k is the wave number, β is the phase constant. Determining the wave number and substituting in the given equation, the wave impedance is 0.55 Ω.
11. The cutoff frequencies of the first two propagating modes of a Teflon on a filled circular waveguide with a=0.5 with ∈r=2.08 is:
a) 12.19 GHz, 15.92 GHz
b) 10 GHz, 12 GHz
c) 12 GHz, 15 GHz
d) 15 GHz, 12 GHz
Answer
Answer: a [Reason:] The cutoff frequencies are given by the expression p*C/2πa√∈. Substituting the given values in the above expression, the cutoff frequencies are 12.19 GHz, 15.92 GHz.
Set 5
1. What are the modes of propagation that a co axial line supports?
a) TM, TE mode
b) TM, TE, TEM mode
c) TM, TEM mode
d) TE, TEM mode
Answer
Answer: b [Reason:] Certain propagating media support specific modes of propagation. Coaxial cables support all the three modes of propagation. They are TM, TE, and TEM modes.
2. The dominant waveguide mode of a co axial line is:
a) TE11 mode
b) TE01 mode
c) TM01 mode
d) TEM mode
Answer
Answer: a [Reason:] Co-axial cable many modes of propagation. Among those supported modes of propagation, the dominant mode, the mode with lowest propagating wavelength is TE11 mode.
3. In a co axial line with inner and outer diameters of 0.0645 and 0.0215 inches and a Teflon di electric with ∈r=2.2. The highest usable frequency before the TE11 waveguide mode starts to propagate is:
a) 16.8 GHz
b) 117.7 GHz
c) 15.3 GHz
d) 8.4 GHz
Answer
Answer: a [Reason:] Before the wave propagation starts, the unstable frequency is given by the expression Ck/2π√∈. Here, C is the velocity of light. Substituting the given values in the above expression, the frequency is 16.8 GHz.
4. If a wave guide has he inner and outer conductor diameters of 0.0645 and 0.0215 inches respectively for a co axial lines then the cut off wave number is:
a) 298
b) 300
c) 285
d) 123
Answer
Answer: a [Reason:] The cutoff wave number for wave propagation is given by 2/ (a + b). a, b are the inner and outer diameter respectively. Substituting in the above expression, cut off wave number is 298.
5. The commercially used co axial cable and connectors used has a characteristic impedance is:
a) 50Ω
b) 100Ω
c) 33.34Ω
d) 66.6Ω
Answer
Answer: a [Reason:] All commercial manufacturer s of coaxial cables and connectors have set a standard for all the manufactured products. The standard value is 50 Ω.
6. In television systems the characteristic impedance of the cables used is:
a) 75Ω
b) 150Ω
c) 100Ω
d) 50Ω
Answer
Answer: a [Reason:] All cable manufacturers of the television system follow a set of standards. As per these set standards, the characteristic impedance of the line is 75 Ω.
7. SWR standing wave ratio has to be ________for co axial connector.
a) Low
b) High
c) Infinite
d) Cannot be calculated
Answer
Answer: a [Reason:] Higher the value of the standing wave ratio more is the reflection which implies mismatch. Hence, standing wave ratio has to be low.
8. What are the connectors used in pairs called?
a) Jack and plug
b) Male and female connectors
c) Both the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] There are certain special connectors that can be used to connect two devices operating devices. These special connectors are to be used in pairs and are with both the names.
9. The frequency range for N type co axial connector is:
a) 8-12 GHz
b) 11-18 GHz
c) 14-20 GHz
d) 2-8 GHz
Answer
Answer: b [Reason:] N type connectors are coaxial connectors used at microwave frequency range. These type of connectors can be used in the frequency range of 11-18 Hz.
10. The frequency range of SMA co-axial connector used most commonly is:
a) 18-25 GHz
b) 25-50 GHz
c) 11-18 GHz
d) 8-12 GHz
Answer
Answer: a [Reason:] SMA coaxial cable connectors are designed to operate at high frequencies. The frequency ranges from 18-25 GHz.
