Multiple choice question for engineering
Set 1
1. _________ is a non linear circuit that converts DC power to an AC waveform of desired frequency based on the oscillator design.
a) Attenuator
b) Amplifier
c) Oscillator
d) None of the mentioned
Answer
Answer: c [Reason:] Oscillator is a non linear circuit that converts DC power to an AC waveform. Most RC oscillators provide sinusoidal outputs, which minimizes undesired harmonics and noise sidebands.
2. The transfer function of an RF oscillator is given by:
a) A/ (1-AH (ω))
b) A/ (1+AH (ω))
c) A/ (-1+AH (ω))
d) 1/ (1-AH (ω))
Answer
Answer: a [Reason:] Transfer function of an RF oscillator is given by A/ (1-AH (ω)). Here, A is the gain of the transistor multiplier used. H(ω) is the function representing the feedback network. In an oscillator, positive feedback is used.
3. The criterion on which oscillations are produced in the oscillator circuit is called:
a) Shannon’s criteria
b) Barkhausen criteria
c) Colpitts criteria
d) None of the mentioned
Answer
Answer: b [Reason:] When the condition 1-AH (ω) =0 occurs, it is possible to achieve non zero output voltage for zero input voltage, thus forming an oscillator. This is called Barkhausen criteria.
4. The necessary condition for oscillation in a Colpitts oscillator is:
a) C2/C1=gm/Gi
b) C1/C2=gm/Gi
c) C2/C1= gm*Gi
d) None of the mentioned
Answer
Answer: a [Reason:] The condition for sustained oscillation in a Colpitts oscillator is C2/C1 = gm/Gi. Here C1 and C2 are the capacitance in the feedback network, gm is the transconductance of the transistor and Gi is the input admittance.
5. Colpitts oscillator operating at 50 MHz has an inductor in the feedback section of value 0.10µH. then the values of the capacitors in the feedback section is:
a) 100 pF, 100 pF
b) 100 pF, 50 pF
c) 70 pF, 130 pF
d) 80 pF, 60 pF
Answer
Answer: a [Reason:] The equivalent value of series combination of the capacitors is given by 1/ω2L. This gives the equivalent capacitance value of 200 pF. C1C2/ (C1+C2) =200 pF. C1 and C2 values can be chosen in several ways. One of the way is C1=C2=100 pF.
6. The necessary condition for oscillation in a Hartley oscillator is:
a) L1/L2 = gm/Gi
b) L1/L2 =Gi /gm
c) L2L/L1 = gm/Gi
d) None of the mentioned
Answer
Answer: a [Reason:] Necessary condition for oscillation in a Hartley oscillator is L1/L2 = gm/Gi. Here, L1 and L2 are the inductances in the feedback network and gm is the transconductance of the transistor and Gi is the input admittance.
7. An inductor is operating at frequency of 50 MHz. Its inductance is 0.1 µH, and then the series resistance associated with the inductor is: (Qo=100)
a) 0.31 Ω
b) 1.32 Ω
c) 1 Ω
d) 1.561 Ω
Answer
Answer: a [Reason:] Series resistance associated with an inductor is given by ωL/Qₒ. Substituting in this equation, the series of an inductor is given by 0.31.
8. Hartley oscillator has inductance values of 12 mH and 4 mH in the feedback section and a capacitor of 4 nF. Then the resonant frequency of the circuit is:
a) 19.89 kHz
b) 25 kHz
c) 45 kHz
d) 12 kHz
Answer
Answer: a [Reason:] Resonant frequency of Hartley oscillator is given by 1/ 2π√(C1 (L1 + L2)). Substituting the given values in the above equation, cut-off frequency is 19.89 kHz.
9. Colpitts oscillator in the feedback section has an inductance of 4 mH and capacitors of 12 nH and 4 nH. Then the resonant frequency of Colpitts oscillator is:
a) 50.4 kHz
b) 35.1 kHz
c) 45.9 kHz
d) None of the mentioned
Answer
Answer: c [Reason:] Resonant frequency of Colpitts oscillator is given by 1/2π√LCₒ, where C0 is the equivalent capacitance given by C1C2/ (C1+C2). Substituting and solving the equation, resonant frequency is 45.9 kHz.
10. For Colpitts oscillator, the capacitors C1 and C2 in the feedback network are 1 µF and 25 µF respectively. Then the β value of the transistor is:
a) 35
b) 000.76
c) 25
d) 0.0025
Answer
Answer: c [Reason:] β for a transistor is defined as the ratio of transconductance of the transistor to the input admittance, which is equal to the ratio of C2/C1. Substituting the given values, β of the transistor is 25.
Set 2
1. 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
Answer
Answer: c [Reason:] 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.
2. 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-].
Answer
Answer: a [Reason:] 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.
3. 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
Answer
Answer: a [Reason:] 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.
4. 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
Answer
Answer: b [Reason:] 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.
5. 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
Answer
Answer: a [Reason:] 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).
6. Scattering matrix for a reciprocal network is:
a) Symmetric
b) Unitary
c) Skew symmetric
d) Identity matrix
Answer
Answer: a [Reason:] 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.
7. 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
Answer
Answer: a [Reason:] 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.
8. Scattering matrix for a lossless matrix is:
a) Unitary
b) Symmetric
c) Identity matrix
d) Null matrix
Answer
Answer: a [Reason:] 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.
9. 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
Answer
Answer: c [Reason:] 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.
10. 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
Answer
Answer: a [Reason:] 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.
Set 3
1. In a series LCR circuit, at resonance point the energy stored in the inductor and capacitor in the form of magnetic and electric energies are equal.
a) True
b) False
Answer
Answer: b [Reason:] At resonant frequency of a series LCR circuit, reactance of the capacitor is equal to the reactance of the inductor. The energy stored in the capacitor in the form of electric energy and the energy stored in the inductor in the form of magnetic energy is both equal.
2. Quality factor-Q of a resonant circuit signifies:
a) Loss in the resonant circuit
b) Gain in the resonant circuit
c) Magnetic energy stored in the circuit
d) Electric energy stored in the circuit
Answer
Answer: a [Reason:] Quality factor of a resonant network is defined as the ratio of average energy stored to the energy loss/ second. Hence, lower loss implies a higher quality factor.
3. The total energy stored in a series RLC circuit is equal to the energy stored in the inductor.
a) True
b) False
Answer
Answer: b [Reason:] The total energy in an RLC circuit is the sum of the energy stored in the magnetic field of the inductor and the electric energy stored in the capacitor. Loss in the circuit occurs due to the resistive component.
4. Higher the quality factor of a series LCR circuit, greater is the operating bandwidth of the resonant circuit.
a) True
b) False
Answer
Answer: b [Reason:] Band width and quality factor of a series RLC circuit are both inversely related. Higher the quality factor, lower the operating bandwidth.
5. The plot of input impedance magnitude v/s frequency has a peak at the resonant frequency.
a) True
b) False
Answer
Answer: b [Reason:] At resonant frequency, the capacitive reactance is equal to the inductive reactance cancelling each other’s effect. Hence, there is a dip at the center frequency in the plot of input impedance magnitude v/s frequency.
6. For a parallel resonance circuit, the plot of input impedance magnitude v/s frequency has a dip at the resonant frequency.
a) True
b) False
Answer
Answer: b [Reason:] In parallel RLC circuit, the input impedance is highest at resonant frequency since the reactive components are in parallel. Hence, there is a peak at the resonant frequency in the plot of input impedance magnitude v/s frequency.
7. The relation between unloaded Q and external Q is:
a) External Q> unloaded Q
b) Unloaded Q> external Q
c) External Q = unloaded Q
d) None of the mentioned
Answer
Answer: b [Reason:] To compute unloaded Q only the resistance in the resonant circuit is considered. But to calculate external Q, the resistance and other load in the external load is also considered. Sine Q and R are inversely proportional, as R increases Q decreases. Since R is greater for external Q computation, unloaded Q> external Q.
8. Loaded Q and External Q are 2 different parameters.
a) True
b) False
Answer
Answer: a [Reason:] loaded Q and external Q are 2 different parameters. They are related by the expression
QL-1=Qe-1+ Q0-1, where QL is the loaded Q, Qe is external Q and Q0 is the unloaded Q.
9. The bandwidth of a series RLC circuit is 200 Hz. Then the quality factor of the circuit is:
a) 200
b) 100
c) 0.005
d) 0.5
Answer
Answer: c [Reason:] The relation between quality factor and bandwidth is given as bandwidth=Q-1. Substituting for bandwidth in this expression, the quality factor of the resonant circuit is 0.005.
10. If a parallel RLC circuit is excited with a source of 8v, 50 Hz and the circuit has an inductor of 1mH, capacitor of 1µF and a resistor of 50Ω, then the power loss that occurs in the circuit is:
a) 6.4mW
b) 3.2mW
c) 12.8mV
d) None of the mentioned
Answer
Answer: a [Reason:] The power loss in a parallel RLC circuit is 0.5│V│2/R. given the values of source voltage and resistance in the circuit, the power loss in the parallel RLC circuit is 6.4mW.
Set 4
1. 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
Answer
Answer: d [Reason:] 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.
2. Shunt stubs are preferred for:
a) Strip and microstrip lines
b) Coplanar waveguides
c) Circular waveguide
d) Circulators
Answer
Answer: a [Reason:] 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.
3. 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
Answer
Answer: a [Reason:] 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.
4. In shunt stub matching, the key parameter used for matching is:
a) Admittance of the line at a point
b) Admittance of the load
c) Impedance of the stub
d) Impedance of the load
Answer
Answer: a [Reason:] 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.
5. 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
d) Admittance of the load
Answer
Answer: a [Reason:] 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.
6. 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
Answer
Answer: b [Reason:] 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.
7. 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
Answer
Answer: b [Reason:] 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.
8. 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
Answer
Answer: c [Reason:] 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.
9. 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
Answer
Answer: b [Reason:] 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.
10. 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
Answer
Answer: b [Reason:] 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.
Set 5
1. Smith chart is based on the polar plot of:
a) Reactance
b) Voltage
c) Current
d) Voltage reflection co-efficient
Answer
Answer: d [Reason:] let the reflection co-efficient be expressed in terms of magnitude and direction as ┌=|┌|ejθ. Magnitude is plotted as radius from the center of the chart, and the angle is measured in counter clockwise direction from the right hand side. Hence, smith chart is based on the polar pot of voltage reflection co-efficient.
2. Any passively realizable reflection coefficient can be plotted as a unique point on the smith chart. This statement implies that:
a) Reflection co-efficient less than or equal to 1 can be plotted
b) Reflection co-efficient greater than or equal to 1 can be plotted
c) Transmission co-efficient has to be less than or equal to one for the point to be located
d) T=Г+1
Answer
Answer: a [Reason:] Reflection co-efficient is defined as the ratio of reflected voltage /current to the incident voltage or current. Hence reflection co-efficient can never be greater than 1. Hence, only reflection co-efficient less than or equal to 1 can be plotted.
3. Reflection coefficient of a transmission line in its polar form can be represented as:
a) ┌=|┌|ejθ
b) ┌=|┌|ejθ-1
c) ┌=|┌|ejθ+1
d) ┌=|┌|ejθ+α
Answer
Answer: a [Reason:] Reflection c co-efficient is defined as the ratio of reflected voltage /current to the incident voltage or current. It is a complex value consisting of both real and imaginary parts. Converting it to polar form, it takes the form of ┌=|┌|ejθ, Consisting of both magnitude and phase θ.
4. If the characteristic impedance of a ƛ/2 transmission line is 50 Ω and reflection coefficient 0.3, then its input impedance
a) 26.92 Ω
b) 30 Ω
c) 40 Ω
d) 34.87 Ω
Answer
Answer: a [Reason:] Given the characteristic impedance and reflection coefficient of a transmission line, input impedance is given by Zₒ (1+Гe-2jβL)/ (1- Гe-2jβL). Substituting the given values, the input impedance of the line is 26.92 Ω
5. If the normalized input impedance of a transmission line is 0.5 Ω, then he reflection coefficient of a ƛ/2 transmission line is
a) 0.3334
b) 0.5
c) 0.6667
d) 1
Answer
Answer: a [Reason:] Given the characteristic impedance and reflection coefficient of a transmission line, input impedance is given by Zₒ (1+Гe-2jβL)/ (1- Гe-2jβL). Substituting the given values in the above equation, reflection coefficient is 0.3334.
6. If the input impedance of a ƛ/2 transmission line is 100 Ω with a voltage reflection coefficient of 0.344, then the characteristic impedance of the transmission line is:
a) 200 Ω
b) 100 Ω
c) 50 Ω
d) None of the mentioned
Answer
Answer: a [Reason:] Given the characteristic impedance and reflection coefficient of a transmission line, input impedance is given by Zₒ (1+Гe-2jβL)/ (1- Гe-2jβL). Substituting the given values in the above equation, characteristic impedance of the transmission line is 200 Ω.
7. Normalized impedance of 0.3+j0.4 lies in the:
a) Upper half of the impedance smith chart
b) Lower half of the impedance smith chart
c) Horizontal line of the chart
d) None of the mentioned
Answer
Answer: a [Reason:] In the impedance smith chart, the upper part of the smith chart refers to positive reactance or inductive reactance. Hence, the given point lies in the upper half of the smith chart corresponding to the intersection of circles r=0.3 and r=0.4
8. Normalized impedance of 1-j is:
a) In the upper half of the impedance smith chart
b) In the Lower half of the impedance smith chart
c) On the outer most circle of the smith chart.
d) On the horizontal line of the smith chart
Answer
Answer: b [Reason:] In the impedance smith chart, the lower half of the smith chart corresponds to negative reactance or capacitive reactance. Hence the given point lies in the lower half of the smith chart.
9. If a transmission line of a characteristic impedance 100 Ω is terminated with a load impedance of 300+j200 Ω, then the normalized load impedance is:
a) 1+j
b) 1-j
c) 3+2j
d) 2-3j
Answer
Answer: c [Reason:] Normalized load impedance is obtained by dividing the load impedance with the characteristic impedance of the transmission line. Dividing 300+200j from 100, we get 3+2j.
10. If the normalized load impedance of a transmission line is 0.3-j0.4 with a characteristic impedance of 50 Ω, then the load impedance is:
a) 15-j20
b) 15+j20
c) 1-j
d) 0.3-0.4j
Answer
Answer: a [Reason:] Load impedance is the product of characteristic impedance and normalized load impedance. Hence taking the product of characteristic impedance and load impedance, we get 15-j20Ω.
11. To get an admittance chart from an impedance chart:
a) Smith chart has to be rotated by 90⁰
b) Smith chart has to be rotated by 180⁰
c) Admittance chart cannot be obtained from the impedance chart anyway.
d) None of the mentioned
Answer
Answer: b [Reason:] Impedance and admittance parameters, both are a reciprocal of one another. Hence one chart can be obtained from the other chart. By rotating the impedance smith chart by an angle of 180⁰, admittance chart is obtained.