All India Electrical Engineering (EE) Group

Phase lead occurs at: 
  • a)
     low frequency regions
  • b)
    high frequency regions
  • c)
    moderate frequencies
  • d)
    very low frequency regions
Correct answer is option 'B'. Can you explain this answer?

Gate Funda answered
Phase lead occurs in phase lead compensator and it occurs at the high-frequency region.
Lead compensator:
Transfer function:
If it is in the form of  then a < 1
If it is in the form of then a > b
Maximum phase lag frequency: ωm = 1√Ta
Maximum phase lag::

ϕm is positive
Pole zero plot:
The zero is nearer to the origin.
Filter: It is a high pass filter (HPF)
Effect on the system:
  • Rise time and settling time decreases and Bandwidth increases
  • The transient response becomes faster
  • The steady-state response is not affected
  • Improves the stability

Which of the following is not correct with respect to a phase-lead compensation network?
  • a)
    It increases system bandwidth
  • b)
    It increases gain at higher frequencies 
  • c)
    It is used when fast transient response is required
  • d)
    It is used when decrease rapidly near crossover frequency
Correct answer is option 'D'. Can you explain this answer?

EduRev GATE answered
Lead compensator:
Transfer function:
If it is in the form of  then a < 1
If it is in the form of then a > b
Maximum phase lag frequency:
ωm = 1√Ta
Maximum phase lag::

ϕm is positive
Pole zero plot:
The zero is nearer to the origin.
Filter: It is a high pass filter (HPF)
Effect on the system:
  • Rise time and settling time decreases and Bandwidth increases
  • The transient response becomes faster
  • The steady-state response is not affected
  • Improves the stability
  • The velocity constant is usually increased
  • Helps to increase the system error constant though to a limited extent
  • The slope of the magnitude curve is reduced at the gain crossover frequency, as a result, relative stability improves
  • The margin of stability of a system (phase margin) increased

For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.
    Correct answer is between '0.30,0.33'. Can you explain this answer?

    EduRev GATE answered
    Given circuit is a lag compensator and transfer function is given as

    On comparison we get, T = 1
    βT = 10 ⇒ β = 10
    The frequency at which maximum lead occurs is ωm = 1/T√β

    The compensator required to improve the steady state response of a system is
    • a)
      Lag 
    • b)
      Lead
    • c)
      Lag-lead
    • d)
      Zero
    Correct answer is option 'A'. Can you explain this answer?

    Lag compensator:
    Transfer function:
    If it is in the form of  then a < 1
    If it is in the form of then a > b
    Maximum phase lag frequency:
    ωm = 1√Ta
    Maximum phase lag::

    ϕm is negative
    Pole zero plot:
    The pole is nearer to the origin.
    Filter: It is a low pass filter (LPF)
    Effect on the system:
    • Rise time and settling time increases and Bandwidth decreases
    • The transient response becomes slower
    • The steady-state response is improved
    • Stability decreases

    The maximum phase shift that can be obtained by using a lead compensator with transfer function Gc(s) = equal to
    • a)
      15° 
    • b)
      30°
    • c)
      45°
    • d)
      60°
    Correct answer is option 'B'. Can you explain this answer?

    The standard T/F of the compensator is 

    Maximum phase lead

    Maximum phase lead frequency, 
    ωm = 1/T√a
    Calculation:
    The given transfer function is,
    By comparing both transfer functions,
    aT = 0.15
    T = 0.05
    a = 3
    Maximum phase lead

    = sin-1 (0.5)
    ϕm = 30° 

    Given a badly underdamped control system, the type of cascade compensator to be used to improve its damping is
    • a)
      phase-lag
    • b)
      phase-lead-lag
    • c)
      phase-lead
    • d)
      notch filter
    Correct answer is option 'C'. Can you explain this answer?

    Phase Lead Compensator:
    • A lead compensator provides a positive phase shift for increasing the value of frequencies from 0 to ∞.
    • It is also known as a differentiator circuit.
    • For a lead network, zero is nearer to the origin.
    • It is used to improve the transient response of the system.
    • It increases the damping of the system.
    Phase Lag Compensator:
    • A lead compensator provides a negative phase shift for increasing the value of frequencies from 0 to ∞.
    • It is also known as an integrator circuit.
    • For a lag network, pole is nearer to the origin.
    • It is used to improve the steady state response of the system.
    • It decreases the steady-state error of the system.

    Which of the following is true for the network shown below -
    • a)
      Lead compensator
    • b)
      Lag compensator
    • c)
      Lead-lag compensator
    • d)
      None of the above
    Correct answer is option 'A'. Can you explain this answer?

    In general, the lead and lag compensator is represented by the below transfer function
     
    If a > b then that is lag compensator because pole comes first.
    If a < b then that is the lead compensator since zero comes first.
    Analysis:
    Lead compensator:
    1) When sinusoidal input applied to this it produces sinusoidal output with the phase lead input.
    2) It speeds up the Transient response and increases the margin for stability.
    A circuit diagram is as shown:

    Response is:

    Lead constant α = 

    A system with impulse response is essentially a _______ compensator and used as a ________ filter.
    • a)
      Integral, Comb
    • b)
      Lead, high-pass
    • c)
      Lag, low-pass
    • d)
      Proportional, all pass
    Correct answer is option 'C'. Can you explain this answer?

    Lag compensator:
    Transfer function:
    If it is in the form of then a < 1
    If it is in the form of  then a > b
    Maximum phase lag frequency: ωm = 1/T√α
    Maximum phase lag: ϕm = sin−1(a−1/a+1)
    ϕm is negative
    Pole zero plot:
    The pole is nearer to the origin.
    Filter: It is a low pass filter (LPF)
    Effect on the system:
    • Rise time and settling time increases and Bandwidth decreases
    • The transient response becomes slower
    • The steady-state response is improved
    • Stability decreases

    Which one of the following gives the V - I characteristic of an ideal voltage source?
    • a)
    • b)
    • c)
    • d)
    Correct answer is option 'C'. Can you explain this answer?

    • In every practical voltage source, there is some electrical resistance inside it. This resistance is called the internal resistance of the source.
    • When the terminal of the source is open-circuited, there is no current flowing through it; hence there is no voltage drop inside the source but when the load is connected with the source, current starts flowing through the load as well as the source itself.
    • But in an ideal voltage source, this difference is considered as zero that means there would not be any voltage drop in it when current flows through it and this implies that the internal resistance of an ideal source must be zero.
    This can be concluded that voltage across the source remains constant for all values of load current.
    The V-I characteristics of an ideal voltage source are shown below.

    Ideal case with noise effect,
    Considering the effect of noise, Ideal Voltage Source delivers constant voltage irrespective of the value of current through it and V-I characteristics are given as:

    The ideal current source will have
    • a)
      high internal resistance
    • b)
      zero internal resistance
    • c)
      an infinite internal resistance
    • d)
      low internal resistance
    Correct answer is option 'C'. Can you explain this answer?

    Ideal Current Source: An ideal current source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.
    Practical Current Source: A practical current source is equivalent to an ideal current source in parallel with a high resistance or low conductance.
    Ideal and practical current sources are represented as shown the below figure.

    To neglect a voltage source, the terminal across the source are-
    • a)
      Open circuited
    • b)
      Short circuited
    • c)
      Replace by inductor
    • d)
      Replace by some resistance
    Correct answer is option 'B'. Can you explain this answer?

    To neglect a voltage source, the terminal across the sources are short-circuited. Because ideally, the internal resistance of the voltage source is zero.
    To neglect a current source, the terminal across the sources are open-circuited. Because ideally, the internal resistance of the current source is infinite.

    Identify the source that has the following symbol.
    • a)
      Current-controlled voltage source
    • b)
      Voltage-controlled voltage source
    • c)
      Voltage-controlled current source
    • d)
      Current-controlled current source
    Correct answer is option 'C'. Can you explain this answer?

    Gate Gurus answered
    Concept:
    Independent sources:
    An independent voltage source maintains a voltage (fixed or varying with time) which is not affected by any other quantity. Similarly, an independent current source maintains a current (fixed or time-varying) which is unaffected by any other quantity.
    Dependent sources:
    • Some voltage (current) sources have their voltage (current) values varying with some other variables.
    • They are called dependent (voltage or current) sources or controlled (voltage or current) sources.
    • It is represented in diamond symbol.
    • Since the control of the dependent source is achieved by a voltage or current of some other element in the circuit, and the source can be voltage or current. 
    • It follows that there are four possible types of dependent sources, namely:
      (i) Voltage-controlled voltage source (VCVS)
      (ii) Current-controlled voltage source (CCVS)
      (iii) Current-controlled current source (CCCS)
      (iv) Voltage-controlled current source (VCCS)
    ... more

    An ideal current source should have
    • a)
      Zero internal resistance
    • b)
      Infinite internal resistance
    • c)
      Large value of e.m.f.
    • d)
      e.m.f.
    Correct answer is option 'B'. Can you explain this answer?

    Gate Gurus answered
    Ideal voltage source: An ideal voltage source have zero internal resistance.
    Practical voltage source: A practical voltage source consists of an ideal voltage source (VS) in series with internal resistance (RS) as follows.
    An ideal voltage source and a practical voltage source can be represented as shown in the figure.
    Ideal current source: An ideal current source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.
    Practical current source: A practical current source is equivalent to an ideal current source in parallel with high resistance or low conductance.
    Ideal and practical current sources are represented as shown in the below figure.

    A non-ideal current source of 25 mA is supplying a resistive load of 5 kΩ. If the actual current flowing through the load is 20 mA, then the internal resistance of the source is:
    • a)
      1 kΩ
    • b)
      10 kΩ 
    • c)
      5 kΩ
    • d)
      20 kΩ
    Correct answer is option 'D'. Can you explain this answer?

    Concept:
    Ideal Current Source: An ideal current source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.
    Practical Current Source: A practical current source is equivalent to an ideal current source in parallel with high resistance or low conductance.
    Ideal and practical current sources are represented as shown in the below figure.

    Calculation:

     
    Resistive load RL = 5 × 103 Ω 
    Current through load IL= 20 × 10-3 A
    Internal resistance = r
    Voltage across load, VL = RL × IL
    = 5 × 103 × 20 × 10-3
    = 100 V
    The net current through internal resistance
    r = 25 mA - 20 mA = 5 mA
    Voltage across internal resistance = current × r
    100 = 5 × 10-3 × r
    r = 20 kΩ

    If an ideal voltage source and ideal current source are connected in series, the combination
    • a)
      Has the same properties as a current source alone
    • b)
      Has the same properties as a voltage source alone
    • c)
      Has the some properties as the source which has a higher value
    • d)
      Results in the branch being redundant
    Correct answer is option 'A'. Can you explain this answer?

    Gate Gurus answered
    Concept:
    Ideal voltage source: An ideal voltage source have zero internal resistance.
    Practical voltage source: A practical voltage source consists of an ideal voltage source (VS) in series with internal resistance (RS) as follows.
    An ideal voltage source and a practical voltage source can be represented as shown in the figure.
    ... more

    In which constant voltage system following operations are performed:
    1. Measure the system current
    2. Compare it with a reference current
    3. Computes and amplifies the error signal
    • a)
      Constant extinction angle δ control
    • b)
      Constant voltage control
    • c)
      Constant lignition angle β control
    • d)
      Constant current control
    Correct answer is option 'D'. Can you explain this answer?

    Gate Gurus answered
    Constant Current Control:
    • This method of current control regulates the current by adjusting the duty cycle to maintain a constant output current regardless of changes to the input voltage and output resistance.
    • A power supply operating in constant current mode will compare its reference against the scaled output current. This can be accomplished through the use of a hall effect sensor, a shunt with a differential amplifier, or any other current to voltage conversion method. 
    • The process of comparing the reference against the scaled output current generates an error signal which is then amplified.

    The open loop transfer function of a unity gain negative feedback system is given as

    The Nyquist contour in the ��-plane encloses the entire right half plane and a small neighbourhood around the origin in the left half plane, as shown in the figure below. The number of encirclements of the point (−1 + j0) by the Nyquist plot of G(s), corresponding to the Nyquist contour, is denoted as N. Then N equals to
    • a)
      0
    • b)
      1
    • c)
      2
    • d)
      3
    Correct answer is option 'B'. Can you explain this answer?

    Concept
    N = P - 2
    N = no. of encirclements of )-1, 0) critical point by the Nyquist plot.
    P = no. of right half of s-plane of G(s) H(s) as F(s)
    z = no. of lright half of s-plane of CLTF as zero of F(s)
    For stability z = 0
    N - P = 0
    N = P 
    for Nyquist stability criteria
    Calculation
    Open loop function:

    Close loop transfer function =

     
    no pole in right hand side
    z = 0, P ⇒ 1
    N = 1 - 0
    N = 1
    no. of oncirclements N = 1

    Consider a closed-loop control system with unity negative feedback and KG(s) in the forward path, where the gain K = 2. The complete Nyquist plot of the transfer function G(s) is shown in the figure. Note that the Nyquist contour has been chosen to have the clockwise sense. Assume G(s) has no poles on the closed right-half of the complex plane. The number of poles of the closed-loop transfer function in the closed right-half of the complex plane is ___________.
    • a)
      0
    • b)
      1
    • c)
      2
    • d)
      3
    Correct answer is option 'B'. Can you explain this answer?

    Concept:
    Nyquist stability criterion:
    N = P – Z
    N is the number of encirclements of (-1+j0) point by the Nyquist contour in an anticlockwise direction.
    P is the open-loop RHP poles
    Z is the closed-loop RHP poles
    Analysis:
    For K = 1,

    For K = 2, the plot will be

     
    N = No. of encirclement about (-1, 0) in anticlockwise.
    P = Total number of open loop poles, in R.H.S.
    Z = P - N
    N = -2, P = 0
    Z = 0 - (-2) = 2
    Z = 2
    Two poles in right side.

    The figure shows the Nyquist plot of the open-loop transfer function G(s)H(s) of a system. If G(s)H(s) has one right-hand pole, the closed-loop system is 
    • a)
      Always stable
    • b)
      Unstable with one closed - loop right hand pole
    • c)
      Unstable with two closed - loop right hand poles
    • d)
      Unstable with three closed - loop right hand poles
    Correct answer is option 'A'. Can you explain this answer?

    Concept:
    The stability from the Nyquist plot is given by:
    N = P - Z
    where N = No. of encirclement of the critical point -1+j0 in an anticlockwise direction
    P = No. of open-loop poles
    Z = No. of zeroes of the characteristic equation
    For a system to be stable, the value of Z = 0.
    Calculation:

     
    Given, P = 1
    From the figure, N =1
    N = P - Z
    1 = 1 - Z
    Z = 0
    Hence, the system is stable.

    If the Nyquist plot cuts the negative real axis at a distance of 0.8, then the gain margin of the system is
    • a)
      −0.8
    • b)
      2.25
    • c)
      0.8
    • d)
      1.25
    Correct answer is option 'D'. Can you explain this answer?

    Concept:
    The gain margin is defined as the amount of change in open-loop gain needed to make a closed-loop system unstable.

    where, |G| = Gain of the system
    ωpc = Phase crossover frequency
    The frequency at which the gain of the system is zero is known as the phase crossover frequency.
    Calculation:

    GM = 1/0.8
    GM = 1.25

    The frequency at which the Nyquist plot of a unity feedback system with the open loop transfer function crosses the negative real axis is
    • a)
      1 rad / s
    • b)
      √2 rad / s
    • c)
      √3 rad / s
    • d)
      5 rad / s
    Correct answer is option 'C'. Can you explain this answer?

    Concept:
    • Nyquist plot crosses the negative real axis at phase cross over frequency (ωpc)
    • The phase crossover frequency is the frequency at which the phase angle first reaches −180°
    • i.e., ∠(G(ωpc)) = -1800
    Calculation:
    Given G(s) =
    ∠(G(ωpc)) = 900 – 0.5 tan-1pc) – 4 tan-1pc) = -1800
    By solving we get ωpc = √3 rad/sec

    In Nyquist plot of a system on adding a pole at s = 0, then plot will -
    • a)
      remain unaltered
    • b)
      rotate clockwise by 90°
    • c)
      rotate anticlockwise by 90°
    • d)
      rotate by 180°
    Correct answer is option 'B'. Can you explain this answer?

    Gate Gurus answered
    Due to adding a pole at s = 0, the angle of shifting in the Nyquist plot of the system will be a shift of 90° clockwise.
    Due to adding a zero at s = 0, the angle of shifting in the Nyquist plot of the system will be a shift of 90° anti-clockwise.
    The shapes of the Nyquist plot for different transfer functions are given below:

    The number and direction of encirclements around the point −1+j0 in the complex plane by the Nyquist plot of 
    • a)
      zero.
    • b)
      one, anti-clockwise.
    • c)
      one, clockwise.
    • d)
      two, clockwise.
    Correct answer is option 'A'. Can you explain this answer?

    Concept:
    The Nyquist plot is equal to the polar plot & mirror Image of the polar plot with respect to the real axis, with opposite direction + semicircle of the infinite radius in a clockwise direction as many as the type of system.
    Calculation:
    Given:

    So,


    Polar plot of Given G(s) is

     
    Now,
    Nyquist plot of Given G(s) is:

    Hence, the number of Encirclements of (-1 + j0) = 0

    The most important technique used for stability and the transient response of the system is
    • a)
      Nyquist plot
    • b)
      Root locus
    • c)
      Bode plot
    • d)
      Routh Hurwitz criteria
    Correct answer is option 'B'. Can you explain this answer?

    Definition: 
    The root locus plots the poles of the closed-loop transfer function in the complex s-plane as K varies from 0 to ∞ 
    The purpose of Root locus is defined as:
    • To find the nature of the system and ‘K’ value for stability.
    • To find the relative stability.
    • If the root locus branches move towards the right, system is less stable and if they move towards the left then the system is more relative stable.
    • It is the best method to find the Relative stability and RH criteria is best to find absolute stability
    The relation between the closed-loop and open-loop poles
    D(s) = 0 gives open-loop poles and N(s) = 0 gives closed-loop poles
    Characteristic equation is D(s) + kN(s) = 0.
    Closed-loop poles are nothing but the sum of open-loop poles and zeroes.

    A single-input single-output feedback system has forward transfer function G(s) and feedback transfer function H(s). It is given that |G(s)H(s)| < 1. Which of the following is true about the stability of the system?
    • a)
      The system is always stable
    • b)
      The system is stable if all zeros of G(s)H(s) are in left half of the s-plane
    • c)
      The system is stable if all poles of G(s)H(s) are in left half of the s-plane
    • d)
      It is not possible to say whether or not the system is stable from the information given
    Correct answer is option 'C'. Can you explain this answer?

    Concept:
    D(s) = 1 + G(s)H(s)
    D(s) gives the roots of characteristic equation i.e. closed-loop poles.
    Nyquist stability criteria state that the number of unstable closed-loop poles is equal to the number of unstable open-loop poles plus the number of encirclements of the origin of the Nyquist plot of the complex function D(s).
    It can be slightly simplified if instead of plotting the function D(s) = 1 + G(s)H(s), we plot only the function G(s)H(s) around the point and count encirclement of the Nyquist plot of around the point (-1, j0).
    From the principal of argument theorem, the number of encirclements about (-1, j0) is
    N = P - Z
    Where
    Where P = Number of open-loop poles on the right half of s plane
    Z = Number of closed-loop poles on the right half of s plane
    Calculation:
    For the given system, we have |G(s)H(s)| < 1
    So, we may easily conclude that the Nyquist-plot intersect the negative real axis between 0 and -1 point, i.e. the Nyquist plot does not enclose the point (-1, 0) or in other words number of encirclements is zero.
    ⇒ N = 0
    For a stable closed loop system, we must have Z = 0.
    To get Z = 0, from the Nyquist criteria, P must be equal to zero.
    Therefore, the system is stable if all poles of G(s)H(s) are in left half of the s-plane.

    From the below given Nyquist plot, calculate the number of open-loop poles on the right-hand side of the s-plane for the closed-loop system to be stable.
    • a)
      1
    • b)
      2
    • c)
      0
    • d)
      -1
    Correct answer is option 'A'. Can you explain this answer?

    Principle arguments
    • It states that if there are “P” poles and “Z” zeroes for a closed, random selected path then the corresponding G(s)H(s) plane encircles the origin with P – Z times.
    • Encirclements in s – plane and GH – plane are shown below.

    • In GH plane Anti clockwise encirclements are taken as positive and clockwise encirclements are taken as negative.
    ... more
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