All questions of Control Systems for Electrical Engineering (EE) Exam

The block diagram of a system is shown in fig.The closed loop transfer function of this system is

a)
b) 
c) 

d) 

Correct answer is 'D'. Can you explain this answer?

Nitya Ahuja answered
Consider the block diagram as SFG. There are two feedback loop -G1G2H1 and -G2G3H2 and one forward path G1G2 G3 . So (D) is correct option.

For the signal flow graph shown in fig. an equivalent graph is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Aim It Academy answered
While writing the transfer function of this signal flow graph,
e2= tae1 + tbe= (ta+ tb) e1 
Then, signal flow graph will lokk like this:

For the system shown in fig. transfer function C(s) R(s) is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Gate Gurus answered
Consider the block diagram as a SFG. Two forward path G1G2 and G3 and three loops -G1G2 H2, -G2H1, -G3 H2
There are no nontouching loop. So (B) is correct.

For the block diagram shown in fig. the numerator of transfer function is
 
  • a)
  • b)
  • c)
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Zoya Sharma answered
SFG

P1 = G2G5G6 , P2 = G3G5G6, P3 = G3G6 , P4 = G4G6
If any path is deleted, there would not be any loop.
Hence Δ1 = Δ2 = Δ3 = Δ4 = 1 

Direction: The following item consists of two statements, one labelled as ‘Statement (I)’ and the other as ‘Statement (II)’. Examine these two statements carefully and select the answers to these items using the code given below:
Statement I: For type-II or higher systems, lead compensator may be used.
Statement II: Lead compensator increases the margin of stability.
  • a)
    Both Statement I and Statement II are individually true and Statement II is the correct explanation of Statement I
  • b)
    Both Statement I and Statement II are individually true but Statement II is not the correct explanation of Statement I
  • c)
    Statement I is true but Statement II is false
  • d)
    Statement I is false but Statement II is true
Correct answer is option 'A'. Can you explain this answer?

In general, there are two situations in which compensation is required.
  • In the first case, the system is absolutely unstable, and the compensation is required to stabilize it as well as to achieve a specified performance.
  • In the second case, the system is stable, but the compensation is required to obtain the desired performance.
     
The systems which are of type-2 or higher are usually absolutely unstable. For type-2 or higher systems, only the lead compensator is required because only the lead compensator improves the margin of stability.
Both Statement I and Statement II are individually true and Statement II is the correct explanation of Statement I
Note: In type-1 and type-0 systems, stable operation is always possible if the gain is sufficiently reduced. In such cases, any of the three compensators, lead, lag, lag-lead may be used to obtain the desired performance.

The output of the feedback control system must be a function of:
  • a)
    Reference input
  • b)
    Reference output
  • c)
    Output and feedback signal
  • d)
    Input and feedback signal 
Correct answer is option 'D'. Can you explain this answer?

Athira Reddy answered
Explanation: In a feedback control system, the output is determined by a combination of the input signal and the feedback signal. The input signal is the desired output or reference, while the feedback signal is a portion of the actual output that is fed back into the system to compare with the input. This comparison helps the system to adjust its output to minimize the error between the input and the actual output. By considering both the input and feedback signals, the system can continuously adapt and achieve the desired output.

The steady-state error of a feedback control system with an acceleration input becomes finite in a
  • a)
    type-3 system
  • b)
    type-2 system
  • c)
    type-1 system
  • d)
    type-0 system
Correct answer is option 'B'. Can you explain this answer?

Raj Choudhary answered
Steady state error with an acceleration input having an amplitude of A is given by

where, 
Hence, if the type of the system = 2, then Ka = some non-zero value or finite value due to which we will get some finite vaiue of Ka.

For which of the following values of K, the feedback system shown in the below figure is stable?
  • a)
    K < 0
  • b)
    K > 0
  • c)
    0 < K < 54
  • d)
    0 < K < 70
Correct answer is option 'D'. Can you explain this answer?

Rhea Reddy answered
The characteristic equation is
1 + G (s) H (s) = 0
or, 

or, s3 + 10s2 + (21 + K )s + 13 K= 0

For stability, 13K > 0 or K > 0
Also, 

or, K < 70
Hence, 0 < K< 70 (For stability).

Consider the List I and List II
The correct match is
  • a)
    2 1 3 4
  • b)
    2 1 4 3
  • c)
    1 2 4 3
  • d)
    1 2 3 4
Correct answer is option 'B'. Can you explain this answer?

Om Pillai answered
P. P1 = ab, Δ = 1, L = 0 ,T = ab
Q1 P1 = a, P2 = 6 , Δ = 1, L = Δk = 0,T  = a+b
R. P1 = a, L1 = b, Δ = 1 - b, Δ1 =1,
S. P1 = a, L1 = ab, Δ = 1 - ab, Δ1 = 1, 

The negative feedback closed-loop system was subjected to 15V. The system has a forward gain of 2 and a feedback gain of 0.5. Determine the output voltage and the error voltage.
  • a)
    15V, 10V 
  • b)
    6V, 5V
  • c)
    15V, 7.5V 
  • d)
    5V, 10V
Correct answer is option 'C'. Can you explain this answer?

Starcoders answered
The negative feedback closed-loop system, subjected to 15V, has a forward gain of 2 and a feedback gain of 0.5.
  • The output voltage is calculated as the product of the forward gain and the effective input voltage.
  • The effective input voltage is the difference between the input voltage and the feedback voltage.
To find the output voltage:
  • Calculate the feedback voltage: Output Voltage multiplied by Feedback Gain.
  • Effective input = Input voltage - Feedback voltage.
  • Output voltage = Forward gain x Effective input voltage.
After solving, the output voltage is 15V and the error voltage is 7.5V.

In the phase lead compensation network, the phase of ______ leads the phase of ______.
  • a)
    input voltage, output voltage
  • b)
    input voltage, input voltage
  • c)
    output voltage, input voltage
  • d)
    output voltage, output voltage
Correct answer is option 'C'. Can you explain this answer?

Lead compensator:
Transfer function:
If it is in the form of  then a < 1
If it is in the form of then a > b
In the frequency domain, 

Phase angle, ∠G (jω) = tan−1ωαT − tan−1ωT
ϕ = tan-1 ωaT – tan-1 ωT
As a > 1 always (from the definition), ϕ is positive
Hence, it is clear that the phase of output voltage leads the phase of the input voltage.

If the constant 'k' is negative, then what would be its contribution to the phase plot:
  • a)
    90 degrees
  • b)
    45 degrees
  • c)
    180 degrees
  • d)
    0 degree
Correct answer is option 'C'. Can you explain this answer?

Polar plot:
  • The polar plot of a transfer function G(jω) is the plot of the magnitude of G(jω) versus the phase angle of G(jω) as ω is varied from 0 to positive infinity.
  • For all pole systems, type indicates the starting point of the polar plot and order indicates the ending point of the polar plot.

 
  • As seen from the above figure, when a zero is added the type decreases, and the end of the polar plot shifts by +90°. 
  • When a pole is added, the type of the system increases, and hence the end of the polar plot shifts by -90°. 
Calculation:
Let the transfer function be:

 
The total phase shift will be:
∠G(jω)H(jω) = -90° - tan-1(ωT) - 180°   ----(1)
From equation (1) we can say that constant K has no contribution to the phase plot.
Hence K contributes 180° to the phase plot.
So option (c) is the correct answer.

Consider a feedback system with gain margin of about 30. At what point does Nyquist plot crosses negative real axis?
  • a)
    -3
  • b)
    -0.3
  • c)
    -30
  • d)
    -0.03
Correct answer is option 'B'. Can you explain this answer?

Nyquist Plot and Gain Margin

Nyquist plot is a graphical representation of a system's frequency response. It is used to analyze the stability of a feedback system. The Nyquist plot is a plot of the frequency response of a system in the complex plane. The gain margin of a feedback system is the amount of gain that the system can handle before it becomes unstable. It is defined as the amount of gain that causes the Nyquist plot to cross the negative real axis.

Answer Explanation

Given that the gain margin of the feedback system is about 30, we can determine at what point the Nyquist plot crosses the negative real axis by using the Nyquist stability criterion. The Nyquist stability criterion states that the number of encirclements of the -1 point in the Nyquist plot is equal to the number of unstable poles of the closed-loop transfer function.

Since the gain margin is about 30, the Nyquist plot will cross the negative real axis at a point where the magnitude of the complex number is equal to 1/30. This is because the gain margin is the reciprocal of the point where the Nyquist plot crosses the negative real axis. Therefore, the Nyquist plot will cross the negative real axis at a frequency of -0.3, which corresponds to a magnitude of 1/30.

Thus, the correct answer is option B, -0.3.

Which of the following is the correct statement?
A minimum phase network is one whose transfer function has
  • a)
    zeros in the right hand s-plane and poles in the left hand s-plane.
  • b)
    zeros and poles in the left hand s-plane.
  • c)
    zeros in the left hand s-plane and poles in the right hand s-plane.
  • d)
    arbitrary distribution of zeros and poles in the S-plane.
Correct answer is option 'B'. Can you explain this answer?

Explanation:

Minimum Phase Network:

A minimum phase network is a type of linear time-invariant (LTI) system in which all the zeroes and poles of the transfer function lie in the left-half of the s-plane. The transfer function of a minimum phase network is a causal and stable function that can be factored into a product of two terms: a minimum phase term and a delay term. The minimum phase term has all its zeroes and poles in the left-half of the s-plane, while the delay term has a pole at the origin.

Transfer Function:

The transfer function of a minimum phase network is given by:

H(s) = e^(-Ds) * G(s)

where D is a positive constant, G(s) is the minimum phase transfer function, and e^(-Ds) is the delay term. The transfer function H(s) can be expressed as a product of the minimum phase term G(s) and the delay term e^(-Ds). The minimum phase term G(s) is a causal and stable function that has all its zeroes and poles in the left-half of the s-plane. The delay term e^(-Ds) is a non-causal and unstable function that has a pole at the origin.

Properties of Minimum Phase Network:

The following are some of the properties of a minimum phase network:

- All the zeroes and poles of the transfer function lie in the left-half of the s-plane.
- The step response of a minimum phase network is faster than that of a non-minimum phase network with the same magnitude response.
- The phase response of a minimum phase network is always less than or equal to the phase response of a non-minimum phase network with the same magnitude response.
- The impulse response of a minimum phase network decays faster than that of a non-minimum phase network with the same magnitude response.

Answer:

Option 'B' is the correct statement. A minimum phase network is one whose transfer function has zeros and poles in the left-hand s-plane.

Which of the following is NOT the disadvantage of lag compensator in a control system?
  • a)
    In lag compensator, the attenuation offered by it shifts the gain crossover frequency to a lower point, thereby decreasing the bandwidth.
  • b)
    The lag network offers a reduction in bandwidth, and this gives shorter rise time and settling time and so the transient response.
  • c)
    A lag compensator somewhat acts as a proportional plus integral controller, hence adversely affects the stability of the system.
  • d)
    Though the system response is longer due to decreased bandwidth, the response is quite slow
Correct answer is option 'B'. Can you explain this answer?

Zoya Sharma answered
Advantages of Lag Compensator:
  • A phase lag network offers high gain at low frequency. Thus, it performs the function of a low pass filter.
  • The introduction of this network increases the steady-state performance of the system.
  • The lag network offers a reduction in bandwidth and this provides longer rise time and settling time and so the transient response.
  • The angular contribution of the pole is more than that of the compensator zero because the pole dominates the zero in the lag compensator.
 
Advantages of Lead Compensator:
  • It improves the damping of the overall system.
  • The enhanced damping of the system supports less overshoot along with less rise time and settling time. Therefore, the transient response gets improved.
  • The addition of a lead network improves the phase margin.
  • A system with a lead network provides a quick response as it increases bandwidth thereby providing a faster response.
  • Lead networks do not disturb the steady-state error of the system.
  • It maximizes the velocity constant of the system.

The inverse polar plot of the open loop transfer function, will be re presented by
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Lavanya Menon answered
Given, G(s) = (1+sT)/(sT)
The inverse polar plot of G(jω) is the polar plot of 1/G(jω)

Thus, 

and

Hence, inverse polar plot will be as show below,

The transfer function C(s)/R(s) for the system described by the block diagram shown below is given by:
  • a)
  • b)
  • c)
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Rhea Reddy answered
On shifting the take-off point beyond block G, we have the reduced block diagram as shown below:

On further reducing the above block diagram, we get the block diagram as shown below.

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

For the system shown in the figure, Y(s)/X(s) = _________. (Answer in integer )
    Correct answer is between '0.95,1.05'. Can you explain this answer?

    The circuit is redrawn as shown:
    [X(s) – Y(s)] G(s) + X(s) = Y(s)
    Given G(s) = 2
    (X(s) – Y(s)) 2 + X(s) = Y(s)
    = 2 X(s) + X(s) = Y(s) + 2Y(s)
    = 3 X(s) = 3 Y(s)

    If a system has an open loop transfer function 1-s / 1+s, then the gain of the system at frequency of 1 rad/s will be
    • a)
      1
    • b)
      1/2
    • c)
      Zero
    • d)
      -1
    Correct answer is option 'D'. Can you explain this answer?

    Anuj Rane answered
    Open Loop Transfer Function
    The open loop transfer function of a system represents the relationship between the input and output of a system without any feedback. It is denoted by G(s) and is expressed as the ratio of the Laplace transform of the output to the Laplace transform of the input.

    G(s) = Y(s) / X(s)

    Where:
    G(s) - open loop transfer function
    Y(s) - Laplace transform of the output
    X(s) - Laplace transform of the input

    Given Transfer Function
    In this question, the given open loop transfer function is 1 - s / 1 + s.

    Frequency Response of a System
    To determine the gain of a system at a specific frequency, we substitute jω for s in the transfer function, where ω represents the angular frequency in radians per second.

    G(jω) = Y(jω) / X(jω)

    To find the gain at a specific frequency, we substitute ω = 1 rad/s in the transfer function and evaluate the expression.

    Gain at Frequency ω = 1 rad/s
    Substituting ω = 1 rad/s in the transfer function:

    G(j1) = Y(j1) / X(j1)

    Calculating the Gain
    G(j1) = (1 - j) / (1 + j)

    To calculate the gain, we need to find the modulus of this complex number.

    |G(j1)| = sqrt((1 - j)^2) / sqrt((1 + j)^2)

    Simplifying the expression:

    |G(j1)| = sqrt(1^2 + (-1)^2) / sqrt(1^2 + 1^2)

    |G(j1)| = sqrt(1 + 1) / sqrt(1 + 1)

    |G(j1)| = sqrt(2) / sqrt(2)

    |G(j1)| = 1

    Hence, the gain of the system at a frequency of 1 rad/s is 1.

    Conclusion
    The correct answer is option 'A' - 1. The gain of the system at a frequency of 1 rad/s is 1.

    What is the full form of PID?
    • a)
      Proportional Integral Derivative
    • b)
      Proportional Integral Device
    • c)
      Programmable Integral Device
    • d)
      Programmable Integral Derivative
    Correct answer is option 'A'. Can you explain this answer?

    Jaya Datta answered
    Understanding PID Control
    PID stands for Proportional Integral Derivative, which is a fundamental control mechanism widely used in industrial control systems. Let’s break down the components:
    1. Proportional (P)
    - The proportional term produces an output value that is proportional to the current error value.
    - The main function is to reduce the overall error by adjusting the control output relative to the error.
    - A higher proportional gain results in a larger change in the output for a given error.
    2. Integral (I)
    - The integral term is concerned with the accumulation of past errors.
    - It integrates the error over time, thus addressing any residual steady-state error that may remain after the proportional response.
    - This helps eliminate the offset that can occur with a purely proportional controller.
    3. Derivative (D)
    - The derivative term predicts future error based on its rate of change.
    - It acts as a damping factor, improving system stability by counteracting the rate of change of the error.
    - This helps to reduce overshoot and oscillations in the control response.
    Application of PID Controllers
    - PID controllers are utilized in various applications, including temperature control, speed control of motors, and pressure control.
    - They are favored for their simplicity and effectiveness in providing a stable control response.
    In summary, the full form of PID is Proportional Integral Derivative, and each component plays a crucial role in achieving precise control in dynamic systems. This method has become a standard in engineering due to its versatility and reliability.

    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?

    Zoya Sharma answered
    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

    The time required for the response curve to reach and stay within the specified 2-5% of final value is referred to as :
    • a)
      Peak time
    • b)
      Rise time
    • c)
      Settling time
    • d)
      Peak overshoot time
    Correct answer is option 'C'. Can you explain this answer?

    Understanding Settling Time
    Settling time is a critical parameter in control systems, particularly in the field of Electronics and Communication Engineering (ECE). It indicates how quickly a system can stabilize after a disturbance or a change in input.
    Definition of Settling Time
    - Settling time is defined as the time required for the system’s response curve to reach and remain within a specified percentage (usually 2-5%) of its final value after a step input is applied.
    Importance of Settling Time
    - Performance Indicator: Settling time is crucial for assessing the performance of control systems. A shorter settling time means the system responds quickly and efficiently to changes.
    - Stability: It indicates how well the system can stabilize after fluctuations or disturbances, ensuring reliability in applications.
    Comparison with Other Time Parameters
    - Peak Time: This is the time it takes for the response to reach its first peak. It does not account for how long it takes to settle.
    - Rise Time: This measures the time required for the response to rise from a specified lower percentage to a specified upper percentage of the final value.
    - Peak Overshoot Time: This refers to the amount by which the response exceeds the final value before settling down.
    Conclusion
    - Settling time is key to evaluating a system's performance and stability. It ensures that after an input change, the system not only responds but also stays within acceptable limits of the final value, providing a clear indication of how effectively the system performs in real-world applications. Understanding this concept is essential for designing efficient control systems in ECE.

    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?

    Yash Patel answered
    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.

    Match the following :-
    • a)
      1 - b, 2 ­- a, 3 ­- d, 4 - c
    • b)
      1 - ­a, 2­ - b, 3­ - c, 4 - ­d
    • c)
      1­ - c, 2 - ­d, 3 - ­b, 4 - ­a
    • d)
      1 - ­a, 2 - ­d, 3 - ­c, 4 - ­b
    Correct answer is option 'A'. Can you explain this answer?

    Sink node:
    • A local sink is a node of a directed graph with no exiting edges, also called a terminal.
    • It is the output node in the signal flow graph. It is a node, which has only incoming branches.
    Lag Compensator: 
    • Phase lag network offers high gain at low frequency.
    • Thus, it performs the function of a low pass filter.
    • The introduction of this network increases the steady-state performance of the system.
    Damping Ratio:
    • The damping ratio gives the level of damping in the control system related to critical damping.
    • The damping ratio is defined as the ratio of actual damping to the critical damping of the system.
    • It is the ratio of the damping coefficient of a differential equation of a system to the damping coefficient of critical damping.
    • ζ = actual damping / critical damping
    Cut-off rate: It is the slope of the log-magnitude curve near the cut-­off region of the Bode-plot.

    Consider the following statements regarding a control system:
    (a) Addition of pole to left half of s-plane reduce the relative stability
    (b) Addition of zero to left half of s-plane increase the damping factor
    (c) Integral controller reduces the steady state error
    (d) Derivate controller cannot be used in isolation
    Which of the above statements are true?
    • a)
      (a) & (c) only
    • b)
      (b) & (d) only
    • c)
      (a), (b), (d) only
    • d)
      All (a), (b), (c), (d)
    Correct answer is option 'D'. Can you explain this answer?

    (a) Addition of pole reduces stability
    Consider system =
    Adding Pole [say at origin]


     
    (b) Addition of zero increase ξ
    Consider system with Transfer function

    Now add one zero to left half say at -2

    (c) Integral controller adds one pole at origin
    As type of system increase steady state error reduce
    (d) Derivative controllers are not used Alone because with sudden changes in the system the derivative controller will compensate the output fast therefore in long term effects the isolated controller will produce huge steady state errors.

    The characteristic equation of a second order discrete-data system is given by:
    F(z) = z2+ z+ 0.25 = 0
    The above system is
    • a)
      stable    
    • b)
      marginally stable
    • c)
      unstable    
    • d)
      asymptotically stable
    Correct answer is option 'A'. Can you explain this answer?

    Nilesh Joshi answered
    For a second order discrete- data system given by:
    F(z) = a2z2 + a1z + a0 = 0
    to be stable, the necessary and sufficient conditions are:
    F(1) > 0
    F(-1) > 0 and |a0| < a2
    Here, F(z) = z2 + z + 0.25
    So,    = 0.25, a1 = 1, a2 = 1
    Thus, F(1) - 12+ 1 + 0.25 = 2.25 > 0
    F(-1) = 1 -1 + 0.25 = 0.25 > 0
    and la0l = 0.25 < a2 = 1 Since ail the conditions are satisfied, therefore given system is stable.

    The characteristic equation of a closed loop control system is given by:
    s+ 2s +10 + K(s2 + 6s + 10) = 0
    The angle of asymptotes for the root loci for K > 0 are given by
    • a)
      180°, 360°    
    • b)
      90°, 270°
    • c)
      90°, 180°    
    • d)
      none of these
    Correct answer is option 'D'. Can you explain this answer?

    Saumya Basak answered
    Given, s2 + 2s + 10 + K(s2 + 6s + 10) = 0
    or, 
    or, 1 + G(s)H(s) = 0
    Thus, G(s)H(s) = 
    Here, number of open loop poles, P = 2. Number of open loop zero, Z = 2
    ∴ P - Z = 0.
    Angle of asymptotes are given by:

    Sines P - Z= 0, therefore there are no angle of asymptotes for the root locus of the given system.

    Which one of the following options correctly describes the locations of the roots of the equation s4 + s2 + 1 = 0 on the complex plane?
    • a)
      Four left half plane (LHP) roots
    • b)
      One right half plane (RHP) root, one LHP root and two roots on the imaginary axis
    • c)
      Two RHP roots and two LHP roots
    • d)
      All four roots are on the imaginary axis
    Correct answer is option 'C'. Can you explain this answer?

    Shlok Sengupta answered
    The roots of the equation s^4 + s^2 + 1 = 0 on the complex plane can be described as follows:

    Explanation:
    To find the roots of the equation s^4 + s^2 + 1 = 0, we can rearrange it as follows:

    s^4 + s^2 + 1 = 0
    (s^2 + 1)^2 - s^2 = 0
    (s^2 + 1 + s)(s^2 + 1 - s) = 0

    This equation can be further simplified into two quadratic equations:

    s^2 + 1 + s = 0 (Equation 1)
    s^2 + 1 - s = 0 (Equation 2)

    Roots of Equation 1:
    To find the roots of Equation 1, we can use the quadratic formula:

    s = (-b ± √(b^2 - 4ac)) / (2a)

    For Equation 1, a = 1, b = 1, and c = 1. Substituting these values into the quadratic formula, we get:

    s = (-1 ± √(1 - 4(1)(1))) / (2(1))
    s = (-1 ± √(-3)) / 2

    Since the discriminant (√(-3)) is imaginary, the roots of Equation 1 will also be imaginary. Therefore, Equation 1 has two roots on the imaginary axis.

    Roots of Equation 2:
    Similarly, using the quadratic formula for Equation 2, we get:

    s = (1 ± √(1 - 4(1)(1))) / (2(1))
    s = (1 ± √(-3)) / 2

    Again, since the discriminant (√(-3)) is imaginary, the roots of Equation 2 will also be imaginary. Therefore, Equation 2 has two roots on the imaginary axis.

    Conclusion:
    In total, the equation s^4 + s^2 + 1 = 0 has four roots on the complex plane. Two of these roots are on the right half plane (RHP) and the other two roots are on the left half plane (LHP).

    Consider the characteristic equation of a control system given by s3 + (K + 0.5)s2 + 4Ks + 50 = 0. Find the value of the frequency if the system has sustained oscillations for a given K.
    • a)
      ω = 50 rad/sec
    • b)
      ω = 25 rad/sec
    • c)
      ω = 3.63 rad/sec
    • d)
      ω = 4.63 rad/sec
    Correct answer is option 'C'. Can you explain this answer?

    Adarsh Chauhan answered
    To find the value of the frequency for sustained oscillations, we need to find the roots of the characteristic equation. The characteristic equation is given by:

    s^3 + (0.5K)s^2 + 4Ks + 50 = 0

    To find the roots, we can use the Routh-Hurwitz stability criterion. The Routh-Hurwitz table for this characteristic equation is as follows:

    1 4K
    0.5K 50
    (-0.5K*50)/(4K) = -12.5K

    From the Routh-Hurwitz table, we can see that for sustained oscillations to occur, we must have at least one sign change in the first column of the table. In this case, since the first column has two positive values (1 and 0.5K), there are no sign changes.

    Therefore, the system does not have sustained oscillations for any value of K.

    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?

    Ravi Singh answered
    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.
    It is applied to the total RH plane by selecting a closed path with r = ∞

     
    Nyquist stability completely deals with the right half of s – plane.
    N(0, 0) = P – Z
    N(0, 0): Number of encirclements around critical point (- 1, 0)
    P: Open loop poles
    Z: Open-loop zeroes.
    Note:
    1) To get the Closed-loop stability we require 1 + GH plane but available is GH plane, hence the origin is shifted to “-1” to get the closed-loop stability.
    2) To become the system stable there should not be any closed-loop pole in the right of s – plane.
    3)  The closed-loop pole is the same as that of the zeroes of Characteristic Equation which must be zero in the right. i.e, Z = 0
    N = P  is the criteria.
    Calculation:
    From the given Nyquist plot there is one encirclement about ( -1, 0 ) in the Anti-clockwise direction.
    So, N = 1

     
    Now to satisfy the stability criteria N should be equal to P.
    N = P = 1
    So the number of open-loop poles in the Right-hand side of the system is 1.

    Introduction of feedback to a system
    • a)
      increases the stability
    • b)
      decreases the stability
    • c)
      has no effect on stability
    • d)
      none of the above
    Correct answer is option 'D'. Can you explain this answer?

    Pranjal Basu answered
    Introduction of feed back to a system will increase the stability if negative feedback is introduced while it will reduce the stability if positive feedback is introduced.

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