Control Systems MCQ Quiz - Objective Question with Answer for Control Systems - Download Free PDF
Last updated on Jun 20, 2025
Latest Control Systems MCQ Objective Questions
Control Systems Question 1:
The transfer function of the given signal flow graph is
Answer (Detailed Solution Below)
Control Systems Question 1 Detailed Solution
Concept:
A signal flow graph represents the relationships between the variables of a set of linear algebraic equations.
Manson's gain formula:
\(T = \mathop \sum \limits_{k = 1}^k \frac{{{P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)
- Where Pk is the forward path transmittance of kth in the path from a specified input to an output node. In arresting Pk no node should be encountered more than once.
- Δ is called the signal flow graph determinant.
- Δ = 1 – (sum of all individual loop transmittances) + (sum of loop transmittance products of all possible pairs of non-touching loops) – (sum of loop transmittance products of all possible triplets of non-touching loops) + (……) – (……)
- Δ k is the factor associated with the concerned path and involves all closed loop in the graph which are isolated from the forward path under consideration.
- The path factor Δk for the kth path is equal to the value of the grab determinant of its signal flow graph, which exists after erasing the Kth path from the graph.
Given:
The transfer function \(\frac{Y(s)}{X(s)} = \frac{1}{1-G}\)
Control Systems Question 2:
In an open loop transfer function the highest power of 's' in the denominator when written as a standard polynomial in 's' is
Answer (Detailed Solution Below)
Control Systems Question 2 Detailed Solution
The correct answer is 1
Explanation:
In an open-loop transfer function, when written as a standard polynomial in 's', the highest power of 's' in the denominator represents the Order of the system.
- Order of the system: This is defined as the highest power of 's' in the denominator polynomial of the transfer function, after cancelling any common factors in the numerator and denominator. It indicates the number of energy storage elements (like capacitors or inductors in electrical systems, or masses and springs in mechanical systems) in the system and is crucial for determining the system's dynamic behavior, such as stability and transient response.
- Type of the system: This refers to the number of poles at the origin () in the open-loop transfer function. It's related to the steady-state error characteristics of the system for various types of inputs.
- Number of differentiators/integrators: These relate to specific components, but the highest power of 's' in the denominator describes the overall system order, not just the number of differentiators or integrators, although integrators contribute to the order by adding poles at the origin.
Control Systems Question 3:
Which of the following statements is 'FALSE' for a second order system with step input?
Answer (Detailed Solution Below)
Control Systems Question 3 Detailed Solution
A second-order control system is characterized by a second-order differential equation. It typically consists of two poles and is widely used in engineering applications to model dynamic systems. When a step input is applied to such a system, its response is analyzed in terms of parameters like damping ratio (𝛿), natural frequency (𝜔n), overshoot, settling time, rise time, and oscillations.
Correct Option Analysis:
The correct option is:
Option 2: For critical damping i.e., 𝛿 = 1, there are oscillations with dying down amplitudes.
This statement is FALSE. In a critically damped system (𝛿 = 1), the system does not exhibit oscillations. Instead, it returns to equilibrium as quickly as possible without overshooting or oscillating. A critically damped system is designed to achieve the fastest response time to a step input without oscillations. Therefore, the assertion that there are "oscillations with dying down amplitudes" for 𝛿 = 1 is incorrect.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: Damping ratio 𝛿 = 0 will give sustained oscillations.
This statement is TRUE. When the damping ratio (𝛿) is zero, the system is undamped, and it will exhibit sustained oscillations. This is because there is no energy dissipation in the system, and the response oscillates indefinitely at the natural frequency (𝜔n).
Option 3: For an overdamped system i.e., 𝛿 > 1, there is no oscillation.
This statement is TRUE. In an overdamped system, the damping ratio (𝛿) is greater than 1. The system response is sluggish, and it approaches equilibrium without oscillations. Overdamping ensures that the system does not overshoot or oscillate, but it takes longer to settle compared to a critically damped system.
Option 4: Settling time is inversely related to the damping ratio.
This statement is TRUE. Settling time, which is the time required for the system to settle within a specific percentage of its final value, is inversely related to the damping ratio for underdamped systems (𝛿 < 1). As the damping ratio increases, the oscillations decrease, and the system settles faster. However, this relationship is not valid for overdamped systems (𝛿 > 1).
Control Systems Question 4:
In system modelling using transfer function, roots of the characteristic equation are
Answer (Detailed Solution Below)
Control Systems Question 4 Detailed Solution
Explanation:
In system modeling using a transfer function, the roots of the characteristic equation are the Poles of the transfer function.
- The characteristic equation of a system is typically obtained by setting the denominator polynomial of the closed-loop transfer function to zero.
- The poles of a transfer function are the values of 's' that make the denominator of the transfer function equal to zero (assuming no common factors with the numerator).
- The locations of the poles in the s-plane are critical for determining the stability and transient response of a system. If any pole lies in the right half of the s-plane, the system is unstable.
Control Systems Question 5:
In the Bode plot of an open loop transfer function with 8 poles and 5 zeros, slope of the Bode plot for very high frequencies
Answer (Detailed Solution Below)
Control Systems Question 5 Detailed Solution
Explanation:
Bode Plot Analysis for an Open Loop Transfer Function
Definition: A Bode plot is a graphical representation of the frequency response of a system. It consists of two plots: one showing the magnitude (in decibels) versus frequency, and the other showing phase versus frequency. The slope of the magnitude plot at high frequencies depends on the number of poles and zeros in the transfer function.
Problem Statement: The given open loop transfer function has 8 poles and 5 zeros. The question asks for the slope of the Bode plot at very high frequencies.
Solution:
The slope of the Bode plot at very high frequencies is determined by the difference between the number of poles (P) and zeros (Z) in the transfer function. The general formula for the slope is:
Slope (in dB/decade) = -20 × (P - Z)
Step-by-Step Calculation:
- Given data:
- Number of poles (P) = 8
- Number of zeros (Z) = 5
- Calculate the difference between the number of poles and zeros:
- P - Z = 8 - 5 = 3
- Apply the formula for the slope:
- Slope = -20 × (P - Z) = -20 × 3 = -60 dB/decade
Conclusion: The slope of the Bode plot at very high frequencies is -60 dB/decade. Hence, Option 1 is correct
Top Control Systems MCQ Objective Questions
Given the differential equation model of a physical system, determine the time constant of the system:
\(40 \frac{dx}{dt}+2x=f(t)\)
Answer (Detailed Solution Below)
Control Systems Question 6 Detailed Solution
Download Solution PDFConcept:
Time constant \(\tau = \frac{{ - 1}}{{{\rm{real\ part\ of\ Dominant\ pole}}}}\)
Calculation:
\(40\frac{{dx}}{{dt}} + 2x = f\left( t \right)\)
Taking Laplace transform, we get
40 s X(s) + 2X(s) = 12(s)
\(\frac{{X\left( s \right)}}{{F\left( s \right)}} = \frac{1}{{40s + 2}}\)
\( = \frac{1}{{40\left( {s + \frac{1}{{20}}} \right)}}\)
Pole will be at -1/20.
Time constant \( = \frac{1}{{pole}} = 20\)
Which of the following methods is the strongest tool to determine the stability and the transient response of the system?
Answer (Detailed Solution Below)
Control Systems Question 7 Detailed Solution
Download Solution PDF- The root locus is the strongest tool for determining stability and the transient response of the system as it gives the exact pole-zero location and also their effect on the response
- A Bode plot is a useful tool that shows the gain and phase response of a given LTI system for different frequencies
- The Nyquist plot in addition to providing absolute stability also gives information on the relative stability of stable systems and degree of instability of the unstable system
- Routh-Hurwitz criterion is used to find the range of the gain for stability and gives information regarding the location of poles
Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation:
\(\frac{{{d^2}y\left( t \right)}}{{d{t^2}}} + 4y\left( t \right) = 6r\left( t \right)\)
The poles of this system are atAnswer (Detailed Solution Below)
Control Systems Question 8 Detailed Solution
Download Solution PDFConcept:
A transfer function is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.
TF = L[output]/L[input]
\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)
For unit impulse input i.e. r(t) = δ(t)
⇒ R(s) = δ(s) = 1
Now transfer function = C(s)
Therefore, the transfer function is also known as the impulse response of the system.
Transfer function = L[IR]
IR = L-1 [TF]
Calculation:
Given the differential equation is,
\(\frac{{{d^2}y\left( t \right)}}{{d{t^2}}} + 4y\left( t \right) = 6r\left( t \right)\)
By applying the Laplace transform,
s2 Y(s) + 4 Y(s) = 6 R(s)
\( \Rightarrow \frac{{Y\left( s \right)}}{{R\left( s \right)}} = \frac{6}{{{s^2} + 4}}\)
Poles are the roots of the denominator in the transfer function.
⇒ s2 + 4 = 0
⇒ s = ±2jLag compensator is a _____
Answer (Detailed Solution Below)
Control Systems Question 9 Detailed Solution
Download Solution PDFLag compensator:
Transfer function:
If it is in the form of
\(\frac{{1 + aTs}}{{1 + Ts}}\), then a < 1
If it is in the form of
\(\frac{{s + a}}{{s + b}}\), then a > b
Pole zero plot:
The pole is nearer to the origin.
Filter: It is a low pass filter (LPF)
Maximum phase lag frequency:
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lag:
\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is negative
Lead compensator:
Transfer function:
If it is in the form of
\(\frac{{1 + aTs}}{{1 + Ts}}\), then a > 1
If it is in the form of
\(\frac{{s + a}}{{s + b}}\), then a < b
Pole zero plot:
The zero is nearer to the origin.
Filter: It is a high pass filter (HPF).
Maximum phase lead frequency:
\({\omega _m} = \frac{1}{{T\sqrt a }}\)
Maximum phase lead:
\({\phi _m} = {\sin ^{ - 1}}\left( {\frac{{a - 1}}{{a + 1}}} \right)\)
ϕm is positive
The steady-state error due to unit step input to a type-1 system is:
Answer (Detailed Solution Below)
Control Systems Question 10 Detailed Solution
Download Solution PDFConcept:
KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)
Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)
Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)
Steady-state error for different inputs is given by
Input |
Type -0 |
Type - 1 |
Type -2 |
Unit step |
\(\frac{1}{{1 + {K_p}}}\) |
0 |
0 |
Unit ramp |
∞ |
\(\frac{1}{{{K_v}}}\) |
0 |
Unit parabolic |
∞ |
∞ |
\(\frac{1}{{{K_a}}}\) |
From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input.
The open loop DC gain of a unity negative feedback system with closed-loop transfer function \(\frac{s+4}{s^2+7s+13}\) is
Answer (Detailed Solution Below)
Control Systems Question 11 Detailed Solution
Download Solution PDFConcept:
Closed-loop transfer function = \(\frac{G(s)}{1+G(s)H(s)}\)
For unity negative feedback system Open-loop transfer function (G(s) H(s)) can be found by subtracting the numerator term from the denominator term
Application:
Open-loop transfer Function
\(= \frac{{s + 4}}{{{s^2} + 7s + 13 - s - 4}} = \frac{{s + 4}}{{{s^2} + 6s + 9}}\)
For DC gain s = 0
∴ open-loop gain \(= \frac{4}{{9}} \)
An open loop system represented by the transfer function \(G\left( s \right) = \frac{{\left( {s - 1} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\) is
Answer (Detailed Solution Below)
Control Systems Question 12 Detailed Solution
Download Solution PDFConcept:
Minimum phase system: It is a system in which poles and zeros will not lie on the right side of the s-plane. In particular, zeros will not lie on the right side of the s-plane.
For a minimum phase system,
\(\mathop {\lim }\limits_{\omega \to \infty } \angle G\left( s \right)H\left( s \right) = \left( {P - Z} \right)\left( { - 90^\circ } \right)\)
Where P & Z are finite no. of poles and zeros of G(s)H(s)
Non-minimum phase system: It is a system in which some of the poles and zeros may lie on the right side of the s-plane. In particular, zeros lie on the right side of the s-plane.
Stable system: A system is said to be stable if all the poles lie on the left side of the s-plane.
Application:
\(G\left( s \right) = \frac{{\left( {s - 1} \right)}}{{\left( {s + 2} \right)\left( {s + 3} \right)}}\)
As one zero lies in the right side of the s-plane, it is a non-minimum phase transfer function.
As there no poles on the right side of the s-plane, it is a stable system.
Slow response of an over-damped system can be made faster with the help of ______ controller.
Answer (Detailed Solution Below)
Control Systems Question 13 Detailed Solution
Download Solution PDFExplanation:
The controller is a device that is used to alter or maintain the transient state & steady-state region performance parameter as per our requirement.
Proportional Controller-
The standard Proportional Controller as shown:
In space-form -
\({G_C}\left( s \right) = \frac{{U\left( s \right)}}{{E\left( s \right)}} = \frac{{{K_p}}}{s(s+1)}\)
In time-domain form -
p(t) = Kp e(t) + po
Where,
po = controller output with zero error
Kp = proportional gain constant.
Some effects of the proportional controller are as follows:
- The P-controller can stabilize a first-order system, can give a near-zero error, and improves the settling time by increasing the bandwidth.
- It also helps in reducing the steady-state error which makes the system more stable.
- The slow response of an over-damped system can be made faster with the help of the proportional controller. Hence option (2) is the correct answer.
Important Points
Effects of Proportional Integral (PI) controllers:
- Increases the type of the system by one
- Rise time and settling time increases and Bandwidth decreases
- The speed of response decreased i.e. transient response becomes slower
- Decreases the steady-state error and steady-state response is improved
- Decreases the stability
Effects of Proportional Derivative (PD) controllers:
- Decreases the type of the system by one
- Reduces the rise time and settling time
- Rise time and settling time decreases and Bandwidth increases
- The speed of response is increased i.e. transient response is improved
- Improves gain margin, phase margin, and resonant peak
- Increases the input noise
- Improves the stability
The transfer function of a system is defined as:
Answer (Detailed Solution Below)
Control Systems Question 14 Detailed Solution
Download Solution PDFThe transfer function of a control system is defined as the ratio of the Laplace transform of the output variable to Laplace transform of the input variable assuming all initial conditions to be zero.
It is also defined as the Laplace transform of the impulse response.
If the input is represented by R(s) and the output is represented by C(s), then the transfer function will be:
\(\frac{T}{F} = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)
Routh Hurwitz criterion is used to determine
Answer (Detailed Solution Below)
Control Systems Question 15 Detailed Solution
Download Solution PDFRouth-Hurwitz criterion:
- Using the Routh-Hurwitz method, the stability information can be obtained without the need to solve the closed-loop system poles. This can be achieved by determining the number of poles that are in the left-half or right-half plane and on the imaginary axis.
- This involves checking the roots of the characteristic polynomial of a linear system to determine its stability.
- It is used to determine the absolute stability of a system.
Other methods of determining stability include:
Root locus:
- This method gives the position of the roots of the characteristic equation as the gain K is varied.
- With Root locus (unlike the case with Routh-Hurwitz criterion), we can do both analysis (i.e., for each gain value we know where the closed-loop poles are) and design (i.e., on the curve we can search for a gain value that results in the desired closed-loop poles).
Nyquist plot:
- This method is mainly used for assessing the stability of a system with feedback.
- While Nyquist is a graphical technique, it only provides a limited amount of intuition for why a system is stable or unstable, or how to modify an unstable system to be stable.