Time Domain Specifications MCQ Quiz - Objective Question with Answer for Time Domain Specifications - Download Free PDF

Last updated on Jun 23, 2025

Latest Time Domain Specifications MCQ Objective Questions

Time Domain Specifications Question 1:

Which of the following statements is 'FALSE' for a second order system with step input? 

  1. Damping ratio δ = 0 will give sustained oscillations
  2. For critical damping ie. δ = 1 there are oscillations with dying down amplitudes
  3. For overdamped system ie. δ > 1 there is no oscillation
  4. Settling time is inversely related to damping ratio

Answer (Detailed Solution Below)

Option 2 : For critical damping ie. δ = 1 there are oscillations with dying down amplitudes

Time Domain Specifications Question 1 Detailed Solution

Explanation:

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).

Time Domain Specifications Question 2:

For a second order system with denominator \(s^{2}+2 \delta w_{n} s+w_{n}^{2} ; w_{n}^{2}>0\) the roots are complex conjugates when

  1. δ ≥ 1
  2. 0 ≤ δ < 1
  3. δ < 0
  4. Roots can be complex conjugates independent of value of δ

Answer (Detailed Solution Below)

Option 2 : 0 ≤ δ < 1

Time Domain Specifications Question 2 Detailed Solution

Concept:

The standard second-order system has the characteristic equation:

\( s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 \)

The roots of this equation determine the nature of the system (overdamped, underdamped, critically damped).

Condition for Complex Conjugate Roots:

Roots are complex conjugates if the discriminant is negative:

\( (2\zeta\omega_n)^2 - 4\omega_n^2 < 0 \Rightarrow \zeta^2 < 1 \)

So, complex roots exist when:

\( 0 \leq \zeta < 1 \)

Such systems are called underdamped systems.

Therefore,  the correct answer is   \( 0 \leq \zeta < 1 \) 

Time Domain Specifications Question 3:

In the circuit shown switch is closed at t = 0. The time constant of the circuit is :

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  1. 12 sec 
  2. 1.2 sec
  3. 10 sec 
  4. 6/7 sec

Answer (Detailed Solution Below)

Option 1 : 12 sec 

Time Domain Specifications Question 3 Detailed Solution

Concept:

The time constant (τ) of an RC circuit is given by the formula:

τ = ReqC

where Req is the equivalent resistance seen by the capacitor, and C is the capacitance.

While finding out the time constant the voltage source will be shorted.

Calculation:

Given:

Resistors: 3Ω and 2Ω will be in parallel
Capacitance: C = 10 F

Solution:

If the resistors are actually in parallel (common in RC circuits):

Req = (3Ω × 2Ω)/(3Ω + 2Ω) = 6/5Ω = 1.2Ω

τ = ReqC = 1.2Ω × 10 F = 12 sec

Final Answer:

The correct time constant is 1) 12 sec when considering parallel resistors.

Time Domain Specifications Question 4:

Consider a causal second-order system with the transfer function \(G(s)=\frac{1}{s^2+2 s+1}\) with a unit-step \(R(s)=\frac{1}{s}\) as an input. Let c(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value \(\displaystyle \lim _{t \rightarrow \infty} c(t)\), rounded off to two decimal places, is

  1. 5.25
  2. 2.81
  3. 4.50
  4. 3.89

Answer (Detailed Solution Below)

Option 3 : 4.50

Time Domain Specifications Question 4 Detailed Solution

Concept:

\(G(s) = \frac{{C\left(s \right)}}{{R\left( s \right)}}= \frac{1}{{s^2} + 2s + 1}\)

\(C\left( s \right) = \frac{1}{{\left( {{s^2} + 2s + 1} \right)}} \times \frac{1}{s}\)

\(C\left( s \right) = \frac{1}{{s{{\left( {s + 1} \right)}^2}}}\)

\(C\left( s \right) = \frac{A}{s} + \frac{B}{{s + 1}} + \frac{C}{{{{\left( {s + 1} \right)}^2}}}\)

A = 1

B = -1

C = -1

\(C\left( s \right) = \frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{{{{\left( {s + 1} \right)}^2}}}\)

Applying inverse Laplace transform:

C(t) = 1 – e-t – te-t

at steady state i.e. at t → ∞

\(\mathop {{\rm{lt}}}\limits_{t \to \infty } C\left( t \right) = 1 - {e^{ - \infty }} - t{e^{ - \infty }}\)

C(t) = 1

94% of steady state \( = \frac{{94}}{{100}} \times 1\) 

= 0.94.

0.94 = 1 – e-t – te-t

Substitute all options

Let us substitute options

option (a) t= 5.25

1 – e-5.25 – 5.25 e-5.25

\(\Rightarrow 1 - \left( {\frac{{1 + 5.25}}{{{e^{5.25}}}}} \right)\)

= 0.967

option (b)

t = 4.50

\(1 - \frac{{\left( {1 + 4.50} \right)}}{{({e^{4.50}})}}\)

1 – 0.938

≈ 0.94

Time Domain Specifications Question 5:

The nature of the response shown below is:

F1 Engineering Mrunal 13.03.2023 D15

  1. Overdamped with ζ = 0.894 and ωn = 20 rad/s
  2. Underdamped with ζ = 1.894 and ωn = 4.472 rad/s
  3. Overdamped with ζ = 1.894 and ωn = 20 rad/s
  4. Underdamped with ζ = 0.894 and ωn = 4.472 rad/s

Answer (Detailed Solution Below)

Option 4 : Underdamped with ζ = 0.894 and ωn = 4.472 rad/s

Time Domain Specifications Question 5 Detailed Solution

Concept

A second-order transfer function is represented as:

\({C(s)\over R(s)}={ω_n^2\over s^2+2ζω_ns+ω_n^2}\)

where, ωn = Natural frequency

ζ = Damping ratio

The nature of damping with respect to the damping ratio is:

Damping ratio

Damping

ζ = 0

Undamped

0 < ζ < 1

Underdamped

ζ = 1

Critically damped

ζ > 1

Overdamped

Calculation

Given, \({C(s)\over R(s)}={20\over s^2+8s+20}\)

\(ω_n^2=20\)

ωn =4.472 rad/s

2ζωn = 8

\(\zeta={8\over 2\times 4.472}\)

ζ = 0.894

Hence, the system is underdamped.

Top Time Domain Specifications MCQ Objective Questions

A second order control system has a damping ratio as 0.6 and natural frequency of oscillations as 11 rad/sec. What will be the Damped frequency of oscillation?

  1. 2.6 rad/sec
  2. 8.8 rad/sec
  3. 6.9 rad/sec
  4. 5.6 rad/sec

Answer (Detailed Solution Below)

Option 2 : 8.8 rad/sec

Time Domain Specifications Question 6 Detailed Solution

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Concept:

Damped natural frequency of a second order system is given by

\({ω _d} = {ω _n}\sqrt {1 - {\xi ^2}} \)

Where,

ωn = Natural frequency

ζ = Damping ratio

Calculation:

Given-

ωn = 11 rad/sec

ζ = 0.6

Now, natural damped frequency can be calculated as

\({ω _d} = {11}\sqrt {1 - {(0.6)^2}} \)

ωd = 11 x 0.8

ωd = 8.8 rad/sec  

Match List-I with List-II and select the correct answer from the following options :

List - I List - II
A. ξ = 0 1. Roots are real and equal
B. 0 < ξ < 1 2. Roots are real and unequal
C. ξ = 1 3. Roots are complex conjugate
D. ξ > 1 4. Roots are purely imaginary

  1. A - 4, B - 1, C - 3, D - 2
  2. A - 2, B - 3, C - 4, D - 1
  3. A - 4, B - 3, C - 1, D - 2
  4. A - 2, B - 1, C - 4, D - 3

Answer (Detailed Solution Below)

Option 3 : A - 4, B - 3, C - 1, D - 2

Time Domain Specifications Question 7 Detailed Solution

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Concept:

The nature of the control system is defined on the basis of the value of the damping ratio.

Damping ratio

Type of system

Nature of root

ξ = 0

Undamped system

Roots are purely imaginary

0 < ξ < 1

Underdamped system

Roots are complex conjugate

ξ = 1

Critically damped

Roots are real and equal

ξ > 1

Over damped system

Roots are real and unequal

 

F1 Savita Engineering 2-8-22 D23

F1 Savita Engineering 2-8-22 D24

F1 Savita Engineering 2-8-22 D25

F1 Savita Engineering 2-8-22 D26

A second order control system exhibits 100% overshoot. Its damping ratio is:

  1. Less than 1
  2. Equal to 1
  3. Greater than 1
  4. Equal to zero

Answer (Detailed Solution Below)

Option 4 : Equal to zero

Time Domain Specifications Question 8 Detailed Solution

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Concept: 

The transfer function of the standard second-order system is:

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)

Characteristic equation: \({s^2} + 2ζ {ω _n} + ω _n^2 = 0\)

ζ is the damping ratio

ωn is the undamped natural frequency

\({M_p} = {e^{\frac{{ - ζ \pi }}{{\sqrt {1 - {ζ ^2}} }}}}\)   ----(1)

Calculation:

Given:

Mp = 100%

From the above equation,

\({M_p} = {e^{\frac{{ - ζ \pi }}{{\sqrt {1 - {ζ ^2}} }}}}\)

\(ln\;1 = \frac{{ - \zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}\) ; (ln 1 = 0)

So, ζ = 0

Note:

Mp is the maximum peak overshoot of the closed-loop transfer function

\(M_p \ \alpha \ \frac{1}{ζ}\)

For what value of K is the time constant of the system of figure given below is less than 0.2 sec?

F2 Shubham  Shraddha 19.8.2021 D3

  1. K > 3
  2. K > 5
  3. K > 7
  4. K > 9

Answer (Detailed Solution Below)

Option 1 : K > 3

Time Domain Specifications Question 9 Detailed Solution

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Concept:

First-order system:

The transfer function of the standard first-order system is given by

\(TF = \frac{{k}}{{\left( {1 + τ s} \right)}}\)

Where,

τ = time constant of the system

The time constant can be defined as the negative reciprocal of the pole of the system.

Calculation:

From the given figure,

\(\frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{{3K}}{{2s + 1}}}}{{1 + \frac{{3K}}{{2s + 1}}}}\)

\( = \frac{{3K}}{{2s + 3K + 1}}\)

\( = \frac{{3K}}{{3K + 1}} \times \frac{1}{{1 + \frac{{2s}}{{3K + 1}}}}\)

From the above transfer function, the time constant is

\(T = \frac{2}{{3K + 1}}\)

The time constant of the system is less than 0.2

\( \Rightarrow \frac{2}{{3K + 1}} < 0.2\;\)

⇒ K > 3   

The characteristic equation of a simple servo system is s2 + 6s + 25 = 0. Damping factor of the system is

  1. 3.2
  2. 2.4
  3. 1.8
  4. 3.0

Answer (Detailed Solution Below)

Option 4 : 3.0

Time Domain Specifications Question 10 Detailed Solution

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Concept:

The general expression of the transfer function of the standard second-order system is:

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)

Where,

ζ is the damping ratio

ωn is the undamped natural frequency

Characteristic equation: \({s^2} + 2ζ {ω _n} + ω _n^2 = 0\)

The roots of the characteristic equation are: 

\(- ζ {ω _n} + j{ω _n}\sqrt {1 - {ζ ^2}} = - \alpha \pm j{ω _d}\)

α is the damping factor

Calculation:

Given, the characteristic equation s2 + 6s + 25 = 0

Comparing this with the general expression of the transfer function of the second-order control system, we can write:

2 ζ ωn = 6

⇒ ζ ωn = 3

Additional Information

 The nature of the system is described by its ‘ζ’ value

ζ

Nature

ζ = 0

Undamped

0 < ζ < 1

Underdamped

ζ = 1

Critically damped

ζ > 1

Overdamped

Closed-loop system has higher _______ than open-loop control system and this implies increased speed of response.

  1. gain
  2. speed
  3. frequency
  4. bandwidth

Answer (Detailed Solution Below)

Option 4 : bandwidth

Time Domain Specifications Question 11 Detailed Solution

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  • The transfer function for a negative closed-loop is given by:

\(T.F. = {G(s) \over 1+G(s)H(s)}\)

where G(s) = Forward path gain

H(s) = Feedback path gain

  • Due to the presence of a feedback path, the gain of the transfer function decreases.
  • As the gain-bandwidth product must be constant, hence bandwidth increases for decreasing value of gain.
  • As the transient response of the system is improved by the use of feedback and it causes the settling time to reduce and the closed-loop system has higher bandwidth than open-loop systems and this implies an increase in the speed of response.

If the natural frequency of oscillation ωn = 13 rad/sec and damping ratio ξ  is 0.8 then find the peak time.

  1. 12 sec
  2. 0.002 sec
  3. 3 sec
  4. 0.4 sec

Answer (Detailed Solution Below)

Option 4 : 0.4 sec

Time Domain Specifications Question 12 Detailed Solution

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Concept:

Time-domain specification (or) transient response parameters:

Peak Time (tp): It is the time taken by the response to reach the maximum value.

\({\left. {\frac{{dc\left( t \right)}}{{dt}}} \right|_{t = {t_p}}} = 0,{\text{}}{t_p} = \frac{\pi }{{{ω _d}}}\)

\(As \ \omega_d=\omega_n\sqrt{ 1-\xi^2}\)

\(=\frac{\pi }{{{ω _n}√ {1 - {\zeta ^2}} }}\)

Where,

ξ = damping ratio

ωn = natural frequency

ωd = damped frequency

Calculation:

Given:

ωn = 13 rad/sec

ξ = 0.8

\(\omega_d=13\sqrt{ 1-0.8^2}\)

ωd = 7.8 rad/sec.

\(t_p=\frac{\pi}{7.8}=0.4 \;sec\)

Hence option(4) is the correct answer.

For a second order system having its transfer function as:

H(s) = 25 / (s+ 8s + 25)

Find the damping ratio.

  1. 0.6
  2. 0.5
  3. 0.4
  4. 0.8

Answer (Detailed Solution Below)

Option 4 : 0.8

Time Domain Specifications Question 13 Detailed Solution

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Concept:

The transfer function of a 2nd order is given by:

\(H(s) = {ω^2_n \over s^2+\space2ζω_ns+\space ω^2_n }\)

where, ωn = Natural frequency

ζ = Damping ratio

Calculation:

Given, \(H(s) = {25 \over s^2\space+\space8s\space+\space 25 }\)

ωn2 = 25

ωn = 5

2ζωn = 8

\(ζ = {8 \over 2\times\omega_n}\)

\(ζ = {8 \over 2\times5}\)

ζ = 0.8

In a unity feedback control system with \(G(s) = \frac{4}{{{s^2} + 0.4s}}\) when subjected to unit step input, it is required that system response should be settled within 2% tolerance band; the system settling time is:

  1. 1 sec
  2. 2 sec
  3. 10 sec
  4. 20 sec

Answer (Detailed Solution Below)

Option 4 : 20 sec

Time Domain Specifications Question 14 Detailed Solution

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Concept:

The transfer function of the standard second-order system is:

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{\omega _n^2}}{{{s^2} + 2\zeta {\omega _n}s + \omega _n^2}}\)

ζ is the damping ratio

ωn is the natural frequency

Characteristic equation: \({s^2} + 2\zeta {\omega _n} + \omega _n^2\)

The roots of the characteristic equation are: \( - \zeta {\omega _n} + j{\omega _n}\sqrt {1 - {\zeta ^2}} = - \alpha \pm j{\omega _d}\)

α is the damping factor

Settling time (Ts): It is the time taken by the response to reach ± 2% tolerance band as shown in the fig above.

\({e^{ - \xi {\omega _n}{t_s}}} = \pm 5\% \;\left( {or} \right) \pm 2\% \)

 \({t_s} \simeq \frac{3}{{\xi {\omega _n}}}\) for a 5% tolerance band.

\({t_s} \simeq \frac{4}{{\xi {\omega _n}}}\) for 2% tolerance band

Calculation:

\(G(s) = \frac{4}{{{s^2} + 0.4s}}\)

By comparing the above transfer function with the standard second-order system,

\(\omega _n^2 = 4 \Rightarrow {\omega _n} = 2\)

⇒ 2ζωn = 0.4 ⇒ ζ = 0.1

As the damping ratio is less than unity, the system is underdamped.

Settling time, \({t_s} = \frac{4}{{\zeta {\omega _n}}} = \frac{4}{{0.1 \times 2}} = 20\;sec\)

Consider a causal second-order system with the transfer function \(G\left( s \right) = \frac{1}{{1 + 2s + {s^2}}}\)  with a unit-step \(R\left( s \right) = \frac{1}{s}\) as an input. Let C(s) be the corresponding output. The time taken by the system output c(t) to reach 94% of its steady-state value \(\mathop {\lim }\limits_{t \to \infty } c\left( t \right)\), rounded off to two decimal places, is

  1. 5.25
  2. 4.50
  3. 3.89
  4. 2.81

Answer (Detailed Solution Below)

Option 2 : 4.50

Time Domain Specifications Question 15 Detailed Solution

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\(G(s) = \frac{{C\left(s \right)}}{{R\left( s \right)}}= \frac{1}{{s^2} + 2s + 1}\)

\(C\left( s \right) = \frac{1}{{\left( {{s^2} + 2s + 1} \right)}} \times \frac{1}{s}\)

\(C\left( s \right) = \frac{1}{{s{{\left( {s + 1} \right)}^2}}}\)

\(C\left( s \right) = \frac{A}{s} + \frac{B}{{s + 1}} + \frac{C}{{{{\left( {s + 1} \right)}^2}}}\)

A = 1

B = -1

C = -1

\(C\left( s \right) = \frac{1}{s} - \frac{1}{{s + 1}} - \frac{1}{{{{\left( {s + 1} \right)}^2}}}\)

Applying inverse laplace transform:

C(t) = 1 – e-t – te-t

at steady state i.e. at t → ∞

\(\mathop {{\rm{lt}}}\limits_{t \to \infty } C\left( t \right) = 1 - {e^{ - \infty }} - t{e^{ - \infty }}\)

C(t) = 1

94% of steady state \( = \frac{{94}}{{100}} \times 1\) 

= 0.94.

0.94 = 1 – e-t – te-t

Substitute all options

Let us substitute options

option (a) t= 5.25

1 – e-5.25 – 5.25 e-5.25

\(\Rightarrow 1 - \left( {\frac{{1 + 5.25}}{{{e^{5.25}}}}} \right)\)

= 0.967

option (b)

t = 4.50

\(1 - \frac{{\left( {1 + 4.50} \right)}}{{({e^{4.50}})}}\)

1 – 0.938

≈ 0.94
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