Laplace Analysis of Networks MCQ Quiz - Objective Question with Answer for Laplace Analysis of Networks - Download Free PDF

Last updated on Jun 19, 2025

Latest Laplace Analysis of Networks MCQ Objective Questions

Laplace Analysis of Networks Question 1:

Laplace transform is applicable to ___________ signals.

  1. integral time
  2. continuous time domain
  3. non linear
  4. digital

Answer (Detailed Solution Below)

Option 2 : continuous time domain

Laplace Analysis of Networks Question 1 Detailed Solution

Explanation:

Laplace Transform and Its Applicability to Signals

Definition: The Laplace transform is a widely used integral transform in mathematics and engineering that converts a time-domain function into a complex frequency-domain representation. It is particularly useful in analyzing and solving linear time-invariant systems such as electrical circuits, mechanical systems, and control systems.

Mathematical Representation:

The Laplace transform of a time-domain function f(t) is defined as:

L{f(t)} = F(s) = ∫0 f(t) e-st dt

Where:

  • t: Time variable (in the time domain).
  • s: Complex frequency variable (s = σ + jω).
  • e-st: Exponential decay factor.

Uses and Advantages:

  • The Laplace transform simplifies the analysis of differential equations by converting them into algebraic equations.
  • It provides a systematic way to handle initial conditions in dynamic systems.
  • It is extensively used in control systems, signal processing, and communication engineering.

Correct Option Analysis:

The correct option is:

Option 2: Continuous time domain.

The Laplace transform is primarily applicable to signals in the continuous time domain. This is because the integral definition of the Laplace transform requires the signal to be defined over a continuous range of time, typically from 0 to infinity. In engineering and physics, most applications of the Laplace transform deal with continuous-time systems, such as analog electrical circuits, mechanical vibrations, and control systems.

Reasoning:

  • In continuous time systems, signals are functions of a continuous variable (time t), and the Laplace transform effectively captures their frequency-domain characteristics.
  • It is especially useful for analyzing linear systems where the system's behavior can be expressed using differential equations.
  • The Laplace transform simplifies the analysis by converting these differential equations into algebraic equations in the s-domain (complex frequency domain).

Application:

  • Analysis of electrical circuits with capacitors and inductors.
  • Modeling and analysis of mechanical systems, such as damped harmonic oscillators.
  • Control system design and stability analysis.
  • Signal processing tasks such as filtering and system identification.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Integral time.

This option is incorrect. The term "integral time" is not a standard term in signal processing or control systems. If it refers to discrete-time signals, then it is unrelated to the Laplace transform, as the Laplace transform is specifically defined for continuous-time signals. For discrete-time signals, the Z-transform is used instead of the Laplace transform.

Option 3: Non-linear.

This option is incorrect. The Laplace transform is primarily applicable to linear systems and signals. Non-linear systems cannot be analyzed directly using the Laplace transform because their behavior does not satisfy the principle of superposition (additivity and homogeneity). For non-linear systems, other mathematical tools such as perturbation methods, numerical simulation, or specific transforms might be needed.

Option 4: Digital.

This option is incorrect. Digital signals are typically discrete-time signals, and the Laplace transform is not applicable to them. Instead, the Z-transform is used for analyzing and processing discrete-time signals in the digital domain. The Z-transform is analogous to the Laplace transform but is specifically designed for signals defined at discrete time intervals.

Option 5: (Blank).

This option is invalid as it does not provide any specific information for analysis. It is not relevant to the question.

Conclusion:

The Laplace transform is an essential mathematical tool for analyzing continuous-time systems and signals. Its ability to convert time-domain differential equations into frequency-domain algebraic equations makes it invaluable in engineering and scientific applications. While the Laplace transform is highly effective for continuous-time systems, it is not applicable to discrete-time or digital signals, nor is it suitable for analyzing non-linear systems. Instead, other transforms like the Z-transform or specific mathematical tools are used in such cases. Understanding the scope and limitations of the Laplace transform is crucial for its correct application in engineering and science.

Laplace Analysis of Networks Question 2:

What will be the current relationship in time domain for a capacitive circuit?

  1. \(\rm C\frac{d^2v}{dt^2}\)
  2. \(\rm i(t)=C\frac{dv}{dt}\)
  3. \(\rm i(t)=C\int_0^tv(t)\)
  4. \(\rm i(t)=C\int_0^tv(t)+i(0)\)

Answer (Detailed Solution Below)

Option 2 : \(\rm i(t)=C\frac{dv}{dt}\)

Laplace Analysis of Networks Question 2 Detailed Solution

Concept

The current through a capacitor is given by:

\(I_C=C{dV_C\over dt}\)

where, \({dV_C\over dt}=\) Rate of change of capacitor voltage

The voltage across a capacitor is given by:

\(V_c(t)={1\over C } \int I_c(t) dt\)

Laplace Analysis of Networks Question 3:

What will be the voltage relationship of frequency domain relation for inductor having time domain v(t) = Ldi/dt?

  1. V(s) = LsI(s) − Li(0)
  2. V(s) = LI(s)
  3. V(s) = LsI(s)
  4. V(s) = LI(s) − Li(0)

Answer (Detailed Solution Below)

Option 1 : V(s) = LsI(s) − Li(0)

Laplace Analysis of Networks Question 3 Detailed Solution

Concept

When an inductor is represented in the frequency domain, it is represented as sL and the initial current across the inductor is also considered.

qImage6812163c91ab4414ddec3333

After source transformation:

qImage6812163c91ab4414ddec3335

V(s) = sL × I(s) − Li(0)

Laplace Analysis of Networks Question 4:

A transfer function model CANNOT be used for the analysis and design of:

  1. non-linear systems
  2. time invariant systems
  3. linear systems
  4. single-input single-output systems

Answer (Detailed Solution Below)

Option 1 : non-linear systems

Laplace Analysis of Networks Question 4 Detailed Solution

Concept:

Transfer function:

  • The transfer function of a control system is defined as the ratio of the Laplace transform of the output variable to the 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:

\({T}{F} = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

Properties of Transfer Function:

  • The transfer function is defined only for a linear time-invariant system. It is not defined for nonlinear or time-variant systems.
  • All initial conditions of the system are set to zero.
  • Transfer function independent of input of the system.
  • The transfer function of a continuous data system is expressed only as a function of the complex variable. 
  • If the system transfer function has no poles and zeros with positive real parts, the system is a minimum phase system.
  • Non-minimum phase functions are the functions that have poles or zeros on the right hand of s plane
  • The stability of a time-invariant linear system can be determined from the characteristic equation.

Laplace Analysis of Networks Question 5:

Choose the option that has the correct spelling.

  1. Foreward
  2. Forword
  3. Forworde
  4. Foreword

Answer (Detailed Solution Below)

Option 4 : Foreword

Laplace Analysis of Networks Question 5 Detailed Solution

The correct answer is "Foreword".

Key Points

  •  Foreword is a term used to describe a preface or introduction to a book, written by someone other than the author of the book.
  • It is typically written by someone who has some connection to the book or the author, such as a friend, colleague, or famous person in the same field.
  • The foreword usually provides background information on the book, the author, or the topic being discussed, and may also offer some comments or reflections on the book's content. The purpose of a foreword is to provide readers with additional context and perspective before they start reading the book itself.
  • Example: In the foreword to "The Art of Cooking," famous chef Julia Child writes about her personal connection to the author and her admiration for the author's culinary skills.
  • Thus, the correct spelling is "Foreword"

Correct answer: Foreword

Top Laplace Analysis of Networks MCQ Objective Questions

The voltage transfer function of the network shown in the figure below is

F2 Madhuri Engineering 20.05.2022 D2

  1. \(\frac{1}{{1 + 2s}}\)
  2. 1 + 4s
  3. 6 - s
  4. \(\frac{1}{{1 + 2{s^2}}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{1}{{1 + 2{s^2}}}\)

Laplace Analysis of Networks Question 6 Detailed Solution

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Concept:

The Laplace transform resistance, inductor, and capacitance are given by:

  1. Resistance: R
  2. Inductor: sL
  3. Capacitor: \({1\over sC}\)

​Calculation:

The circuit diagram in the Laplace domain is given below:

F2 Madhuri Engineering 20.05.2022 D3

Applying voltage division rule across capacitor:

\(V_{out}(s) = V_{in}\times{{1 \over s}\over {1 \over s}+2s}\)

\({V_{out}(s)\over V_{in}(s)} ={1\over{1 +2s^2}}\)

Find the transfer function of the given network.

F30 Shubham B 27-4-2021 Swati D8

  1. \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{RC}{s+RC}\)
  2. \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{1+\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)
  3. \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)
  4. \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{1+\dfrac{1}{RC}}{s-\dfrac{1}{RC}}\)

Answer (Detailed Solution Below)

Option 3 : \(\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)

Laplace Analysis of Networks Question 7 Detailed Solution

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Concept:

A transfer function is defined as the ratio of the 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:

The Laplace transform equivalent network for the given circuit is,

By applying KVL in the Laplace equivalent circuit, we get

Vi (s) - R I(s) - \({{1} \over cS}\) I(s) = 0

Vi (s) = I(s) (R + \({{1} \over cS}\))

I(s) = Vi (s) / (R + \({{1} \over cS}\))    ......(1)

Now output voltage or voltage across the capacitor is

Vo (s) = \({{1} \over cS}\) I(s)       ...... (2)

Put the value of I(s) from equation (1) in equation(2)

∴ \(V_o(s)= {\frac {1} {sC}} {{V_i (s)} \over {(R+\frac {1} {sC})}}\)

Now the transfer function of the given circuit is the ratio of the Laplace transform of output voltage to the Laplace transform of input voltage.

∴ \(TF =\dfrac{V_0\left( s\right)}{V_i\left( s\right)}=\dfrac{\dfrac{1}{RC}}{s+\dfrac{1}{RC}}\)

Important Points

Element

It’s equivalent in Laplace domain

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What will be the transfer function of the system shown below?

F2 Rohit Shraddha 13-10-2021 D14

  1. \(\frac {1}{\left(LCs^2 + RCs + 1\right)}\)
  2. RCs2 + LCs + 1
  3. \(\frac {1}{(RCs^2 + LCs + 1)}\)
  4. LCs2 + RCs + 1

Answer (Detailed Solution Below)

Option 1 : \(\frac {1}{\left(LCs^2 + RCs + 1\right)}\)

Laplace Analysis of Networks Question 8 Detailed Solution

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Concept:

Transfer function:

The transfer function is defined as the ratio of the Laplace transform of the output variable to the input variable with all initial conditions zero.

F1 Uday.S 23-12-20 Savita D1

TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\) Initial conditions = 0

TF = C(s) / R(s)

The transfer function of a linear time-invariant system can also be defined as the Laplace transform of the impulse response, with all the initial conditions set to zero. 

Application:

The given circuit can be drawn as in Laplace domain as shown,

F1 Nakshatra 02-11-21 Savita D15

For the circuit above,

Transfer Function = \(\frac{V_o(S)}{V_i(S)}\)    ---(1)

We have,

V(S) = I(S) (R + LS + \(\frac{1}{Cs}\))

And, V(S) = I(S) (\(\frac{1}{Cs}\))

From equation (1),

\(TF=\frac{1/Cs}{(R+Ls+1/Cs)}\)

⇒ Transfer Function = \(\frac {1}{\left(LCs^2 + RCs + 1\right)}\)

The transfer function V2(s) / V1(s) is 

quesOptionImage481

  1. 1/ (s + 1)
  2. 1/s
  3. (s + 1) / s
  4. s / (s + 1)
  5. s / (s + 2)

Answer (Detailed Solution Below)

Option 1 : 1/ (s + 1)

Laplace Analysis of Networks Question 9 Detailed Solution

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Transform the network given in the question into Laplace Domain. we get,

F18 Jai Prakash 26-3-2021 Swati D1

Voltage division rule:

If two resistors R1 and R2 are connected in series across the supply voltage V then the voltage across the resistor R1 is given by,

\(V_1=\frac{{{\rm{V\;}}{{\rm{R}}_1}}}{{{{\rm{R}}_1} + {R_2}}}\)

And voltage across the resistor R2 is given by,

\(V_2=\frac{{{\rm{V\;}}{{\rm{R}}_2}}}{{{{\rm{R}}_1} + {R_2}}}\)

Calculation:

Now apply voltage division to the above circuit then we get,

\(V_{2}(s)=\frac{{{{\rm{V}}_1}\left( s \right) \times 1/s}}{{1 + 1/s}} = \frac{{{{\rm{V}}_1}\left( s \right)}}{{s + 1}}\)

\(\frac{{{{\rm{V}}_2}\left( s \right)}}{{{{\rm{V}}_1}\left( s \right)}} = \frac{1}{{s + 1}}\)

Note:

F18 Jai Prakash 26-3-2021 Swati D2

Which of the following is NOT one of the properties of transfer function?

  1. All initial conditions of the system are set to zero.
  2. The transfer function is dependent on the input of the system.
  3. It is defined only for a linear time-invariant system.
  4. The transfer function between an input variable and an output variable of a system is defined as the Laplace transform of the impulse response.

Answer (Detailed Solution Below)

Option 2 : The transfer function is dependent on the input of the system.

Laplace Analysis of Networks Question 10 Detailed Solution

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Transfer function:

The transfer function is defined as the ratio of the Laplace transform of the output variable to the input variable with all initial conditions zero.

F1 Uday.S 23-12-20 Savita D1

TF = \(\left. {\frac{{L\left[ {output} \right]}}{{L\left[ {input} \right]}}} \right|\) Initial conditions = 0

TF = C(s) / R(s)

 The transfer function of a linear time-invariant system can also be defined as the Laplace transform of the impulse response, with all the initial conditions set to zero. 

Properties of transfer function:

  • The transfer function is defined only for a linear time-invariant system. It is not defined for nonlinear systems.
  • The transfer function is independent of the input and output.
  • Because the transfer function of the system depends on the governing dynamic equation of the system only.
  • If the transfer function is dependent on the input means the system will come under non-linear systems, but actually, the transfer function is defined for linear systems only.
  • Transfer function analysis is not valid for the system that contains variables having initial values.

Choose the option that has the correct spelling.

  1. Foreward
  2. Forword
  3. Forworde
  4. Foreword

Answer (Detailed Solution Below)

Option 4 : Foreword

Laplace Analysis of Networks Question 11 Detailed Solution

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The correct answer is "Foreword".

Key Points

  •  Foreword is a term used to describe a preface or introduction to a book, written by someone other than the author of the book.
  • It is typically written by someone who has some connection to the book or the author, such as a friend, colleague, or famous person in the same field.
  • The foreword usually provides background information on the book, the author, or the topic being discussed, and may also offer some comments or reflections on the book's content. The purpose of a foreword is to provide readers with additional context and perspective before they start reading the book itself.
  • Example: In the foreword to "The Art of Cooking," famous chef Julia Child writes about her personal connection to the author and her admiration for the author's culinary skills.
  • Thus, the correct spelling is "Foreword"

Correct answer: Foreword

If the step response to the input step amplitude of 1 V is given by Vo(t) = (1 - e-t / RC), the network can be represented by:

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  2. F1 Tapesh.S 21-01-21 Savita D5
  3. F1 Tapesh.S 21-01-21 Savita D6
  4. F1 Tapesh.S 21-01-21 Savita D7

Answer (Detailed Solution Below)

Option 1 : F1 Tapesh.S 21-01-21 Savita D4

Laplace Analysis of Networks Question 12 Detailed Solution

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Concept:

A transfer function is defined as the ratio of the Laplace transform of output to the Laplace transform of input when initial conditions are zero.

\(T.F. = \frac{{{V_O}\left( s \right)}}{{{V_i}\left( s \right)}}\)

Where, V= Output voltage

Vi = input voltage

Calculation:

Vo(t) = (1 - e-t / RC)   ---(1)

Vi(t) = 1 V  ---(2)

Applying Laplace transform to the (1) equation:

\(V_0(s)=\frac{1}{s}-\frac{1}{s \ + \ \frac{1}{RC}}\)

\(V_0(s)=\frac{1}{s(RCs \ + \ 1)}\)  ---(3)

Now taking Laplace transform of equation (2):

\(V_i(s)=\frac{1}{s}\)

Now dividing V0(s) by Vi(s) we get:

\(\frac{V_0(s)}{V_i(s)}=\frac{1}{1 \ + \ sRC}\)  ---(4)

Now consider the diagram given in option(1):

F1 Tapesh.S 21-01-21 Savita D4

Finding V0 by using voltage division rule:

\(V_0(s)=\frac{V_i(s) \ \times \ \frac{1}{sC} }{R \ + \ \frac{1}{sC}}\)

\(\frac{V_0(s)}{V_i(s)}=\frac{1}{1 \ + \ sRC}\)

Hence it gives the same transfer function as calculated in equation (4)

Hence option (1) is the correct answer.

Important Points

Voltage Division Rule:

The voltage across a resistor in a series circuit is equal to the value of that resistor times the total voltage across the series elements divided by the total resistance of the series elements.

F12 Jai Prakash 2-2-2021 Swati D35

\( {V_{R1}} = \;\frac{{{R_1}}}{{{R_1} + {R_2}}}{V_s}\;\; \ and\)

\({V_{R2}} = \;\frac{{{R_2}}}{{{R_1} + {R_2}}}{V_s}\)

Find the Transfer function of the network given below.

F1 Shubham B 26.4.21 Pallavi D13

  1. RCs - 1
  2. RCs + 1
  3. \(\frac{1}{RCs + 1}\)
  4. Cs

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{RCs + 1}\)

Laplace Analysis of Networks Question 13 Detailed Solution

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Concept:

A transfer function is defined as the ratio of the 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:

The Laplace transform equivalent network for the given circuit is,

F1 Shubham B 26.4.21 Pallavi D14

By applying KVL in the Laplace equivalent circuit, we get

Vi (s) - R I(s) - \({{1} \over cS}\) I(s) = 0

Vi (s) = I(s) (R + \({{1} \over cS}\))

I(s) = Vi (s) / (R + \({{1} \over cS}\))    ......(1)

Now output voltage or voltage across the capacitor is

Vo (s) = \({{1} \over cS}\) I(s)       ...... (2)

Put the value of I(s) from equation (1) in equation(2)

∴ \(V_o(s)= {\frac {1} {sC}} {{V_i (s)} \over {(R+\frac {1} {sC})}}\)

Now the transfer function of the given circuit is the ratio of the Laplace transform of output voltage to the Laplace transform of input voltage.

∴ \(TF =\frac{V_0\left( s\right)}{V_i\left( s\right)}=\frac{\frac{1}{RC}}{s+\frac{1}{RC}}\)

\(TF =\frac{V_0\left( s\right)}{V_i\left( s\right)}=\frac{1}{sRC + 1}\)

Important Points

Element

It’s equivalent in the Laplace domain

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For the RC parallel circuit, determine the voltage across the capacitor using Laplace transform, assume capacitor is initially relaxed.
F6 Vinanti Engineering 21.04.23 D02

  1. 50(1 − e−t/5)
  2. 50(1 + e−t/5)
  3. 5(1 − e−t/5)
  4. 5(1 + e−t/5)

Answer (Detailed Solution Below)

Option 1 : 50(1 − e−t/5)

Laplace Analysis of Networks Question 14 Detailed Solution

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Concept:

Element Laplace Transformation
R R
L sL
C 1/sC

 

If V0 is the initial voltage across capacitor

F1  Madhuri Engineering 24.06.2022 D29

Calculation:

Circuit diagram at laplace domain

As capacitor was initially relaxed: V0 = 0 (Voltage source will be replaced by a short circuit)

F1  Madhuri Engineering 24.06.2022 D30

Given;

C = 1μF

R= 5 Ω 

Use nodal analysis at node VC

⇒ VC/R + VC/(1/sC) = 10/s

⇒ VC( 1/R + sC) = 10/s

⇒ VC( 1sRC)/R = 10/s

⇒ VC= 10R/s( 1sRC)

⇒ VC= 10R/sRC( 1/RCs)

⇒ VC= (10/C)/s( s+ 1/RC)

⇒ VC = A/s + B/(s + 1/RC)

Determine the values of A and B using partial fraction

As=0 = (10/C)/(s + 1/RC) = (10/C)( 0 +1/RC) = 10R

Bs= -1/RC = (10/C)/s = -10R

⇒ VC = (10R)/s + (-10R)/(s + 1/RC)

Using inverse laplace transform:

⇒ VC = 10R -10Re-t/RC

⇒ VC = 10R(1 -e-t/RC)

∴  VC = 50(1 -e-t/5)

The final value of \(X(s) = \frac {12(s + 2)}{\{s(s^2 + 1)\})}\) will be

  1. 0
  2. 6
  3. cannot be determined
  4. 24

Answer (Detailed Solution Below)

Option 3 : cannot be determined

Laplace Analysis of Networks Question 15 Detailed Solution

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Concept:

1. Final value theorem:

A final value theorem allows the time domain behavior to be directly calculated by taking a limit of a frequency domain expression

The final value theorem states that the final value of a system can be calculated by

\(f\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} sF\left( s \right)\)

 Where F(s) is the Laplace transform of the function.

For the final value theorem to be applicable system should be stable in a steady-state and for that real part of the poles should lie on the left side of s plane.

2. Initial value theorem:

\(C\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} c\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sC\left( s \right)\)

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

Calculation:

Given that, \(X(s) = \frac {12(s + 2)}{\{s(s^2 + 1)\})}\)

Poles lies at s = 0, ±j 1

Roots lies on the imaginary axis, so it is marginally stable.

So the Final value theorem is not applicable as the system is oscillatory in nature. 

∴ The correct answer is option C 'cannot be determined'.

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