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

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.

Concept

The current through a capacitor is given by:

$I_C=C\frac{d{V}_C}{dt}$

where, $\frac{d{V}_C}{dt}=$ Rate of change of capacitor voltage

The voltage across a capacitor is given by:

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

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.

After source transformation:

V(s) = sL × I(s) − Li(0)

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:

$TF=\frac{C\left(s\right)}{R\left(s\right)}$

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.

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

Concept:

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

1. Resistance: R
2. Inductor: sL
3. Capacitor: $\frac{1}{sC}$

​Calculation:

The circuit diagram in the Laplace domain is given below:

Applying voltage division rule across capacitor:

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

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

V_i (s) - R I(s) - $\frac{1}{cS}$ I(s) = 0

Vi (s) = I(s) (R + $\frac{1}{cS}$)

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

Now output voltage or voltage across the capacitor is

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

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

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

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.

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

Important Points

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.

TF = $\frac{L\left[output\right]}{L\left[input\right]}|$ 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,

For the circuit above,

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

We have,

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

And, V_o (S) = I(S) ($\frac{1}{Cs}$)

From equation (1),

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

⇒ Transfer Function = $\frac{1}{(LC{s}^2+RCs+1)}$

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

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,

$V_1=\frac{\mathrm{V}{\mathrm{R}}_1}{{\mathrm{R}}_1+R_2}$

And voltage across the resistor R₂ is given by,

$V_2=\frac{\mathrm{V}{\mathrm{R}}_2}{{\mathrm{R}}_1+R_2}$

Calculation:

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

$V_2(s)=\frac{{\mathrm{V}}_1\left(s\right)\times 1/s}{1+1/s}=\frac{{\mathrm{V}}_1\left(s\right)}{s+1}$

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

Note:

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.

TF = $\frac{L\left[output\right]}{L\left[input\right]}|$ 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.

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

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, Vo = Output voltage

Vi = input voltage

Calculation:

Vo(t) = (1 - e-t / RC)   ---(1)

V_i(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 V_i(s) we get:

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

Now consider the diagram given in option(1):

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.

$V_{R1}=\frac{R_1}{R_1+R_2}{V}_{s}and$

$V_{R2}=\frac{R_2}{R_1+R_2}{V}_s$

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) - $\frac{1}{cS}$ I(s) = 0

Vi (s) = I(s) (R + $\frac{1}{cS}$)

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

Now output voltage or voltage across the capacitor is

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

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

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

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

Concept:

If V₀ is the initial voltage across capacitor

Calculation:

Circuit diagram at laplace domain

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

Given;

C = 1μF

R= 5 Ω 

Use nodal analysis at node V_C

⇒ V_C/R + V_C/(1/sC) = 10/s

⇒ VC( 1/R + sC) = 10/s

⇒ VC( 1+ sRC)/R = 10/s

⇒ VC= 10R/s( 1+ sRC)

⇒ VC= 10R/sRC( 1/RC+ s)

⇒ VC= (10/C)/s( s+ 1/RC)

⇒ V_C = A/s + B/(s + 1/RC)

Determine the values of A and B using partial fraction

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

B_{s= -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)

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

$f\left(\mathrm{\infty }\right)=\underset{s\to 0}{lim}sF\left(s\right)$

 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:

$C\left(0\right)=\underset{t\to 0}{lim}c\left(t\right)=\underset{s\to \mathrm{\infty }}{lim}sC\left(s\right)$

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'.