Laplace Analysis of Networks MCQ Quiz in తెలుగు - Objective Question with Answer for Laplace Analysis of Networks - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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.

Key Points

•  The 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.

G(s) = C(s)/ R(s)

here,

C(s) → Laplace transform of output (Voltage or current)

R(s) → Laplace transform of input (Voltage or current)

• A proper rational transfer function is a minimum phase transfer function if all of its zeroes lie in the left half of s-plane. It is a nonminimum phase transfer function if it has one or more zeros in the right half of the s-plane.
• As transfer function is just a ratio of output and input expression it does matter whether it is minimum or nonminimum phase system.

Hence, option (2) is correct.

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:

\({V_2(s)\over V_1(s)}={Z_1\over Z_1+Z_2}\)

From diagram, \(C_1=C_2=C\)

\(Z_1={R+{1\over sC}}={1+sRC\over sC}\)

\(Z_2={1\over sC}\)

\(\)\(Z_1+Z_2={1+sRC_1\over sC}+{1\over sC}={2+sRC\over sC}\)

\({V_2(s)\over V_1(s)}={{1+sRC\over sC}\over {2+sRC\over sC}}\)

RC = (100 × 103) × (10 × 10-6) = 1

\({V_2(s)\over V_1(s)}={{s+1}\over {s+2}}\)

Concept:

The transfer function of a linear time-invariant function is defined to be the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable under the assumption that all initial conditions are zero.

Order: The highest power of the complex variable ‘s’ in the denominator of the transfer function determines the order of the system.

In the Laplace domain, inductor impedance is written as “sL”

and capacitance impedance as:

\(\frac{1}{{sC}}\) .

For a resistor, there is no change.

Calculation:

From the given circuit capacitor is of 100 μ F value and resistor of 10 KΩ  

Capacitive impedance is

 \(\frac{1}{{sC}} = \frac{1}{{s100 \times {{10}^{ - 6}}}}\) 

\( = \frac{1}{{s{{10}^{ - 4}}}}\)

\( = \frac{{{{10}^4}}}{s}\)

Assume current I is flowing in a circuit and the circuit in Laplace domain will be

Output is

\({V_2}\left( s \right) = I\left( s \right)\left( {10 \times {{10}^3} + \frac{{{{10}^4}}}{s}} \right)\)

Applying KVL to the loop we get

 \( - {V_1}\left( s \right) + I\left( s \right)\left( {\frac{{{{10}^4}}}{s}} \right) + I\left( s \right)\left( {{{10}^4} + \frac{{{{10}^4}}}{s}} \right) = 0\)

\({V_1}\left( s \right) = I\left( s \right)\left( {2 \times \frac{{{{10}^4}}}{s} + {{10}^4}} \right)\)

Now the ratio of Laplace transform is:

 \(\frac{{{V_2}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{I\left( s \right)\left( {\frac{{s + 1}}{s}} \right){{10}^4}}}{{I\left( s \right)\left( {\frac{{s + 2}}{s}} \right){{10}^4}}} \) 

\( = \frac{{s + 1}}{{s + 2\;}}\)

 Important points:

NOTE: poles and zeroes are nothing but the negative of the inverse of time constant

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.

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

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

Modify

For a driving point function, the degree of numerator polynomial and the degree of denominator polynomial must be equal.

\(i.e\;\left| {n - m} \right| = 0\)  

The driving point function of a lossless network has the following properties:

• They are the ratio of odd to even or even to odd polynomials in the s-domain.
• They do satisfy the separation property.
• They have either a pole or a zero at the origin and infinity.
• Their degree of the numerator and denominator polynomials in the s-domain may differ by either zero or unity only.