Laplace Transform and Its Application

Contents

**Define the term Laplace Transform and Complex frequency** 1

**List properties of Laplace Transform** 2

**Determine Final value and Initial value theorem** 3

**Recall Inverse Laplace Transform** 4

**Recall Convolution Integral and Graphical Convolution** 4

**Evaluate Laplace Transform of various Fundamental Signals: a) Step b) Ramp c) Impulse d) Parabolic** 5

**Explain Partial Fraction Expansion method to find the solution of Integrodifferential equations** 6

**Recall Heaviside’s Expansion Theorem** 7

**Calculate the Sinusoidal Steady State response using Laplace transform** 8

**Find the solution of Differential equation using Laplace Transform** 9

**Calculate the Transient response for R-L circuit for Step signal by using Laplace transform** 9

**Calculate the Transient response for RC circuit for Impulse signal by using Laplace transform** 10

**Calculate the Transient response for RLC circuit for Step signal by using Laplace transform** 11

**Define the term Laplace Transform and Complex frequency**

The Laplace transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is particularly useful in solving differential equations, especially those that involve initial and boundary conditions.

Given a function f(t), the Laplace transform is defined as:

F(s) = ∫[0,∞] e^(-st) f(t) dt

where s is a complex number.

The Laplace transform is essentially a weighted sum of the function f(t), where the weight at each point in time t is given by e^(-st). The Laplace transform of a function is often denoted by a capital letter, such as F(s), to distinguish it from the original function.

The complex frequency s can be written as s = σ + jω, where σ is the real part and jω is the imaginary part. The Laplace transform can be expressed in terms of the complex frequency as:

F(s) = ∫[0,∞] e^(-(σ+jω)t) f(t) dt

The Laplace transform has many important properties, such as linearity, time shifting, frequency shifting, differentiation, integration, convolution, and initial and final value theorems. These properties make it a powerful tool in solving differential equations and analyzing systems.

The complex frequency s is also known as Laplace variable or Laplace parameter. It allows us to represent the frequency domain behavior of a system in terms of a single complex variable. The real part of s, σ, determines the decay or growth rate of the system, while the imaginary part, ω, determines the oscillatory behavior of the system. The Laplace transform provides a convenient way to analyze the behavior of linear systems in the frequency domain.

**List properties of Laplace Transform**

The Laplace transform has several properties that make it a useful tool for solving differential equations and analyzing systems. Here are some of the important properties of Laplace transform:

- Linearity: The Laplace transform is a linear operator, which means that for any constants a and b, and any functions f(t) and g(t), we have:

L{a f(t) + b g(t)} = a L{f(t)} + b L{g(t)} - Time shifting: If F(s) is the Laplace transform of f(t), then the Laplace transform of f(t-a) is e^(-as) F(s).
- Frequency shifting: If F(s) is the Laplace transform of f(t), then the Laplace transform of e^(at) f(t) is F(s-a).
- Differentiation: If F(s) is the Laplace transform of f(t), then the Laplace transform of f'(t) is s F(s) – f(0).
- Integration: If F(s) is the Laplace transform of f(t), then the Laplace transform of the integral of f(t) from 0 to t is 1/s F(s).
- Convolution: If F(s) and G(s) are the Laplace transforms of f(t) and g(t), respectively, then the Laplace transform of the convolution of f(t) and g(t) is F(s) G(s).
- Initial and final value theorems: The initial value theorem states that the limit of s F(s) as s approaches infinity is equal to f(0), while the final value theorem states that the limit of s F(s) as s approaches zero is equal to the final value of f(t) as t approaches infinity, provided that the limit exists.

These properties of Laplace transform make it a powerful tool in solving differential equations and analyzing linear systems in the frequency domain.

**Determine Final value and Initial value theorem**

The final value theorem and the initial value theorem are important properties of the Laplace transform that are commonly used in engineering and physics.

The initial value theorem relates the initial conditions of a system to the Laplace transform of its output. Specifically, it states that the limit of s times the Laplace transform of the output function as s approaches infinity is equal to the initial value of the output function:

lim_{(s->∞)} s Y(s) = y(0+)

where Y(s) is the Laplace transform of the output function y(t) and y(0+) is the initial value of y(t) at time t=0+.

In other words, the initial value theorem tells us that we can determine the initial condition of a system by examining the high frequency behavior of its Laplace transform.

The final value theorem relates the steady-state behavior of a system to the Laplace transform of its input. Specifically, it states that the limit of the Laplace transform of the output function as s approaches zero is equal to the steady-state value of the output function:

lim_{(s->0)} s Y(s) = lim_{(t->∞)} y(t)

where Y(s) is the Laplace transform of the output function y(t) and lim_{(t->∞)} y(t) is the steady-state value of y(t).

**Recall Inverse Laplace Transform**

The Laplace transform is a powerful tool for solving differential equations and analyzing linear systems in the frequency domain. However, in order to fully understand the behavior of a system, we often need to convert the Laplace transform back into the time domain. This process is known as the inverse Laplace transform.

The inverse Laplace transform is defined as follows:

f(t) = (1/2πi) F(s) e^{(st)} ds

where F(s) is the Laplace transform of f(t), c is a constant greater than the real part of all singularities of F(s), and the integral is taken along a contour in the complex plane that encloses all singularities of F(s).

In practice, the inverse Laplace transform can be difficult to compute directly using this formula. Instead, we often use tables of Laplace transforms and their inverses, or numerical methods such as partial fraction expansion or the method of residues to compute the inverse Laplace transform.

The inverse Laplace transform is a crucial tool in analyzing linear systems in the time domain. By converting the Laplace transform back into the time domain, we can examine the behavior of a system over time and determine its response to various inputs and initial conditions.

**Recall Convolution Integral and Graphical Convolution**

The convolution integral is a mathematical operation that combines two functions to produce a third function that describes how one function modifies the other. In the context of the Laplace transform, convolution is used to compute the output of a linear time-invariant system when it is driven by an input signal.

The convolution of two functions f(t) and g(t) is defined as follows:

h(t) = (f * g)(t) = f(τ) g(t – τ) dτ

where h(t) is the convolution of f(t) and g(t).

In other words, to compute the value of h(t) at a particular time t, we integrate the product of f(τ) and g(t-τ) over all values of τ from 0 to t.

Graphical convolution is a visual method for computing the convolution of two functions. To perform graphical convolution, we plot the two functions on a graph and then “slide” one function over the other, computing the area of overlap at each point. The resulting curve represents the convolution of the two functions.

Graphical convolution can be a useful tool for gaining intuition about the behavior of linear systems, particularly in cases where the functions involved do not have simple closed-form expressions. It can also be used to check the accuracy of numerical or analytical computations of the convolution integral.

**Evaluate Laplace Transform of various Fundamental Signals: a) Step b) Ramp c) Impulse d) Parabolic**

Here are the Laplace transforms of some common fundamental signals:

a) Step Function: u(t) = { 0 for t < 0, 1 for t >= 0 } , L{u(t)} = e^(-st) dt = 1/s

b) Ramp Function: r(t) = { 0 for t < 0, t for t >= 0 },L{r(t)} = t e^(-st) dt = 1/s^2

c) Impulse Function: δ(t) = { 0 for t ≠ 0, ∞ for t = 0 },L{δ(t)} = δ(t) e^(-st) dt = 1

d) Parabolic Function: p(t) = { 0 for t < 0, t^2/2 for t >= 0 },L{p(t)} = (t^2/2) e^(-st) dt = 2/s^3

Note that these Laplace transforms are derived using the definition of the Laplace transform and the properties of integrals. There are also tables of Laplace transforms available that can be used to look up the Laplace transform of a particular function without having to perform the integral.

**Explain Partial Fraction Expansion method to find the solution of Integrodifferential equations**

Partial fraction expansion is a technique used to simplify rational expressions and can be used to find the solution of integro-differential equations in Laplace domain. The idea behind partial fraction expansion is to express a rational function as a sum of simpler fractions.

The partial fraction expansion of a rational function can be found by first factoring the denominator into linear and/or quadratic factors. Then, each factor is written as a separate fraction with a coefficient in the numerator. For example, the partial fraction expansion of the rational function (3s+2)/(s^2+2s+1) is:

(3s+2)/(s^2+2s+1) = A/(s+1) + B/(s+1)^2

where A and B are constants that we need to determine.

To find the values of A and B, we can multiply both sides of the equation by the denominator and simplify. This gives:

3s+2 = A(s+1)^2 + B(s+1)

Next, we can substitute specific values of s into this equation to obtain a system of linear equations that can be solved for A and B. For example, if we set s=-1, we get:

-1 = B(0) ,which implies that B=-1. If we differentiate both sides of the equation with respect to s and set s=-1, we get:

3 = 2A,which implies that A=3/2.Once we have found the values of A and B, we can write the original rational function as:

(3s+2)/(s^2+2s+1) = 3/2/(s+1) – 1/(s+1)^2

**Recall Heaviside’s Expansion Theorem**

Heaviside’s Expansion Theorem is a method for finding the partial fraction expansion of a rational function with repeated or multiple roots. This theorem states that if the denominator of the rational function can be factored into linear factors, some of which are repeated, then the partial fraction expansion can be written in the form:

F(s) = ∑(i=1 to n) [∑(j=1 to m_{i}) P_{j},i/(s-a_{i})^j]

where F(s) is the Laplace transform of the function, a_i are the distinct roots of the denominator, m_i is the multiplicity of each root, and P_{j},i are constants to be determined.

The formula for the coefficients P_{j},i is given by:

P_{j},i = lim_{(s->ai)}[1/((m_{i}-j)!) d^((m_{i}-j))(s-a_{i})^(m_{i}) F(s)/ds^(m_{i}-j)]

where d^((m_{i}-j)) denotes the (m_{i}-j)th derivative of the function with respect to s.

The method involves first factoring the denominator of the rational function and determining the multiplicity of each root. Then, we write the partial fraction expansion in the form given above and solve for the unknown coefficients using the formula above. Finally, we can use inverse Laplace transform to transform the expression back to the time domain.

Heaviside’s Expansion Theorem is particularly useful when the denominator has multiple repeated roots, as it can simplify the process of finding the partial fraction expansion.

**Calculate the Sinusoidal Steady State response using Laplace transform **

To calculate the sinusoidal steady-state response using Laplace transform, we need to first find the Laplace transform of the input signal and the transfer function of the system. Then, we can multiply these Laplace transforms to obtain the Laplace transform of the output signal. Finally, we can use inverse Laplace transform to transform the expression back to the time domain and obtain the sinusoidal steady-state response.

For example, let’s consider a system with transfer function H(s) and input signal x(t) = A sin(ωt). The Laplace transform of x(t) is:

X(s) = A/(s^2+ω^2)

The Laplace transform of the output signal y(t) is given by:

Y(s) = H(s) X(s)

To find the sinusoidal steady-state response, we can substitute s = jω into the expression for Y(s):

Y(jω) = H(jω) X(jω)

The sinusoidal steady-state response y(t) is then given by the inverse Laplace transform of Y(jω):

y(t) = (1/2π) ∫[j(ω)] [H(jω) X(jω) e^(jωt) dω]

where [j(ω)] denotes the contour of integration along the imaginary axis.

**Find the solution of Differential equation using Laplace Transform**

To find the solution of a differential equation using Laplace transform, we can take the Laplace transform of both sides of the equation and use the properties of Laplace transform to simplify the resulting algebraic equation. Then, we can solve for the Laplace transform of the unknown function and use inverse Laplace transform to obtain the solution in the time domain.

For example, let’s consider the differential equation:

y”(t) + 2y'(t) + 5y(t) = f(t)

where y(t) is the unknown function and f(t) is the input signal. Taking the Laplace transform of both sides of the equation, we get:

s^2 Y(s) – s y(0) – y'(0) + 2s Y(s) – 2y(0) + 5Y(s) = F(s)

where Y(s) and F(s) denote the Laplace transforms of y(t) and f(t), respectively, and y(0) and y'(0) are the initial conditions. Solving for Y(s), we get:

Y(s) = [s y(0) + y'(0) + F(s)] / (s^2 + 2s + 5)

We can then use inverse Laplace transform to obtain the solution y(t) in the time domain.

**Calculate the Transient response for R-L circuit for Step signal by using Laplace transform**

To calculate the transient response of an R-L circuit for a step input signal using Laplace transform, we can follow these steps:

- Write the differential equation governing the circuit. For an R-L circuit, the differential equation is:

L(di/dt) + R*i = Vstep(t)

where i is the current through the circuit, L is the inductance, R is the resistance, and Vstep(t) is the step input signal. - Take the Laplace transform of both sides of the differential equation. Using the property of linearity of Laplace transform, we get:

L(sI(s) – i(0)) + R*I(s) = Vstep(s)

where I(s) and Vstep(s) are the Laplace transforms of i(t) and Vstep(t), respectively, and i(0) is the initial condition of the circuit. - Solve for the Laplace transform of the current I(s):

I(s) = [Vstep(s) + i(0)*s] / [Ls + R] - Use partial fraction expansion to simplify the expression for I(s). If the denominator of the expression has distinct real roots, we can write:

I(s) = [A/(s + α)] + [B/(s + β)]

where α and β are the roots of the denominator, and A and B are constants to be determined. - Use inverse Laplace transform to find the current i(t) in the time domain. If the roots α and β are real and distinct, the inverse Laplace transform of I(s) can be written as:

i(t) = A*e^(-αt) + B*e^(-βt)

where A and B are the constants obtained from partial fraction expansion.

**Calculate the Transient response for RC circuit for Impulse signal by using Laplace transform**

To calculate the transient response of an RC circuit for an impulse input signal using Laplace transform, we can follow these steps:

- Write the differential equation governing the circuit. For an RC circuit, the differential equation is:

RC(di/dt) + i = δ(t)

where i is the current through the circuit, R is the resistance, C is the capacitance, and δ(t) is the impulse input signal. - Take the Laplace transform of both sides of the differential equation. Using the property of linearity of Laplace transform, we get:

RC(sI(s) – i(0)) + I(s) = 1

where I(s) and i(0) are the Laplace transforms of i(t) and the initial condition of the circuit, respectively. - Solve for the Laplace transform of the current I(s):

I(s) = 1 / [RCs + 1] - Use partial fraction expansion to simplify the expression for I(s). Since the denominator of the expression has only one root, we can write:

I(s) = A / (RCs + 1)

where A is a constant to be determined. - Use inverse Laplace transform to find the current i(t) in the time domain. The inverse Laplace transform of I(s) can be written as:

i(t) = A*e^(-t/RC)

where A is the constant obtained from partial fraction expansion.

**Calculate the Transient response for RLC circuit for Step signal by using Laplace transform**

To calculate the transient response of an RLC circuit for a step input signal using Laplace transform, we can follow these steps:

- Write the differential equation governing the circuit. For an RLC circuit, the differential equation is:

L(d^2i/dt^2) + R(di/dt) + i/C = u(t)

where i is the current through the circuit, L is the inductance, R is the resistance, C is the capacitance, and u(t) is the step input signal. - Take the Laplace transform of both sides of the differential equation. Using the property of linearity of Laplace transform, we get:

Ls^2I(s) + RsI(s) + 1/C I(s) = 1/s

where I(s) is the Laplace transform of i(t). - Solve for the Laplace transform of the current I(s):

I(s) = 1 / [Ls^2 + Rs + 1/C] - Use partial fraction expansion to simplify the expression for I(s). Since the denominator of the expression has two distinct roots, we can write:

I(s) = (As + B) / [(Ls+α)(Ls+β)]

where α and β are the roots of the denominator, A and B are constants to be determined. - Use inverse Laplace transform to find the current i(t) in the time domain. The inverse Laplace transform of I(s) can be written as:

i(t) = (Ae^(-αt) + Be^(-βt))/sqrt((β-α)^2/(4LC) – R^2/(4L^2))

where A and B are the constants obtained from partial fraction expansion.