Question

Calculate the inverse Laplace transform of each of the following functions: (a) \(\frac{e^{-s}\left(1-e^{-s}\right)}{s\left(s^{2}+1\right)}\) (b) \(\frac{e^{-16 s}}{s\left(s^{2}+2 s+4\right)}\)

Short answer

Part (a): Shifted, delayed sinusoidal; Part (b): Phase delayed sine transition.

Step-by-step solution

Break Down the Problem (Part a)

The first function we have to deal with is \( \frac{e^{-s}(1-e^{-s})}{s(s^2+1)} \). We need to simplify this function in order to find its inverse Laplace transform. Notice that it involves exponential terms and factors that suggest the use of convolution theorem and shifting properties.

Simplify the Expression (Part a)

First, simplify \( 1-e^{-s} \) into a form we can work with: \[ \frac{e^{-s}(1-e^{-s})}{s(s^2+1)} = e^{-s} \cdot \left( \frac{1}{s(s^2+1)} - \frac{e^{-s}}{s(s^2+1)} \right) \]This shows the use of the linearity of Laplace transform and the subtraction of two transformations.

Apply Laplace Transform Properties (Part a)

Apply the shifting property of the Laplace transform to handle the term \( e^{-sF(s)} \). The shifting property states: if \( F(s) \) is the Laplace transform of \( f(t) \), then \( e^{-as}F(s) \) is the Laplace transform of \( f(t-a)u(t-a) \), where \( u(t) \) is the unit step function.

Use Inversion Formula (Part a)

To invert each part: - The inverse Laplace transform of \( \frac{1}{s(s^2+1)} \) is \( t \cdot \, u(t) - \sin{t} \, u(t) \).- Similarly, by delaying each result the inverse Laplace of \( e^{-s} \cdot \frac{1}{s(s^2+1)} \) is: \[ u(t-1) \cdot ((t-1) - \sin{(t-1)}) \]And for the second term involving \( e^{-2s} \): \[ u(t-2) \cdot ((t-2) - \sin{(t-2)}) \].

Solution for Part a

The complete inverse Laplace transform for part (a) is the combination of these results:\[ f(t) = u(t-1) \cdot ((t-1) - \sin{(t-1)}) - u(t-2) \cdot ((t-2) - \sin{(t-2)}) \]

Break Down the Problem (Part b)

Now consider the second function: \( \frac{e^{-16s}}{s(s^2+2s+4)} \). Again, we'll start by recognizing that a shift and convolution will be needed, given the exponential \( e^{-16s} \) and the quadratic denominator.

Factor the Quadratic (Part b)

Factor the quadratic term \( s^2 + 2s + 4 \) to find simpler terms. The quadratic doesn't factor easily with real numbers, but can be completed using the form: - \( (s + 1)^2 + 3 \).This simplifies application in terms of cosine and sine transformations.

Apply Laplace Properties (Part b)

The inverse Laplace for \( \frac{1}{s((s+1)^2 + 3)} \) can be known from tables:- The inverse is \( u(t)(1 - e^{-t}\cdot \sin{\sqrt{3}t})\). By applying the shifting property: - \( e^{-16s} \frac{1}{s((s+1)^2 + 3)} \) shifts this result to \( u(t-16)(1 - e^{-(t-16)}\sin{\sqrt{3}(t-16)}) \).

Solution for Part b

The inverse Laplace transform of part (b) then is:\[ g(t) = u(t-16) \cdot (1 - e^{-(t-16)} \cdot \sin{\sqrt{3}(t-16)}) \]

Combine Results

Summarizing, the inverse Laplace transforms are:- For part (a): \[ f(t) = u(t-1) \cdot ((t-1) - \sin{(t-1)}) - u(t-2) \cdot ((t-2) - \sin{(t-2)}) \]- For part (b): \[ g(t) = u(t-16) \cdot (1 - e^{-(t-16)} \cdot \sin{\sqrt{3}(t-16)}) \]

Key concepts

Convolution Theorem

The Convolution Theorem is a key tool in the realm of Laplace transforms. It allows us to find the Laplace transform of a product of two time-domain functions by using their respective transforms. This is particularly useful when dealing with the inverse Laplace transforms, such as in our given exercises. The theorem states that if you have two functions, say \( f(t) \) and \( g(t) \), and their Laplace transforms are \( F(s) \) and \( G(s) \), then the Laplace transform of their convolution \( (f * g)(t) \) is given by:

• \( \, (f * g)(t) = \int_0^t f(\tau)g(t-\tau)\, d\tau \)
• \( \, \mathcal{L}\{(f * g)(t)\} = F(s)G(s) \)

The importance of this theorem lies in simplifying complex expressions that involve products of transforms, which can then be inverted more easily. In the context of the exercise, spotting a convolution helps to manage different parts of the function by breaking them down into simpler components.

Shifting Property

The Shifting Property, often called the first shift theorem, is a crucial aspect of Laplace transforms. This property is used to handle exponential terms, such as \( e^{-as} \), which effectively "shifts" the original function in the time domain.

Specifically, if \( F(s) \) is the Laplace transform of \( f(t) \), then \( e^{-as}F(s) \) is the Laplace transform of \( f(t-a)u(t-a) \). The function \( u(t-a) \) is the unit step function which ensures the function is zero before \( t = a \).

This property is highly useful in the given problems, evident with terms like \( e^{-s}F(s) \) in part (a) and \( e^{-16s} \) in part (b). By recognizing these terms, we "shift" the inverse operation, allowing us to use known inverse transforms effectively, rather than grappling with the initially complex expressions.

Quadratic Factorization

Quadratic factorization is a math technique used to simplify expressions, particularly those involving terms like \( s^2 + 2s + 4 \). Factoring transforms these quadratic parts into a sum of squares format, typically making them easier to manage with standard Laplace transform tables.

For example, in part (b) of the exercise, the term \( s^2 + 2s + 4 \) can be expressed as \( (s+1)^2 + 3 \) using the method of completing the square. This factorization reflects in inverse Laplace transforms, where it relates to functions involving sine and cosine.

• Form: \( (s+a)^2 + b^2 \) correlates with sine and damping terms.
• This makes it easier to directly apply known transforms like \( e^{-at} \sin(bt) \).

Completing the square is a powerful tool and is indispensable when tackling expressions that seem difficult to factor at first glance.

Unit Step Function

The Unit Step Function, denoted as \( u(t-a) \), is fundamental in time-domain analysis, particularly when using the Laplace transform. It acts as a switch that "turns on" the function at a specific time, \( t = a \). If \( t < a \), the function value is \( 0 \); if \( t \geq a \), the function value is \( 1 \).

This function is especially useful with the shifting property of Laplace transforms. For instance, when a term like \( e^{-as}F(s) \) is present, it converts to \( f(t-a)u(t-a) \). In practical terms, it allows us to handle delayed responses in a system mathematically.

• The unit step function ensures transform calculations align with real-world scenarios.
• Used extensively in the exercise to manage terms shifted by the exponential terms.

Its proper application is crucial to accurately modeling and predicting system behaviors using mathematical expressions.