Problem 20 - Advanced Engineering Mathematics

Question

Find either \(F(s)\) or \(f(t)\), as indicated. $$ \mathscr{L}^{-1}\left\\{\frac{(s+1)^{2}}{(s+2)^{4}}\right\\} $$

Step-by-Step Solution

Answer

The inverse Laplace transform is \( \frac{t^2}{2} e^{-2t} \).

Step 1: Identify the Problem

The task is to find the inverse Laplace transform of the given function \( \mathscr{L}^{-1}\left\{\frac{(s+1)^{2}}{(s+2)^{4}}\right\} \).

Step 2: Apply Shifting Theorem

Recognize that the Laplace function can be expressed with a shifting property \( \frac{(s+1)^2}{(s+2)^4} = \left( \frac{s^2}{(s+2)^4} \right) \), allowing us to identify a shift of \( s \to s+1 \). The inverse Laplace transform of \( \frac{s^2}{s^4} \) is already known by the power formula.

Step 3: Use Inverse Laplace Formula

The shifting theorem informs us to use the formula for finding inverse Laplace. The Laplace transform for \( \frac{s^2}{s^4} \) corresponds to \( t^2 \) by the standard table. Therefore, the inverse Laplace of \( \frac{s^2}{s^4} \) is \( \frac{t^2}{2!} = \frac{t^2}{2} \).

Step 4: Apply Time Shifting Property

Given the shift \((s + 2)\), apply the time-shifting property \( f(t-a) U(t-a) \) where \( U(t-a) \) is the Heaviside step function. The transformed function is \( e^{-2t} \frac{(t^2)}{2} \).

Step 5: Construct the Final Solution

After applying the time shift, the inverse Laplace transform is \( \frac{(t^2)}{2} e^{-2t}\). This is the time-domain function \( f(t) \) corresponding to the given Laplace transform.

Key Concepts

Shifting Theorem

Understanding the shifting theorem is crucial when working with Laplace Transforms, especially in moving from frequency to time domain. The Shifting Theorem or First Shifting Theorem tells us how to handle a sub in the Laplace domain. When you have a transformed function like \( \frac{f(s+a)} {g(s+a)} \) instead of \( \frac{f(s)}{g(s)} \), the shifting theorem simplifies this process. The concept here is to shift every occurrence of \(s\) in your original function by a constant \(a\). Hence, for our example \( \frac{(s+1)^{2}}{(s+2)^{4}} \), this is equivalent to handling a shift from \( s \) to \( s+2 \). Recognizing this helps us transform the given function into a simpler form which can directly use known inverse Laplace values.

Heaviside Step Function

The Heaviside Step Function, often denoted as \( U(t-a) \), is crucial when dealing with time shifts in inverse Laplace transforms. It's a mathematical function that switches on at a given time \(a\). For time-shifting properties, the Heaviside function ensures the transformation is correctly aligned with time shifts defined in initial equations. In the given exercise, after applying the shift for \(s\) terms, we adopt the Heaviside function to mirror these shifts back into time terms. If the function is transformed with \( e^{-as} \), its inverse expression results in multiplying the inverse by \( U(t-a) \). This is how we ensure our inverse transform adheres strictly to the mathematical shifts in the problem.

Time Shifting Property

The Time Shifting Property is an essential concept when applying the inverse Laplace Transform, especially when handling systems or processes with delays. It suggests if you move a function in the Laplace domain by \( e^{-as} \), its corresponding time domain function should reflect a delay of \(a\) time units. In time domain, this indicates a multiplication by \(U(t-a)\) and substituting \(t-a\) in place of \(t\). For our solution, where \( s \) is shifted in the domain as \( s+2 \), it mirrors in the time domain with a \(2\)-unit delay with the form \( \frac{t^2}{2}e^{-2t} \). Understanding this property helps to map frequency shifts to correct time delays or advances, translating into the accurate inverse result.