Question

Find \(f(t)\) for each of the following functions a) \(\quad F(s)=\frac{100(s+1)}{s^{2}\left(s^{2}+2 s+5\right)}\) b) \(\quad F(s)=\frac{20 s^{2}}{(s+1)^{3}}\) c) \(\quad F(s)=\frac{40(s+2)}{s(s+1)^{3}}\) d) \(\quad F(s)=\frac{5(s+2)^{2}}{s^{4}(s+1)}\)

Short answer

Apply partial fraction decomposition and inverse Laplace transform on each part's terms, using known tables for specific cases.

Step-by-step solution

Determine Method for Part a

For part (a), we have the function \[ F(s) = \frac{100(s+1)}{s^2(s^2 + 2s + 5)}. \]This requires partial fraction decomposition as it involves a polynomial of higher order in the denominator.

Partial Fraction Decomposition for Part a

Decompose \[ \frac{100(s+1)}{s^2(s^2 + 2s + 5)} \]into terms:\[ \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 2s + 5}. \]Solve for constants by setting common denominators and equating coefficients.

Solve for Constants Part a

Setting up the equation, solve for A, B, C, and D by comparing coefficients. Substitute back into the partial fraction form to find:\[ F(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 2s + 5}. \]

Apply Inverse Laplace for Part a

Use the inverse Laplace transform on each term. Recognize standard transforms:- \( \frac{A}{s} \rightarrow A \)- \( \frac{B}{s^2} \rightarrow Bt \)- \( \frac{Cs+D}{s^2+2s+5} \) will involve a shift and exponential decay.

Determine Method for Part b

For part (b), the function\[ F(s) = \frac{20s^2}{(s+1)^3} \]suggests using a known Laplace transform and properties of transforms like shifting and derivatives.

Use known Laplace Transform for Part b

Recognize that the term \( \frac{20s^2}{(s+1)^3} \) corresponds to the Laplace transform of a polynomial multiplied by an exponential decay term due to \((s+1)^n\). Use the table of Laplace transforms to convert.

Application of Final Inverse for Part b

Utilize the inverse Laplace transformation:- Recognizing the derivative nature of \( s^2 \) term,- The presence of \((s+1)^3\) indicates exponential decay modification.

Break Down Part c

For part (c), with\[ F(s) = \frac{40(s+2)}{s(s+1)^3}, \]use partial fraction decomposition due to polynomial form in the denominator.

Decompose for Part c

Express as sum of simpler fractions:\[ \frac{A}{s} + \frac{B}{(s+1)} + \frac{C}{(s+1)^2} + \frac{D}{(s+1)^3}. \]Set equation, find A, B, C, D by polynomial comparison.

Inverse Laplace for Part c

After solving for constants, convert each term using inverse Laplace transforms for terms with \(s\) and \((s+1)^n\).

Part d Breakdown

For part (d), function is \[ F(s) = \frac{5(s+2)^2}{s^4(s+1)}, \]involving higher powers and needing decomposition/exponentials.

Decomposition and Analysis for Part d

Decompose into partial fractions:\[ \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s^4} + \frac{E}{s+1}. \]Find constants, consider how each inverse impacts (time shift, derivatives).

Final Inverse for Part d

Complete inverse Laplace on each component, applying polynomial, shift, and decay effects. Create final \( f(t) \).

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex rational expressions. When we have a higher order polynomial in the denominator, it can be helpful to express this complex fraction as a sum of simpler fractions. Doing this makes finding inverse Laplace transforms easier.
For instance, consider the expression \( \frac{100(s+1)}{s^2(s^2 + 2s + 5)} \). This expression can be decomposed into simpler terms like \( \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 2s + 5} \).

• The term \( \frac{A}{s} \) might represent a simple scaling factor.
• \( \frac{B}{s^2} \) is a polynomial multiplied by \( t \) when transformed back.
• \( \frac{Cs+D}{s^2 + 2s + 5} \) introduces a complexity with shifts and exponential decay, requiring a thoughtful look at the constant and linear terms in the numerator.

By equating coefficients from both sides of the decomposed equation, we solve for unknown constants like \( A, B, C, \) and \( D \). This step-by-step simplification leverages algebraic techniques to facilitate a deeper understanding of the function and make subsequent calculations more manageable.

Laplace Transforms Table

The Laplace Transforms Table is an explorer's map in the world of Laplace transforms. It guides us in converting complex functions of \( s \) into time domain functions. This is critical for both simplification and solving differential equations.
When working with expressions like \( \frac{20s^2}{(s+1)^3} \), it's useful to recall standard transformations listed in these tables. Each known form relates to a specific transformation, helping us translate algebraic terms directly to time functions.

• The entry \( \frac{1}{s-a} \rightarrow e^{at} \) indicates that a shift in the \( s \) domain translates to exponential growth or decay in time.
• Polynomial terms such as \( s^n \) correspond to higher order derivatives in the time domain.

Using this table lets us quickly ascertain the behavior of a system, recognize patterns, and reduce time spent calculating manually. Understanding these transforms means you can quickly apply the inverse Laplace transform and gain insights into systems described by these functions.

Polynomial Functions

Polynomial functions play a vital role in the structure of Laplace transform problems. They are formed by expressions including terms like \( s^2, s^3, \) etc. In transformation problems, especially in inverse Laplace transforms, these polynomials can represent derivatives or integrations.
Consider the term \( s^2 \) in \( \frac{20s^2}{(s+1)^3} \). Here, \( s^2 \) in the numerator points to a differentiation operation, which can be seen as successive derivatives in the time domain.

• Higher powers indicate more complex behavior when transformed back to the time domain.
• Each degree has unique implications for how the function behaves over time.

For instance, a term like \( \frac{1}{s^n} \) when inverted could represent a time function that increases in complexity as \( n \) increases, illustrating increasing derivatives or integrations in time. Inverse transformations involving these polynomial terms help model various natural and engineered systems.

Exponential Decay

Exponential decay in the context of Laplace transforms refers to the way functions diminish over time, often modeled with expressions like \( e^{-at} \).
This is particularly noted in transformations involving terms such as \( (s+1)^{-n} \). As seen in problems like part (b) and (d), where \( (s+1)^3 \) or \( (s+1) \) parts appear, these indicate an exponential decay behavior that affects how the solution transforms back to time domain functions.

• The term influences the damping and stability of the system you're analyzing.
• It provides insight into how quickly or slowly a system returns to equilibrium.

Exponential decay is fundamental in oscillatory systems, control systems, and natural phenomena like radioactive decay. Recognizing this transformation lets you predict behavior over time, crucial for applications requiring stability predictions or system analysis.