Chapter 22: Problem 2
Find the Laplace transform of the following: (a) \(0.6 t\) (b) \(6-6 t\) (c) \(2(t+1)\) (d) \((t+1)(t-1)\) (e) \(3 t^{4}-2\)
Short Answer
Expert verified
(a) \( \frac{0.6}{s^2} \); (b) \( \frac{6}{s} - \frac{6}{s^2} \); (c) \( \frac{2}{s^2} + \frac{2}{s} \); (d) \( \frac{2}{s^3} - \frac{1}{s} \); (e) \( \frac{72}{s^5} - \frac{2}{s} \)."
Step by step solution
01
Laplace Transform of 0.6t
The formula for the Laplace transform of a function is \( \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \, dt \). For \( f(t) = 0.6t \), we have:\[ \mathcal{L}\{0.6t\} = \int_0^{\infty} e^{-st} (0.6t) \, dt \]Using the known property \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \), with \( n = 1 \), we get:\[ \mathcal{L}\{0.6t\} = 0.6 \cdot \frac{1}{s^2} = \frac{0.6}{s^2} \]
02
Laplace Transform of 6-6t
For \( f(t) = 6 - 6t \), we apply the linearity of the Laplace transform:\[ \mathcal{L}\{6 - 6t\} = 6 \mathcal{L}\{1\} - 6 \mathcal{L}\{t\} \]\[ \mathcal{L}\{1\} = \frac{1}{s} \] and \( \mathcal{L}\{t\} = \frac{1}{s^2} \), so:\[ \mathcal{L}\{6\} = 6 \cdot \frac{1}{s} = \frac{6}{s} \]\[ \mathcal{L}\{6t\} = 6 \cdot \frac{1}{s^2} = \frac{6}{s^2} \]Hence:\[ \mathcal{L}\{6 - 6t\} = \frac{6}{s} - \frac{6}{s^2} \]
03
Laplace Transform of 2(t+1)
For \( f(t) = 2(t+1) \), again use linearity:\[ \mathcal{L}\{2(t+1)\} = 2 \mathcal{L}\{t\} + 2 \mathcal{L}\{1\} \]We know:\[ \mathcal{L}\{t\} = \frac{1}{s^2} \] and \( \mathcal{L}\{1\} = \frac{1}{s} \).Thus:\[ \mathcal{L}\{2t\} = 2 \cdot \frac{1}{s^2} = \frac{2}{s^2} \]\[ \mathcal{L}\{2\} = 2 \cdot \frac{1}{s} = \frac{2}{s} \]Therefore:\[ \mathcal{L}\{2(t+1)\} = \frac{2}{s^2} + \frac{2}{s} \]
04
Laplace Transform of (t+1)(t-1)
First, expand the function:\[ (t+1)(t-1) = t^2 - 1 \]Then apply the Laplace transform:\[ \mathcal{L}\{t^2 - 1\} = \mathcal{L}\{t^2\} - \mathcal{L}\{1\} \]Using \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \):\[ \mathcal{L}\{t^2\} = \frac{2!}{s^3} = \frac{2}{s^3} \]Therefore:\[ \mathcal{L}\{(t+1)(t-1)\} = \frac{2}{s^3} - \frac{1}{s} \]
05
Laplace Transform of 3t^4 - 2
For \( f(t) = 3t^4 - 2 \), use linearity again:\[ \mathcal{L}\{3t^4 - 2\} = 3 \mathcal{L}\{t^4\} - 2 \mathcal{L}\{1\} \]With \( n = 4 \), we know:\[ \mathcal{L}\{t^4\} = \frac{4!}{s^5} = \frac{24}{s^5} \]So:\[ \mathcal{L}\{3t^4\} = 3 \cdot \frac{24}{s^5} = \frac{72}{s^5} \]And:\[ \mathcal{L}\{2\} = 2 \cdot \frac{1}{s} = \frac{2}{s} \]Thus:\[ \mathcal{L}\{3t^4 - 2\} = \frac{72}{s^5} - \frac{2}{s} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linearity of Laplace Transform
The linearity of the Laplace transform is a remarkable property that simplifies the transformation of composite functions. Simply put, if you have two functions, say \( f(t) \) and \( g(t) \), their Laplace transforms \( \mathcal{L}\{f(t)\} \) and \( \mathcal{L}\{g(t)\} \), and constants \( a \) and \( b \), then the transform of a linear combination is: \[ \mathcal{L}\{af(t) + bg(t)\} = a \cdot \mathcal{L}\{f(t)\} + b \cdot \mathcal{L}\{g(t)\} \] This rule is akin to the distributive property of arithmetic, allowing us to handle each term separately and then combine the results.
- When applied to the exercise examples, like for \(6 - 6t\), we compute \( \mathcal{L}\{6\} \) and \( \mathcal{L}\{6t\} \) independently, and subtract the results to find the Laplace transform of the entire expression.
Laplace Transform of Polynomial Functions
Polynomial functions, those like \( t^n \), learn to shine under the Laplace transform, thanks to a convenient formula. The Laplace transform of \( t^n \) is given by: \[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \]in which \( n! \) repeats the multiplication of all natural numbers up to \( n \).
- You can see this clearly when looking at simpler polynomial functions such as \( f(t) = 0.6t \). In this instance, \( n = 1 \), and the transform simplifies to \( 0.6 \cdot \frac{1}{s^2} \).
Integration in Laplace Transform Calculations
The backbone of the Laplace transform is integration. Particularly, transforming a function \( f(t) \) involves an integral with an exponential decay factor: \[ \mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \, dt \] This decay, created by \( e^{-st} \), ensures the integration is convergent, meaning it settles into a finite value even as \( t \) stretches to infinity. When handling functions like polynomials, the integral takes advantage of the exponential term to scale down the function over time, providing a more robust representation in the \( s \)-domain.
- For example, consider \( f(t) = 0.6t \), where the integration simplifies thanks to known formulas for polynomial functions.
- Each power of \( t \) integrates separately, maintaining order and accuracy in calculations.