Problem 57
Question
In Problems, write each function in terms of unit step functions. Find the Laplace transform of the given function. $$ f(t)=\left\\{\begin{array}{lr} 0, & 0 \leq t<1 \\ t^{2}, & t \geq 1 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The Laplace transform is \( \frac{2e^{-s}}{s^3} \).
1Step 1: Express the Function Using Unit Step Functions
The function \( f(t) \) is piecewise defined. First, identify the intervals: from 0 to 1, the function is 0; from 1 onward, the function is \( t^2 \). Using the unit step function \( u(t-a) \), where \( u(t-a) = 0 \) for \( t
2Step 2: Write the Entire Function Using the Definition of Unit Step
The function \( f(t) \) is then: \( f(t) = 0 + t^2 u(t-1) \). Simplifying, we have \( f(t) = t^2 u(t-1) \). This expression correctly sets \( f(t) = 0 \) for \( 0 \leq t < 1 \) and \( f(t) = t^2 \) for \( t \geq 1 \).
3Step 3: Prepare to Find the Laplace Transform
The Laplace transform of a function \( f(t) \cdot u(t-a) \) can be found using the formula \( L\{f(t-a)u(t-a)\} = e^{-as}L\{f(t-a)\} \). First, replace \( t \) with \( (t-1) \) in \( t^2 \) to get \( (t-1)^2 \), then apply the transform.
4Step 4: Rewrite \( f(t) \) for Laplace Transformation
Substitute \( t^2 \) in the expression as \( (t-1+1)^2 \), which becomes \( (t-1 + 1)^2 = (t-1)^2 + 2(t-1) + 1 \). Here, we focus on \( (t-1)^2 \) for \( t \geq 1 \), which represents the shifted function to transform.
5Step 5: Calculate the Laplace Transform
Apply the Laplace transform using \( L\{(t-a)^n\} = \frac{n!}{s^{n+1}} \) for \( n=2 \), where \( a = 1 \): \[ L\{(t-1)^2 u(t-1) \} = e^{-s} \cdot L\{ (t-1)^2 \} = e^{-s} \cdot \frac{2!}{s^{3}} = \frac{2e^{-s}}{s^3} \].
Key Concepts
Unit Step FunctionPiecewise FunctionsTransform Techniques
Unit Step Function
The unit step function is a fundamental building block for expressing functions that change abruptly at certain points. Often denoted by \( u(t-a) \), this function switches from 0 to 1 at the point \( t = a \). It has the following key properties:
- For \( t < a \), \( u(t-a) = 0 \).
- For \( t \geq a \), \( u(t-a) = 1 \).
Piecewise Functions
Piecewise functions allow us to express a function over different intervals with different expressions. They are particularly useful in situations where a function behaves differently across segments of its domain.
Take the given exercise with \( f(t) \). Its piecewise representation is:
Take the given exercise with \( f(t) \). Its piecewise representation is:
- \( 0 \) for the interval \( 0 \leq t < 1 \)
- \( t^2 \) for \( t \geq 1 \)
Transform Techniques
The Laplace transform technique is a powerful tool for solving differential equations and analyzing linear systems. It transforms time-domain functions into a complex frequency domain, making operations like differentiation and convolution much easier.
When using the Laplace transform on functions involving the unit step function, we utilize the formula:
\[ L\{f(t-a)u(t-a)\} = e^{-as}L\{f(t-a)\} \]
This involves shifting the function \( f(t) \) by \( a \) and applying the transform to the shifted function. In our example, the function \( f(t) \) after incorporating the step function is \( t^2 u(t-1) \), leading to the transform focusing on \( (t-1)^2 \). By applying the formula \( L\{(t-a)^n\} = \frac{n!}{s^{n+1}} \), we find:
\[ L\{(t-1)^2 u(t-1) \} = e^{-s} \cdot \frac{2}{s^3} = \frac{2e^{-s}}{s^3} \]
Understanding how to manipulate functions with these transform techniques can greatly enhance one's capability to solve complex problems across various fields.
When using the Laplace transform on functions involving the unit step function, we utilize the formula:
\[ L\{f(t-a)u(t-a)\} = e^{-as}L\{f(t-a)\} \]
This involves shifting the function \( f(t) \) by \( a \) and applying the transform to the shifted function. In our example, the function \( f(t) \) after incorporating the step function is \( t^2 u(t-1) \), leading to the transform focusing on \( (t-1)^2 \). By applying the formula \( L\{(t-a)^n\} = \frac{n!}{s^{n+1}} \), we find:
\[ L\{(t-1)^2 u(t-1) \} = e^{-s} \cdot \frac{2}{s^3} = \frac{2e^{-s}}{s^3} \]
Understanding how to manipulate functions with these transform techniques can greatly enhance one's capability to solve complex problems across various fields.
Other exercises in this chapter
Problem 56
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