Let $f$ be a function defined on $[0, \infty)$. Then the function $F$ defined by
$F(s)=\int_0^{\infty} e^{-s t} f(t) d t$
is said to be the Laplace transform of $f$. The domain of $F(s)$ is the set of values of $s$ for which the given improper integral converges.
Use the definition of a Laplace transform to find $L \{f(t)\}$. (Write your answer as a function of $s$.)
$\begin{array}{c}
f(t)=\left\{\begin{array}{rr}
-1, & 0 \leq t<1 \\
1, & t \geq 1
\end{array}\right. \\
c\{f(t)\}=\square$
$\end{array}$
Answer (2)
Define the function f ( t ) piecewise and apply the Laplace transform definition.
Split the integral into two parts: from 0 to 1 and from 1 to infinity.
Evaluate each integral separately using the fundamental theorem of calculus.
Combine the results to obtain the Laplace transform: s 2 e − s − 1 .
Explanation
Problem Analysis We are given a function f ( t ) defined as follows:
Function Definition f ( t ) = { − 1 , 1 , 0 ≤ t < 1 t ≥ 1
Laplace Transform Definition We are asked to find the Laplace transform F ( s ) of f ( t ) , which is defined as:
Laplace Transform Formula F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t
Splitting the Integral To find the Laplace transform, we need to split the integral into two parts based on the definition of f ( t ) :
Integral Setup F ( s ) = ∫ 0 1 e − s t ( − 1 ) d t + ∫ 1 ∞ e − s t ( 1 ) d t
Evaluating First Integral Let's evaluate the first integral:
Result of First Integral ∫ 0 1 − e − s t d t = s e − s t 0 1 = s e − s − s 1
Evaluating Second Integral Now, let's evaluate the second integral:
Result of Second Integral ∫ 1 ∞ e − s t d t = − s e − s t 1 ∞ = 0 − ( − s e − s ) = s e − s , assuming 0"> s > 0
Combining Results Finally, let's combine the results:
Final Laplace Transform F ( s ) = s e − s − s 1 + s e − s = s 2 e − s − 1
Final Answer Therefore, the Laplace transform of f ( t ) is:
Final Answer F ( s ) = s 2 e − s − 1 for 0"> s > 0
Examples
Laplace transforms are used in electrical engineering to analyze circuits. For example, if f ( t ) represents the voltage signal applied to a circuit, then F ( s ) can be used to determine the circuit's response to that signal. This allows engineers to design circuits that behave in a predictable way. They are also used in control systems to analyze the stability and performance of systems.
The Laplace transform of the piecewise function f ( t ) is F ( s ) = s 1 for 0"> s > 0 .
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