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 $s$.)
$\begin{array}{r}
f(t)=\left\{\begin{array}{lr}
0, & 0 \leq t<\pi / 2 \\
\cos (t), & t \geq \pi / 2
\end{array}\right. \\
L \{f(t)\}=\square \\
x
\end{array}$
Answer (2)
The Laplace transform is defined as L { f ( t )} = ∫ 0 ∞ e − s t f ( t ) d t .
Substitute the given piecewise function into the Laplace transform integral, which simplifies to ∫ π /2 ∞ e − s t cos ( t ) d t .
Evaluate the integral using integration by parts twice.
The final Laplace transform is − s 2 + 1 e − 2 π s .
Explanation
Setting up the Laplace Transform We are asked to find the Laplace transform of the function f ( t ) = { 0 , cos ( t ) , 0 ≤ t < π /2 t ≥ π /2 By definition, the Laplace transform is given by L { f ( t )} = ∫ 0 ∞ e − s t f ( t ) d t Since f ( t ) = 0 for 0 ≤ t < π /2 , the integral becomes L { f ( t )} = ∫ π /2 ∞ e − s t cos ( t ) d t
First Integration by Parts To evaluate the integral ∫ π /2 ∞ e − s t cos ( t ) d t , we will use integration by parts twice. Let u = cos ( t ) and d v = e − s t d t . Then d u = − sin ( t ) d t and v = − s 1 e − s t .
∫ e − s t cos ( t ) d t = − s 1 e − s t cos ( t ) − s 1 ∫ e − s t sin ( t ) d t
Second Integration by Parts Now, integrate ∫ e − s t sin ( t ) d t by parts. Let u = sin ( t ) and d v = e − s t d t . Then d u = cos ( t ) d t and v = − s 1 e − s t .
∫ e − s t sin ( t ) d t = − s 1 e − s t sin ( t ) + s 1 ∫ e − s t cos ( t ) d t
Solving for the Integral Substitute this back into the first integration by parts: ∫ e − s t cos ( t ) d t = − s 1 e − s t cos ( t ) − s 1 ( − s 1 e − s t sin ( t ) + s 1 ∫ e − s t cos ( t ) d t ) Rearrange to solve for the integral: ∫ e − s t cos ( t ) d t = − s 1 e − s t cos ( t ) + s 2 1 e − s t sin ( t ) − s 2 1 ∫ e − s t cos ( t ) d t ( 1 + s 2 1 ) ∫ e − s t cos ( t ) d t = − s 1 e − s t cos ( t ) + s 2 1 e − s t sin ( t ) ∫ e − s t cos ( t ) d t = 1 + s 2 e − s t ( − s cos ( t ) + sin ( t ) )
Evaluating the Definite Integral Evaluate the definite integral: ∫ π /2 ∞ e − s t cos ( t ) d t = b → ∞ lim [ s 2 + 1 e − s t ( − s cos ( t ) + sin ( t )) ] π /2 b = b → ∞ lim s 2 + 1 e − s b ( − s cos ( b ) + sin ( b )) − s 2 + 1 e − s ( π /2 ) ( − s cos ( π /2 ) + sin ( π /2 )) = 0 − s 2 + 1 e − s π /2 ( 0 + 1 ) = − s 2 + 1 e − s π /2 assuming 0"> s > 0 .
Final Result Therefore, the Laplace transform of the given function is L { f ( t )} = − s 2 + 1 e − s π /2 However, since we are looking for L { f ( t )} = ∫ π /2 ∞ e − s t cos ( t ) d t , and we know that cos ( t + π /2 ) = − sin ( t ) , we can rewrite the integral as ∫ π /2 ∞ e − s t cos ( t ) d t = ∫ 0 ∞ e − s ( u + π /2 ) cos ( u + π /2 ) d u = ∫ 0 ∞ e − s ( u + π /2 ) ( − sin ( u )) d u = − e − s π /2 ∫ 0 ∞ e − s u sin ( u ) d u = − e − s π /2 s 2 + 1 1 Thus, L { f ( t )} = − s 2 + 1 e − 2 π s
Examples
Laplace transforms are used in electrical engineering to analyze circuits. For example, if f ( t ) represents the voltage signal in a circuit, its Laplace transform F ( s ) can be used to determine the circuit's response to that signal. This is particularly useful for analyzing circuits with piecewise-defined inputs, like the one in this problem, where the input signal changes abruptly at a certain time.
The Laplace transform of the given function f ( t ) is L { f ( t )} = − s 2 + 1 e − s 2 π for 0"> s > 0 . This was obtained by evaluating the integral of the piecewise definition of the function and using integration by parts.
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