Let f(t) be a function on [0,∞). The Laplace transform of fis the function F defined by the integral F(s)=∫0∞e−stf(t)dt . Use this definition to determine the Laplace transform of the following function. f(t)={1−t0&lt;t&lt;101&lt;t

Step 1 We know that , F(s)=∫0∞e−stf(t)dt Here f(t)={1−t0&lt;t&lt;101&lt;t from F(s)=∫0∞(f(t))e−stdt=∫01(1−t)e−stdt+∫1∞0⋅e−stdt =∫01(1−t)e−stdt+0 F(s)=[(1−t)e−st−s−(−1)e−sts2]01 F(s)=1s+e−ss−1s2 Step 2 This is required Laplace Transform.

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Elleanor Mckenzie

Answered question

2020-12-27

Let f(t) be a function on $[0, \infty)$ . The Laplace transform of fis the function F defined by the integral $F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ . Use this definition to determine the Laplace transform of the following function.
$f (t) = {\begin{matrix} 1 - t & 0 < t < 1 \\ 0 & 1 < t \end{matrix}$

Answer & Explanation

grbavit

Skilled2020-12-28Added 109 answers

Step 1
We know that , $F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$
Here $f (t) = {\begin{matrix} 1 - t & 0 < t < 1 \\ 0 & 1 < t \end{matrix}$
from $F (s) = \int_{0}^{\infty} (f (t)) e^{- s t} d t = \int_{0}^{1} (1 - t) e^{- s t} d t + \int_{1}^{\infty} 0 \cdot e^{- s t} d t$
$= \int_{0}^{1} (1 - t) e^{- s t} d t + 0$
$F (s) = {[\frac{(1 - t) e^{- s t}}{- s} - \frac{(- 1) e^{- s t}}{s^{2}}]}_{0}^{1}$
$F (s) = \frac{1}{s} + \frac{e^{- s}}{s} - \frac{1}{s^{2}}$
Step 2
This is required Laplace Transform.

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