For f(t)=e−t−1 , find the Laplace transform of {f(t)t}. Then , for g(t)=e−t−2 , examine if the Laplace transform of {g(t)t} exists.

Name: For f(t)=e−t−1 , find the Laplace transform of {f(t)t}. Then , for g(t)=e−t−2 , examine if the Laplace transform of {g(t)t} exists.
Uploaded: 2022-02-23T17:22:57+00:00
Description: For f(t)=e−t−1 , find the Laplace transform of {f(t)t}. Then , for g(t)=e−t−2 , examine if the Laplace transform of {g(t)t} exists.

Step 1 Given that, f(t)=e−t−1 Now , we know laplace transformation of f(t) is L(f(t)) is given by L{f(t)}=∫0∞f(t)e−stdt Also we know , property of unilafenal Linear transformation is L[f(t)t]=∫0∞ F(w)dw : where F(w) is the Laplace transformation of f(t) Now , Laplace transformation of f(t)=e−t−1 is F(s)=L{f(t)}=L{e−t−1}=∫0∞(e−t−1)e−stdt =∫0∞(e−(s+1)t−e−st)dt =[e−(s+1)t−(s+1)+e−sts]0∞ =1s+1−1s.(since e−∞=0) Now , linear transformation of {f(t)t} is L{f(t)t}=L{e−t−1t}=∫s∞F(w)dw =∫s∞[1w+1−1w]dw =[ln⁡(w+1)−ln⁡(w)]s∞ =[ln⁡(w+1w)]s∞ =−ln⁡(s+1s).⟨limw→∞ln⁡(w+1w)=0⟩ =ln⁡(ss+1) But for g(t)=e−t−2 G(s)=L{g(t)}=1s+1−2s . Simply as previous Now , L{g(t)t}=L{e−t−2t}=∫s∞G(w)dw

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geduiwelh

Answered question

2021-09-17

For $f (t) = e^{- t} - 1$ , find the Laplace transform of ${\frac{f (t)}{t}}$ . Then , for $g (t) = e^{- t} - 2$ , examine if the Laplace transform of ${\frac{g (t)}{t}}$ exists.

Answer & Explanation

tabuordg

Skilled2021-09-18Added 99 answers

Step 1
Given that, $f (t) = e^{- t} - 1$
Now , we know laplace transformation of $f (t)$ is $L (f (t))$ is given by $L {f (t)} = \int_{0}^{\infty} f (t) e^{- s t} d t$
Also we know , property of unilafenal Linear transformation is $L [\frac{f (t)}{t}] = \int_{0}^{\infty}$
F(w)dw : where F(w) is the Laplace transformation of f(t)
Now , Laplace transformation of $f (t) = e^{- t} - 1$ is
$F (s) = L {f (t)} = L {e^{- t} - 1} = \int_{0}^{\infty} (e^{- t} - 1) e^{- s t} d t$
$= \int_{0}^{\infty} (e^{- (s + 1) t} - e^{- s t}) d t$
$= {[\frac{e^{- (s + 1) t}}{- (s + 1)} + \frac{e^{- s t}}{s}]}_{0}^{\infty}$
$= \frac{1}{s + 1} - \frac{1}{s} . (since e^{- \infty} = 0)$
Now , linear transformation of ${\frac{f (t)}{t}}$ is
$L {\frac{f (t)}{t}} = L {\frac{e^{- t} - 1}{t}} = \int_{s}^{\infty} F (w) d w$
$= \int_{s}^{\infty} [\frac{1}{w + 1} - \frac{1}{w}] d w$
$= {[\ln (w + 1) - \ln (w)]}_{s}^{\infty}$
$= {[\ln (\frac{w + 1}{w})]}_{s}^{\infty}$
$= - \ln (\frac{s + 1}{s}) . ⟨ lim_{w \to \infty} \ln (\frac{w + 1}{w}) = 0 ⟩$
$= \ln (\frac{s}{s + 1})$
But for $g (t) = e^{- t} - 2$
$G (s) = L {g (t)} = \frac{1}{s + 1} - \frac{2}{s}$ . Simply as previous

Now , $L {\frac{g (t)}{t}} = L {\frac{e^{- t} - 2}{t}} = \int_{s}^{\infty} G (w) d w$

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