Expert answers to "Derive Laplace Transform of the following: a)f(t)=e−at b)..."

permaneceerc

Answered question

2021-09-19

Derive Laplace Transform of the following:
a) $f (t) = e^{- a t}$
b) $f (t) = t$ c) $f (t) = t^{2}$
d) $\cosh t$

Answer & Explanation

Sally Cresswell

Skilled2021-09-20Added 91 answers

Step 1
we have given f(t).
We have to derive the following laplace transform
Step 2
1) Derive Laplace transform of the following
$f (t) = e^{- a t}$
We know that
$L {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t$
Now, $L {e^{- d t}} = \int_{0}^{\infty} e^{- s t} \cdot e^{- 2 t} d t = \int_{0}^{\infty} e^{- (s + a) t} d t = {[{\frac{e^{- (s + a) t}}{- (s + 2)}}_{0}^{\infty}]}_{0}^{\infty}$
Hence, $L {e^{- d t}} = [0 - \frac{e^{0}}{- (s + 2)}] = \frac{1}{(s + 2)}$
b) f(t)=t
Now, $L {f (t)} = L {t} = \int_{0}^{\infty} e^{- s t} t d t$
$= {[t \frac{e^{- s t}}{- s}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{d t}{d t} \frac{e^{- s t}}{- s} d t$
$= (0 + 0) + \frac{1}{s} {[\frac{e^{- s t}}{(- s)}]}_{0}^{\infty} = \frac{1}{s^{2}} (0 + 1) = \frac{1}{s^{2}}$
Hence, $L {f (t)} = L {t} = \frac{1}{s^{2}}$
c) $f (t) = t^{2}$
Now, $L {f (t)} = L {t^{2}} = \int_{0}^{\infty} e^{- s t} t^{2} d t$
$= {[t^{2} \frac{e^{- s t}}{(- s)}]}_{0}^{\infty} - \int_{0}^{\infty} 2 t \frac{e^{- s t}}{(- s)} d t$
$= (0 - 0) + \frac{2}{s} [{[t \frac{e^{- s t}}{(- s)}]}_{0}^{\infty} - \int_{0}^{\infty} \frac{d t}{d t} \frac{e^{- s t}}{(- s)} d t]$
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