The Laplace transform of t^2 u(t-1) is equal to e-s[2/s3 + 2/s2 +1/s]. Here we find the Laplace of t2u(t-1) using the second shifting property of Laplace transforms.
The formula of the Laplace of t2u(t-1) is given as
L{t2u(t-1)} = $e^{-s}\Big[ \dfrac{2}{s^3} + \dfrac{2}{s^2} +\dfrac{1}{s}\Big]$.
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What is the Laplace of t2u(t-1)?
| Answer: The Laplace of t2u(t-1) is e-s[2/s3 + 2/s2 +1/s]. |
Explanation:
To find the Laplace of t2u(t-1) we will use the second shifting theorem of Laplace transforms which says that if L{f(t)} = F(s), then
L{f(t-a) u(t-a)} = e-as F(s) for a>0 …(∗)
Comparing f(t-a) u(t-a) with the given function t2u(t-1), we get that
a=1 and f(t-1)=t2.
So f(t) = (t+1)2 = t2+2t+1.
Hence, F(s) = L{t2+2t+1} = 2/s3 + 2/s2 +1/s, by the formula L{tn} = n!/sn+1.
Now, using the formula (∗), we obtain that
L{t2u(t-1)} = $e^{-s}\Big[ \dfrac{2}{s^3} + \dfrac{2}{s^2} +\dfrac{1}{s}\Big]$.
So the Laplace transform of t2u(t-1) is equal to e-s[2/s3 + 2/s2 +1/s], and this is proved by using the second shifting property of Laplace transforms.
Related Topics:
Laplace transform of unit step function, L{u(t)}
Find the Laplace transform of u(t-1)
Find the Laplace transform of u(t-2)
FAQs
Answer: The Laplace transform of t2u(t-1) is equal to e-s[2/s3 + 2/s2 +1/s], that is, L{t2u(t-1)} = e-s[2/s3 + 2/s2 +1/s].
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