The laplace transform of f(t) is
F (s) = 2/s³ - 2e^{-2s}/s³ + e^{-2s}/s² + 2e^{-2s}/s - 3e^{-3s}/s² + 7e^{-3s}/s
How to express f(t) in unit step function?
Since f(t) = t² 0 < t < 2,
= t - 1 for 2 < t < 3 and
= 7 for t > 3
Since f(t) has discontinuities at t = 0, t = 2, t = 3, expressing f(t) in terms of unit step, we have
t² = t²u(t) - t²u(t - 2)
t - 1 = (t - 1)u(t - 2) - (t - 1)u(t - 3)
= (t - 1 + 1 - 1)u(t - 2) - [(t - 1 + 3 - 3)u(t - 3)]
= (t - 2 + 1)u(t - 2) - [(t - 3 + 2)u(t - 3)]
= (t - 2)u(t - 2) + u(t - 2) - (t - 3)u(t - 3) + 2u(t - 3)
f(t) = 7 = 7u(t - 3)
So, f(t) = t²u(t) - t²u(t - 2) + (t - 2)u(t - 2) + u(t - 2) - (t - 3)u(t - 3) - 2u(t - 3) + 7u(t - 3)
How to find the laplace transform of f(t)
[tex]L[u(t - c)f(t - c)}] = e^{-cs} F(s)[/tex]
and [tex]L[t^{n}] = \frac{n!}{s^{n + 1} }[/tex]
Taking the laplace transform, we have
L{f(t)} = L{t²u(t)} - L{t²u(t - 2)} + L{(t - 2)u(t - 2)} + L{u(t - 2)} - L{(t - 3)u(t - 3)} - L{2u(t - 3)} + L{7u(t - 3)}
[tex]F(s) = \frac{2!}{s^{3} } - \frac{e^{-2s} 2!}{s^{3} } + \frac{e^{-2s} 1!}{s^{2} } + \frac{e^{-2s} 1!}{s } - \frac{e^{-3s} 1!}{s^{2} } - \frac{2e^{-3s} 1!}{s^{2} } + \frac{7e^{-3s} 1!}{s} \\= \frac{2!}{s^{3} } - \frac{2e^{-2s}}{s^{3} } + \frac{e^{-2s} }{s^{2} } + \frac{2e^{-2s}}{s} - \frac{e^{-3s}}{s^{2} } - \frac{2e^{-3s}}{s^{2} } + \frac{7e^{-3s}}{s}\\[/tex]
[tex]F (s) = \frac{2}{s^{3} } - \frac{2e^{-2s}}{s^{3} } + \frac{e^{-2s} }{s^{2} } + \frac{2e^{-2s}}{s} - \frac{e^{-3s}}{s^{2} } - \frac{2e^{-3s}}{s^{2} } + \frac{7e^{-3s}}{s}\\[/tex]
So, the laplace transform of f(t) is [tex]F (s) = \frac{2}{s^{3} } - \frac{2e^{-2s}}{s^{3} } + \frac{e^{-2s} }{s^{2} } + \frac{2e^{-2s}}{s} - \frac{3e^{-3s}}{s^{2} } + \frac{7e^{-3s}}{s}\\[/tex]
Learn more about Laplace transform here:
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