Question

Write the function in terms of unit step functions. find the laplace transform of the given function. f(t) = t, 0 ≤ t < 5 0, t ≥ 5

Answer 1

The Laplace transform of the given function is $f(s)=\frac{1-e^{-5s}(1-5s) }{s^{2} }$

Given,

$f(t)=\left \{ {{t, 0\leq t < 5} \atop {0,t\geq 5}} \right.$

Write the function in unit step function.

$f(t)=t(u_{0}(t)-u_{5} (t))$

      = $t(1-u_{5}(t))$                       ∵$u_{0} (t)=1$

      = $t-tu_{5} (t)$

$f(t)=t-tu_{5} (t)$

In order to make it easier to take the Laplace transform of the function, we can follow the steps:

$f(t)=t-tu_{5} (t)$

      = $t-(t-5+5)u_{5} (t)$

      = $t-(t-5)u_{5} (t)+5u_{5} (t)$

$f(t)=t-(t-5)u_{5} (t)+5u_{5} (t)$

We need to find the Laplace transform of f(t).

Apply Laplace transform on both sides,

£$[f(t)=$£$(t)-$£$(t-5)u_{5} (t))+5$£$(u_{5} (t))$

         = $\frac{1}{s^{2} } -e^{-s}$£$(t)+5(\frac{e^{-5s} }{s} )$

         = $\frac{1}{s^{2} } -e^{-5s} (\frac{1}{s^{2} } )+\frac{5se^{-5s} }{s^{2} }$

        = $\frac{1-e^{-5s}+5se^{-5s} }{s^{2} }$

       = $\frac{1-e^{-5s}(1-5s) }{s^{2} }$

$f(s) =\frac{1-e^{-5s}(1-5s) }{s^{2} }$

Learn more about Laplace transform here: brainly.com/question/17190535

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