6.5 Second Shifting Theorem: the \(t\)-Shift
Definition 6.5.1 (Unit step function). For \(a\geq 0\) the unit step, or Heaviside, function is \[ U(t-a) = \begin {cases} 0, & \text {if} \hspace {0.4cm} t<a\\ 1, & \text {if}\hspace {0.4cm} t>a .\\ \end {cases} \]
It is the switch of the subject: multiplying by \(U(t-a)\) turns a function off until time \(a\) and on thereafter. In particular \[f(t-a)\,U(t-a) = \begin {cases} 0 & \text {if} \hspace {0.2cm} t<a\\ f(t-a) & \text {if}\hspace {0.2cm} t>a ,\\ \end {cases} \] which is the graph of \(f\) delayed by \(a\) and held at zero before it. Written this way the next theorem needs no cases at all.
Theorem 6.5.2 (Second shifting theorem). Let \(f\) have transform \(F(s)\), and let \(a\geq 0\). Then the delayed function \(f(t-a)\,U(t-a)\) has transform \[\mathcal {L}\big \{f(t-a)\,U(t-a)\big \} = e^{-as}F(s) .\]
A delay in \(t\) therefore costs a factor \(e^{-as}\) in \(s\), just as the first shifting theorem showed that a factor \(e^{at}\) in \(t\) costs a shift in \(s\). The two theorems are the same statement read in opposite directions, which is why they are usually met together.
It \(\mathcal {L}\{f(t)\} = F(s),\) then \(\displaystyle {\mathcal {L}\{f(t-a)U(t-a)\}e^{-as}F(s)}\)
Solution. \[\mathcal {L}^{-1} \Big \{e^{-as}F(s)\Big \} = f(t-a) U(t-a)\]
Here, \(F(s) = \dfrac {1}{s^3},\hspace {1cm} a = 3\).
\[f(t) = \mathcal {L}^{-1}\big \{\frac {1}{s^3}\big \} = \frac {1}{2}t^2\]
\[\implies \hspace {0.5cm} f(t-3) = \frac {1}{2}(t-3)^2\]
\[\mathcal {L}^{-1}\Big \{ \frac {1}{s^3}e^{-3s}\Big \} = \frac {1}{2}(t-3)^2U(t-3)\]
\[ \mathcal {L}^{-1}\Big \{ \frac {1}{s^3}e^{-3s}\Big \} = \begin {cases} 0 & t<3\\\\ \dfrac {1}{2}(t -3)^2 & t>3\\ \end {cases} \]
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