\[\mathscr{L}\left\{ f(t) \right\}(s) := \int_{t=0^-}^{+\infty} e^{-st} f(t) \, dt = F(s) \ .\]
13.2. Properties
Linearity.
\[\mathscr{L}\{ a f(t) + b g(t) \}(s) = a F(s) + b G(s)\]
Dirac delta.
\[\mathscr{L}\left\{ \delta(t) \right\} = \int_{t=0^-}^{+\infty} \delta(t) \, e^{st} \, dt = 1 \]
Time delay. If \(f(t) = 0\) for \(t < 0\) (“causality”), for \(\tau > 0\),
\[\mathscr{L}\{ f(t-\tau) \}(s) = e^{-s \tau} F(s)\]
Proof readily follows direct computation with change of variable \(z = t - \tau\), \(dt = dz\)
\[\mathscr{L}\{ f(t - \tau) \}(s) = \int_{t=0^-}^{+\infty} f(t-\tau) e^{-s t} \, dt = \int_{z = - \tau}^{+\infty} f(z) e^{-s z } \, dz \, e^{-s \tau} = \int_{z = 0}^{+\infty} f(z) e^{-s z } \, dz \, e^{-s \tau} = e^{-s \tau} F(s) \ . \]
“Frequency shift”
\[\mathscr{L}\{ f(t) e^{a t} \}(s) = F(s-a)\]
Direct computation gives
\[\mathscr{L}\{ f(t) e^{a t} \}(s) = \int_{t=0^-}^{+\infty} f(t) e^{a t} e^{-st} \, dt = \int_{t=0^-}^{+\infty} f(t) e^{-(s-a)t} \, dt = F(s-a)\]
Derivative.
\[\mathscr{L}\{ f'(t) \}(s) = s F(s) - f(0^-) \ .\]
Proof readily follows direct computation, with integration by parts
\[\mathscr{L}\{ f'(t) \}(s) = \int_{t=0^-}^{+\infty} f'(t) e^{-s t} \, dt = \left[ f(t) e^{-s t} \right]|_{t = 0^-}^{+\infty} + s \int_{t=0^-}^{+\infty} f(t) e^{-s t} \, dt = s F(s) - f(0^-) \ ,\]
provided that \(\lim_{s \rightarrow +\infty} f(t) e^{-s t} = 0\).
Integral.
\[\mathscr{L}\left\{ \int_{\tau=0}^{t} f(\tau) \, d \tau \right\}(s) = \frac{1}{s} F(s) \ .\]
Proof readily follows direct computation, with integration by parts
\[\mathscr{L}\left\{ \int_{\tau=0^-}^{t} f(\tau) \, d \tau \right\}(s) = \int_{t=0^-}^{+\infty} \int_{\tau=0^-}^{t} f(\tau) \, d \tau e^{-s t} \, dt = \left[ -\frac{e^{-st}}{s} \int_{\tau=0^-}^{t} f(\tau) \, d\tau \right]_{t=0}^{+\infty} + \frac{1}{s} \int_{t=0}^{+\infty} f(t) e^{-s t} \, dt = \frac{1}{s} F(s) \ ,\]
provided that \(\int_{\tau=0^-}^{0} f(\tau) d \tau = 0\) and \(\lim_{t \rightarrow +\infty}\frac{e^{-st}}{s} \int_{\tau=0^-}^{+\infty} f(\tau) \, d \tau = 0\).
Convolution.
(13.1)\[\begin{split}\begin{aligned}
\mathscr{L}\left\{ f(t) \ast g(t) \right\}
& = \int_{t=0^-}^{+\infty} \int_{\tau=-\infty}^{+\infty} f(t-\tau) g(\tau) \, d \tau \, e^{-s t }\, dt = && (1) \\
& = \int_{\tau=-\infty}^{+\infty} \int_{z=-\tau^-}^{+\infty} f(z) g(\tau) \, e^{-s (z + \tau)}\, d\tau \, dz = (2) \\
& = \int_{z=0^-}^{+\infty} f(z) \, e^{-s z }\, dz \int_{\tau=0^-}^{+\infty} g(\tau) e^{-s \tau} = \\
& = \mathscr{L}\{ f(t) \}(s) \, \mathscr{L}\{ g(t) \} (s) \ .
\end{aligned}\end{split}\]
having performed the change of coordinates \(z = t - \tau\), \(\tau = \tau\), with unitary Jacobian,
\[\frac{\partial(t,\tau)}{\partial(z,\tau)} = \partial_z t \partial_{\tau} \tau - \partial_z \tau \partial_z t = 1 \cdot 1 - 1 \cdot 0 = 1 ,\]
given the proper description of the domain of integration summarised in the extremes of integration in (1), and causality - i.e. all the functions \(f(t)\) are identically zero for \(t < 0\) - in (2).
Initial value. If …
\[f(0^+) = \lim_{s \rightarrow + \infty} s F(s)\]
From direct computation,
\[\begin{split}\begin{aligned}
\lim_{s \rightarrow +\infty} s F(s)
& = \lim_{s \rightarrow +\infty} s \int_{t = 0^-}^{+\infty} f(t) \, e^{-st} \, dt = \\
& = \lim_{s\rightarrow + \infty} \left\{ \left[s \left(-\frac{e^{-st}}{s}\right)f(t) \right]\bigg|_{t=0}^{+\infty} + \int_{t=0}^{+\infty} e^{-st} f'(t) \, dt \right\} = \\
& = \lim_{s \rightarrow +\infty} \left\{ \left[-e^{-st} f(t) \right]\bigg|_{t=0}^{+\infty} + \int_{t=0}^{+\infty} e^{-st} f'(t) \, dt \right\} = \\
& = f(0) \ ,
\end{aligned}\end{split}\]
provided that \(\lim_{s \rightarrow +\infty} \lim_{t \rightarrow +\infty} e^{-s t} f(t) = 0\) and \(\lim_{s \rightarrow + \infty} \int_{t=0}^{+\infty} e^{-st} f'(t) \, dt = 0\).
Final value. If …
\[f(+\infty) = \lim_{s \rightarrow 0} s F(s)\]
From direct computation (todo check and/or explain proof),
\[\begin{split}\begin{aligned}
\lim_{s \rightarrow 0} s F(s)
& = \lim_{s \rightarrow 0} s \int_{t = 0^-}^{+\infty} f(t) \, e^{-st} \, dt = \\
& = \lim_{s \rightarrow 0} \left\{ \left[s \left(-\frac{e^{-st}}{s}\right)f(t) \right]\bigg|_{t=0}^{+\infty} + \int_{t=0}^{+\infty} e^{-st} f'(t) \, dt \right\} = \\
& = \lim_{s \rightarrow 0} \left\{ \left[-e^{-st} f(t) \right]\bigg|_{t=0}^{+\infty} + \int_{t=0}^{+\infty} e^{-st} f'(t) \, dt \right\} = \\
& = f(0) + f(+\infty) - f(0) = f(+\infty) \ ,
\end{aligned}\end{split}\]
provided that \(\lim_{s \rightarrow 0} \lim_{t \rightarrow +\infty} e^{-s t} f(t) = 0\).
