Integral transforms, operational calculus Laplace transform: operational rules Theorem

Time shift theorem

Builds on Laplace transform Heaviside step function

Statement

Let $f \colon [0, \infty) \to \C$ be integrable on each interval $[0, T]$ , and let $F$ be its Laplace transform. Let $a \ge 0$ be real, and let $u$ be the Heaviside step function. Let $g(t) = u(t - a) f(t - a)$ , that is, $g(t) = 0$ for $0 \le t < a$ and $g(t) = f(t - a)$ for $t \ge a$ . If the Laplace integral of $f$ converges at the complex number $s$ , then so does the Laplace integral of $g$ , and $\mathcal{L}\{g\}(s) = e^{-a s} F(s).$

Remarks

The substitution $t = \tau + a$ gives $\int_a^\infty f(t - a) e^{-s t} \, dt = e^{-a s} \int_0^\infty f(\tau) e^{-s \tau} \, d\tau$ . Example: with $f = 1$ one gets $\mathcal{L}\{u(t - a)\}(s) = e^{-a s} / s$ for $\operatorname{Re} s > 0$ . A rectangular pulse $u(t - 1) - u(t - 2)$ has transform $(e^{-s} - e^{-2 s}) / s$ . The condition $a \ge 0$ matters: for $a < 0$ the shifted function would need values of $f$ at negative times, which the one-sided transform does not see.

Going deeper

For a student

This rule is also called the second shifting theorem. It is how one handles forcing terms that switch on and off in the Laplace method for initial value problems. Read backwards, a factor $e^{-a s}$ in a transform means: invert the rest, then delay by $a$ . Compare the frequency shift theorem, where a factor $e^{a t}$ becomes a shift in $s$ .

Reading path

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Tag	`M4VJHPZJMQ`
URI	`m2://M4VJHPZJMQ`
Kind	statement, theorem
Writing	W0 stub
MSC	44A10
Status	This copy of the site has no build data. Run `m2 status M4VJHPZJMQ`.
About	Laplace transform
Uses	Laplace transform, Heaviside step function