Some Interesting Properties of Laplace Transforms
Stretching
The stretching property is
$$ \mathcal{L}(f(at)) = \frac{1}{a}F\left(\frac{s}{a}\right) $$To show this,
$$ \mathcal{L}(f(at)) = \int_{0}^{\infty} f(at) \, e^{-st}\,dt $$Let \( u = at, \) so that \( \displaystyle \frac{du}{dt} = a \), then
$$ \mathcal{L}(f(at)) = \frac{1}{a} \int_{0}^{\infty} f(u) e^{- (s/a) u} \, du $$ $$ = \frac{1}{a}F\left(\frac{s}{a}\right) $$Differentiation II
Another useful property is
$$ \mathcal{L}((-t)^n f(t)) = \frac{d^nF(s)}{ds^n} = F^n(s) $$eg. Suppose that \( \displaystyle \color{#00f0ff} Y(s) = \frac{6}{s^4}. \) Find \( \color{#00f0ff} y(t) \)
We write
$$ \frac{6}{s^4} = -\frac{d}{ds}\left(\frac{2}{s^3}\right) $$ $$ = \frac{d^2}{ds^2}\left(\frac{1}{s^2}\right) $$ $$ = -\frac{d^3}{ds^3}\left(\frac{1}{s}\right) $$ $$ F(s) = \frac{1}{s} \, \Rightarrow \, f(t) = u_0(t) $$where \( u_0(t) \) is the unit step function:
$$ u_0(t) = \begin{cases} 0 & t < 0 \\ 1 & t > 0 \end{cases} $$Therefore \( \displaystyle -\frac{d^3}{ds^3}\left(\frac{1}{s}\right) \) is the Laplace transform of \( -(-t)^3u_0(t) \), so
$$ y(t) = t^3u_0(t) $$Shifting I
The first shifting property is
$$ \mathcal{L}(e^{at}f(t)) = F(s-a) $$This shifts \( F(s) \) to the right by \( a \) units.
To prove this,
$$ \mathcal{L}(e^{at}f(t)) = \int_{0}^{\infty} e^{at} f(t) e^{-st}\,dt $$ $$ = \int_{0}^{\infty} f(t) e^{-(s-a)t}\,dt $$ $$ = F(s-a) $$eg. Suppose \( \color{#00f0ff} g(t) = \sin(4t) \) has Laplace transform \( \displaystyle \color{#00f0ff} G(s) = \frac{4}{s^2 + 16}. \) What is the inverse transform of \( \displaystyle \color{#00f0ff} F(s) = \frac{4}{s^2 - 6s + 25} \)?
Complete the square in the denominator:
$$ F(s) = \frac{4}{(s - 3)^2 + 16} $$Therefore,
$$ F(s) = G(s - 3) $$So by the shifting property,
$$ f(t) = e^{3t}g(t) = e^{3t}\sin(4t) $$The unit step function can be defined more generally as
$$ u_c(t) = \begin{cases} 0 & 0 \leq t \leq c \\ 1 & t > c \end{cases} $$We compute its Laplace transform:
$$ \mathcal{L}(u_c(t)) = \int_{0}^{\infty} u_c(t) e^{-st}\,dt $$ $$ = \int_{c}^{\infty} e^{-st}\,dt $$ $$ = \frac{e^{-cs}}{s} \quad s > 0 $$For a given function \( f(t) \), suppose we shift the function to the right by an amount \( c \), defining \( f(t) = 0 \) for \( t < 0 \). Define a new function \( g(t) \) by
$$ g(t) = \begin{cases} \quad \; 0 & 0 \leq t \leq c \\ f(t - c) & t > c \end{cases} $$This corresponds to turning on the function \( f(t) \) at time \( t = c \). We can write this more compactly as
$$ g(t) = u_c(t)\,f(t - c) $$Shifting II
The Laplace transform of a shifted function satisfies
$$ \mathcal{L}\big(u_c(t)f(t - c)\big) = e^{-cs}F(s) $$where \( F(s) = \mathcal{L}(f(t)) \).
Proof:
$$ \mathcal{L}\big(u_c(t)f(t - c)\big) = \int_{0}^{\infty} u_c(t)f(t - c) e^{-st}\,dt $$ $$ = \int_{c}^{\infty} f(t - c) e^{-st}\,dt $$Let \( u = t - c \), so \( du = dt \)
$$ = \int_{0}^{\infty} f(u) e^{-s(u + c)}\,du $$ $$ = e^{-cs} \int_{0}^{\infty} f(u) e^{-su}\,du $$ $$ = e^{-cs}F(s) $$eg. Find the inverse Laplace transform of \( \displaystyle \color{#00f0ff} F(s) = \frac{e^{2s}}{(s - 3)^3} \)
$$ F(s) = e^{2s}G(s) \quad \text{where} \quad G(s) = \frac{1}{(s - 3)^3} $$First consider
$$ H(s) = \frac{1}{s^3} $$We can write
$$ H(s) = -\frac{1}{2}\frac{d}{ds}\left(\frac{1}{s^2}\right) = -\frac{1}{2}\frac{d}{ds} K(s) $$ $$ \text{where} \quad K(s) = \frac{1}{s^2} \quad \text{and so} \quad k(t) = t $$Using the second differentiation property,
$$ h(t) = -t\left(-\frac{1}{2}\right)k(t) $$ $$ h(t) = \frac{t^2}{2} $$Therefore,
$$ H(s) = \frac{1}{s^3} \quad \Leftrightarrow \quad h(t) = \frac{t^2}{2} $$Now
$$ G(s) = H(s - 3) = \frac{1}{(s - 3)^3} $$so by the shifting property,
$$ g(t) = e^{3t}h(t) = e^{3t}\frac{t^2}{2} $$Since
$$ F(s) = e^{2s}G(s) $$we apply the time-shifting property to obtain
$$ f(t) = g(t + 2) $$Therefore,
$$ f(t) = \frac{1}{2} \, e^{3(t+2)} \, (t+2)^2 $$