Proof of the First Shifting Theorem

The first shifting theorem (or $s$ -shifting theorem) is one of the most useful properties of the Laplace transform, relating multiplication by an exponential in the time domain to a shift in the $s$ -domain.

Statement

Theorem4.12First shifting theorem

If $\mathcal{L}\{f(t)\}(s) = F(s)$ for $s > \alpha$ , then

$\mathcal{L}\{e^{at}f(t)\}(s) = F(s - a) \quad \text{for } s > \alpha + a.$

Proof

By definition of the Laplace transform:

$\mathcal{L}\{e^{at}f(t)\}(s) = \int_0^\infty e^{-st}\cdot e^{at}f(t)\,dt = \int_0^\infty e^{-(s-a)t}f(t)\,dt.$

This is precisely $F(s - a)$ , the Laplace transform of $f$ evaluated at $s - a$ instead of $s$ . The integral converges when $s - a > \alpha$ , i.e., $s > \alpha + a$ . $\blacksquare$

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Applications

ExampleUsing the first shifting theorem

$\mathcal{L}\{e^{3t}t^2\} = \frac{2!}{(s-3)^3} = \frac{2}{(s-3)^3}$ (shifting $2/s^3$ by $a = 3$ ).
$\mathcal{L}\{e^{-t}\cos 2t\} = \frac{s+1}{(s+1)^2+4}$ (shifting $s/(s^2+4)$ by $a = -1$ ).
$\mathcal{L}\{e^{2t}\sin 3t\} = \frac{3}{(s-2)^2+9}$ (shifting $3/(s^2+9)$ by $a = 2$ ).

ExampleDamped harmonic oscillator

Solve $y'' + 2y' + 5y = 0$ , $y(0) = 1$ , $y'(0) = 0$ .

Transform: $(s^2+2s+5)Y = s + 2$ , so $Y = \frac{s+2}{s^2+2s+5} = \frac{s+2}{(s+1)^2+4}$ .

Complete the square: $Y = \frac{(s+1)+1}{(s+1)^2+4} = \frac{s+1}{(s+1)^2+4} + \frac{1}{2}\cdot\frac{2}{(s+1)^2+4}$ .

By the first shifting theorem: $y(t) = e^{-t}\cos 2t + \frac{1}{2}e^{-t}\sin 2t = e^{-t}\left(\cos 2t + \frac{1}{2}\sin 2t\right)$ .

The Second Shifting Theorem (Comparison)

Theorem4.13Second shifting theorem (restated)

If $\mathcal{L}\{f(t)\} = F(s)$ , then $\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s)$ .

RemarkComparison of shifting theorems

| | First shifting | Second shifting | |--|---------------|-----------------| | Time domain | Multiply by $e^{at}$ | Delay by $c$ and multiply by $u_c(t)$ | | $s$ -domain | Replace $s$ by $s-a$ | Multiply by $e^{-cs}$ | | Effect | Shift in $s$ | Shift in $t$ | | Used for | Damped systems | Discontinuous inputs |

ExampleCombined shifting

Find $\mathcal{L}^{-1}\left\{\frac{e^{-2s}}{(s-3)^2}\right\}$ .

By second shifting: this is $u_2(t) \cdot g(t-2)$ where $G(s) = 1/(s-3)^2$ .

By first shifting: $g(t) = te^{3t}$ .

Therefore: $f(t) = u_2(t)(t-2)e^{3(t-2)}$ .