inverse laplace e^(-2s)/(s+1) — step-by-step solution

The factor

e^{-2 s}

delays the original function by

2

: invert

\frac{1}{s + 1}

, then replace

t

with

t - 2

Find the causal inverse transform, for t ≥ 0. By linearity, invert each partial fraction

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

\frac{1}{s + 1}

\frac{\exp (-1 \cdot t)}{0!}

⇥ⓘ Multiplying by -1 gives the opposite:

-1 \cdot t

- t

■By convention, the factorial of 0 is 1:

0!

1

⇥

\frac{\exp (- t)}{1}

⇥ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

\exp (- t)

\frac{1}{\exp (t)}

⇥

\frac{1}{\exp (t)}

Delayed by

2

, the original function is zero for

t < 2

: replace

t

with

t - 2

and multiply by the unit step

u (t - 2)

, which is 0 for

t < 2

and 1 for

t \geq 2

▶Inverse Laplace transform:

\frac{u (t - 2)}{\exp (t - 2)}