inverse oflaplace (e^{-s})/(s^2+3s+2)

⚠ (e^{-s})/(s^2+3s+2) is read as (e^(-s))/(s^2+3s+2): use ^(...) to group an exponent, use * to multiply

The factor

e^{- s}

delays the original function by

1

: invert

\frac{1}{s^{2} + 3 s + 2}

, then replace

t

with

t - 1

⇥1) Let's factor in

ℝ

s^{2} + 3 s + 2

⇥Given the quadratic form

a s^{2} + b s + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 3

and

c = 2

⇥

Δ = 3^{2} - 4 \cdot 1 \cdot 2

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

3^{2}

3 \cdot 3

9

⇥

Δ = 9 - 4 \cdot 2

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 2

8

⇥

Δ = 9 - 8

⇥ⓘ Start from

9

and move back

8

on the number line:

9 - 8

1

⇥

Δ = 1

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

s_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 3 - \sqrt{1}}{2 \cdot 1}

⇥

s_{1} = \frac{- 3 - 1}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

- 3 - 1

-4

⇥

s_{1} = - \frac{4}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{4}{2}

-2

⇥

s_{1} = -2

⇥The quadratic formula gives

s_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 3 + \sqrt{1}}{2 \cdot 1}

⇥

s_{2} = \frac{- 3 + 1}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

- 3 + 1

-2

⇥

s_{2} = - \frac{2}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{2}

-1

⇥

s_{2} = -1

⇥Factors identified:

(s + 1) \cdot (s + 2)

⇥2) You can perform a partial fraction decomposition:

⇥

\frac{1}{(s + 1) \cdot (s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2}

⇥For the simple factor:

s + 1

and its root:

-1

, the unknown A can be easily computed by multiplying by the denominator:

s^{2} + 3 s + 2

and evaluating at the root because the other terms cancel out:

⇥3) Let's solve in

ℝ

(- 1 + 2) \cdot A = 1

⇥ⓘ Parentheses are computed first:

- 1 + 2

1

■Multiplying by 1 changes nothing:

1 \cdot A

A

⇥Solution found:

A = 1

⇥For the simple factor:

s + 2

and its root:

-2

, compute B by multiplying by the denominator:

s^{2} + 3 s + 2

and evaluating at the root (the other terms cancel out):

⇥4) Let's solve in

ℝ

(- 2 + 1) \cdot B = 1

⇥ⓘ Parentheses are computed first:

- 2 + 1

-1

■Multiplying by -1 gives the opposite:

-1 \cdot B

- B

⇥

- B = 1

⇥ⓘ The sign of both sides is changed

⇥Solution found:

B = -1

⇥5) Finally, you can substitute the values of the unknowns into the initial partial fraction decomposition:

⇥

\frac{1}{(s + 1) \cdot (s + 2)} = \frac{1}{s + 1} - \frac{1}{s + 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!}

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 + 2}

\frac{-1 \cdot \exp (-2 t)}{0!}

⇥

\frac{\exp (-1 \cdot t)}{0!} + \frac{-1 \cdot \exp (-2 t)}{0!}

⇥ⓘ Multiplying by -1 gives the opposite:

-1 \cdot t

- t

-1 \cdot \exp (-2 t)

- \exp (-2 t)

■By convention, the factorial of 0 is 1:

0!

1

⇥

\frac{\exp (- t)}{1} - \frac{\exp (-2 t)}{1}

⇥ⓘ A negative exponent moves the power to the other side of the fraction bar, for a nonzero base:

\exp (- t)

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

⇥

\frac{1}{\exp (t)} - \exp (- 2 t)

⇥ⓘ A negative exponent moves the power to the other side of the fraction bar, for a nonzero base:

\exp (- 2 t)

\frac{1}{\exp (2 t)}

⇥

\frac{1}{\exp (t)} - \frac{1}{\exp (2 t)}

Delayed by

1

, the original function is zero for

t < 1

; for

t \geq 1

, replace

t

with

t - 1

▶Inverse Laplace transform, for

t \geq 1

\frac{1}{\exp (t - 1)} - \frac{1}{\exp (2 \cdot (t - 1))}