inverse Laplace of 1/((2*s+1)*s) — all steps free

⇥1) You can perform a partial fraction decomposition:

⇥

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

⇥For the simple factor:

s

and its root:

0

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

(2 s + 1) \cdot s

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

⇥2) Let's solve in

ℝ

(2 \cdot 0 + 1) \cdot A = 1

⇥ⓘ If a factor is zero, the product is zero:

2 \cdot 0

0

⇥

(0 + 1) \cdot A = 1

⇥ⓘ Adding 0 changes nothing:

0 + 1

1

⇥Solution found:

A = 1

⇥For the simple factor:

2 s + 1

and its root:

- \frac{1}{2}

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

(2 s + 1) \cdot s

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

⇥3) Let's solve in

ℝ

- \frac{1}{2} \cdot B = 1

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

- \frac{1}{2} \cdot B

- \frac{1 \cdot B}{2}

- \frac{B}{2}

⇥

- \frac{B}{2} = 1

⇥

- B = 2

⇥ⓘ The sign of both sides is changed

⇥Solution found:

B = -2

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

⇥

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

Let's 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)!}

, so for

\frac{1}{s}

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

Before applying the formula, normalize the pole:

- \frac{2}{2 s + 1} = - \frac{1}{s + \frac{1}{2}}

The inverse Laplace formula for

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

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

, so for

- \frac{1}{s + \frac{1}{2}}

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

⇥

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

⇥ⓘ If a factor is zero, the product is zero:

0 \cdot t

0

■Raising a nonzero number to the power 0 gives 1:

\exp (0)

1

■By convention, the factorial of 0 is 1:

0!

1

■Multiplication by -1 gives the opposite value:

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

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

⇥

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

ⓘ Multiplying a fraction by a term multiplies its numerator:

- \frac{1}{2} \cdot t

- \frac{1 \cdot t}{2}

- \frac{t}{2}

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

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

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

▶Inverse Laplace transform:

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

inverse Laplace of 1/((2s+1)s)