What is inverse oflaplace e^{-s} 1/s ?

The answer to inverse oflaplace e^{-s} 1/s is \H(t-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace e^{-s} 1/s

inverse laplace

e^{- s} \frac{1}{s}

H (t - 1)

Solution steps

L^{- 1} {e^{- s} \frac{1}{s}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 1) L^{- 1} {\frac{1}{s}} (t - 1)

L^{- 1} {\frac{1}{s}} : H (t)

= H (t - 1) H (t - 1)

= H (t - 1)

4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x

\int_{0}^{100} e^{- \frac{x}{0.033}} d x

f (x) = (x^{3} - 2)^{3}

\frac{d}{d x} (x - 5 x^{2} + x^{3})

3 x^{2} + 3 x - 8, 2 x^{2} + x + 7

What is inverse oflaplace e^{-s} 1/s ?
The answer to inverse oflaplace e^{-s} 1/s is \H(t-1)