What is inverse oflaplace 1/((s^2+0.5s+1)) ?

The answer to inverse oflaplace 1/((s^2+0.5s+1)) is 1.03279…e^{-0.25t}sin(0.96824…t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace 1/((s^2+0.5s+1))

inverse laplace

\frac{1}{( s ^{2} + 0.5 s + 1 )}

1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

Solution steps

L^{- 1} {\frac{1}{( s ^{2} + 0.5 s + 1 )}}

\frac{1}{s ^{2} + 0.5 s + 1} = \frac{1}{( s + 0.25 ) ^{2} + 0.9375}

= L^{- 1} {\frac{1}{( s + 0.25 ) ^{2} + 0.9375}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s + 0.25 ) ^{2} + 0.9375} : a = - 0.25, F (s) = \frac{1}{s ^{2} + 0.9375}

= e^{- 0.25 t} L^{- 1} {\frac{1}{s ^{2} + 0.9375}}

L^{- 1} {\frac{1}{s ^{2} + 0.9375}} : \frac{1}{0.96824 \dots} sin (0.96824 \dots t)

Simplify

e^{- 0.25 t} \frac{1}{0.96824 \dots} sin (0.96824 \dots t) : 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

= 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

What is inverse oflaplace 1/((s^2+0.5s+1)) ?
The answer to inverse oflaplace 1/((s^2+0.5s+1)) is 1.03279…e^{-0.25t}sin(0.96824…t)