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Mathematics

Complex Analysis

The Inverse Laplace Transform

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F(s) = \frac{1}{s - a}

f(t) = e^{at}

F(s) = \frac{2s}{s^2 + 4s + 13}

e^{-2t}(2\cos(3t) - \sin(3t))

F(s) = \frac{s}{s^2 + 4}

\cosh(2t)

F(s) = \frac{1}{s(s^2+1)}

1 - \cos(t)

F(s) = \frac{1}{s}

f(t) = 1

F(s) = \frac{s + 4}{(s + 2)^2 + 1}

e^{-2t}(\cos(t) + 4\sin(t))

F(s) = \frac{2}{(s + 1)^2}

2te^{-t}

F(s) = \frac{1}{(s-a)^2}

te^{at}

F(s) = \frac{s}{(s-a)^2}

e^{at}(1 + at)

F(s) = \frac{4s + 5}{(s+1)(s+3)}

3e^{-t} - 2e^{-3t}

F(s) = \frac{s - 6}{s^2 + 2s + 10}

e^{-t}(\cos(3t) - 2\sin(3t))

F(s) = \frac{2}{(s - 2)^2}

2te^{2t}

F(s) = \frac{1}{s^2 + a^2}

f(t) = \frac{1}{a} \sin(at)

F(s) = \frac{s}{s^2 + a^2}

f(t) = \cos(at)

F(s) = \frac{3}{s^3}

\frac{1}{2}t^2

F(s) = \frac{1}{s(s - a)}

1 - e^{at}

F(s) = \frac{s}{(s + 4)^2}

e^{-4t}(1 - 4t)

F(s) = \frac{4}{(s + 1)^3}

\frac{t^2e^{-t}}{2}

F(s) = \frac{1}{s^2}

f(t) = t

F(s) = \frac{s + 2}{(s + 2)^2 + 9}

e^{-2t}\cos(3t)

F(s) = \frac{s - a}{s^2 + b^2}

\cos(bt) - \frac{a}{b} \sin(bt)

F(s) = \frac{2s + 3}{s^2 + 4s + 5}

2e^{-2t} \cos(t) + e^{-2t} \sin(t)

F(s) = \frac{10}{s(s^2 + 9)}

1 - 10\frac{\sin(3t)}{3}

F(s) = \frac{s^2}{s^2 + 16}

\cos(4t)

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

2 e^{-\frac{t}{2}}\sin\left(\frac{\sqrt{3}t}{2}\right) + e^{-\frac{t}{2}}\cos\left(\frac{\sqrt{3}t}{2}\right)

F(s) = \frac{6s}{s^2 - 4s + 13}

e^{2t}(3\cos(3t) + 2\sin(3t))

F(s) = \frac{a}{(s-a)(s-b)}

\frac{a}{b-a} (e^{at} - e^{bt})

F(s) = \frac{5s - 3}{s^2 + 12s + 36}

5e^{-6t} - 3te^{-6t}

F(s) = \frac{2}{s^2 + 3s + 2}

2e^{-t} - 2e^{-2t}

F(s) = \frac{6}{(s + 2)^3}

t^2 e^{-2t}

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