Das ist einer meiner Rechenwege.. ich habe mehrere versucht, aber die kamen nie zum selben Ergebnis
Partialbrüche:
\(\frac{8}{s (s - 2)^3} = \frac{A}{s} + \frac{B}{s - 2} + \frac{C}{(s - 2)^2} + \frac{D}{(s - 2)^3}\)
Multipliziere \( s(s - 2)^3 \):
\(8 = A(s - 2)^3 + Bs(s - 2)^2 + Cs(s - 2) + Ds\)
\( s = 0 \):
\(8 = A(-2)^3 \implies A = -1\)
\( s = 2 \):
\(8 = D(2) \implies D = 4\)
\( s = 1 \):
\(8 = 1 + B - C + 4 \implies B - C = 3\)
\( s = -1 \)
\(8 = 27 + 9B + 3C - 4 \implies 9B + 3C = -15\)
Gleichungssystem:
\(B - C = 3\)
\(9B + 3C = -15\)
\(B = -\frac{1}{2}, \quad C = -\frac{7}{2}\)
Partialbruchzerlegung:
\(F(s) = \frac{-1}{s} - \frac{1/2}{s - 2} - \frac{7/2}{(s - 2)^2} + \frac{4}{(s - 2)^3}\)
Inverse :
\(\mathcal{L}^{-1}\left(\frac{-1}{s}\right) = -1\)
\(\mathcal{L}^{-1}\left(\frac{-\frac{1}{2}}{s - 2}\right) = -\frac{1}{2} e^{2t}\)
\(\mathcal{L}^{-1}\left(\frac{-\frac{7}{2}}{(s - 2)^2}\right) = -\frac{7}{2} te^{2t}\)
\(\mathcal{L}^{-1}\left(\frac{4}{(s - 2)^3}\right) = 2 t^2 e^{2t}\)
\(f(t) = -1 - \frac{1}{2} e^{2t} - \frac{7}{2} te^{2t} + 2 t^2 e^{2t}\)
