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chisigma
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As You can see in…
laplace transform of y(t) whole power 3
… on mathhelpforum.com a ‘newbie’ user asked to know how compute, given an $y(t)$ the L-Transform $\mathcal {L} \{y^{3}(t)\}$. At first, without any knowledege of y(t), it seems that no other chance exists apart the direct definition of L-Transform. Surprisinghly enough, if the L-Transform $\mathcal{L} \{y(t)\}$ is known, a 'magic and forgotten formula' conducts to the result. The formula can be found in Appendix two of the electrical engineering textbook 'N. Balabanian, T.A. Bickart, Electrical Network Theory, 1969, Wiley & Sons, New York' where You can read... Let be $f_{1}(t)$ and $f_2(t)$ two functions and their L-Transform $F_{1}(s)= \mathcal{L} \{f_{1}(t)\}$ and $F_{2}(s)= \mathcal{L} \{f_{2}(t)\}$ converge for $\text{Re} (s)> \sigma_{1}$ and $\text{Re} (s)> \sigma_{2}$ respectively. In this case is... $\displaystyle \mathcal{L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{c - i \infty}^{c + i \infty} F_{1}(z)\ F_{2} (s-z)\ dz$ (1)... where $\sigma_{1} < c < \sigma - \sigma_{2}\ , \sigma > \sigma_{1} + \sigma_{2}$ ...
As in the famous James Bond's film : never say never again!...
Kind regards
$\chi$ $\sigma$
laplace transform of y(t) whole power 3
… on mathhelpforum.com a ‘newbie’ user asked to know how compute, given an $y(t)$ the L-Transform $\mathcal {L} \{y^{3}(t)\}$. At first, without any knowledege of y(t), it seems that no other chance exists apart the direct definition of L-Transform. Surprisinghly enough, if the L-Transform $\mathcal{L} \{y(t)\}$ is known, a 'magic and forgotten formula' conducts to the result. The formula can be found in Appendix two of the electrical engineering textbook 'N. Balabanian, T.A. Bickart, Electrical Network Theory, 1969, Wiley & Sons, New York' where You can read... Let be $f_{1}(t)$ and $f_2(t)$ two functions and their L-Transform $F_{1}(s)= \mathcal{L} \{f_{1}(t)\}$ and $F_{2}(s)= \mathcal{L} \{f_{2}(t)\}$ converge for $\text{Re} (s)> \sigma_{1}$ and $\text{Re} (s)> \sigma_{2}$ respectively. In this case is... $\displaystyle \mathcal{L} \{f_{1}(t)\ f_{2}(t)\} = \frac{1}{2\ \pi\ i}\ \int_{c - i \infty}^{c + i \infty} F_{1}(z)\ F_{2} (s-z)\ dz$ (1)... where $\sigma_{1} < c < \sigma - \sigma_{2}\ , \sigma > \sigma_{1} + \sigma_{2}$ ...
As in the famous James Bond's film : never say never again!...
Kind regards
$\chi$ $\sigma$