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Laplace solution : verify correctness of full demonstration

Hi, I followed this very well written article completely and found the same formulas... EXCEPT for the very last one for the case of a sinusoidal function as input. After careful examination, I perform the same transform and arrive to the conclusion that certain factors in front of the sinusoidal functions in the time domain expression namely 1/omega0 and 1/omegaf are there only if we replace the nominator of the summands 1 by omega_0/omega_0 and omega_f/omega_f in order to be able to perform the Laplace transform. Is that correct ?

$\operatorname {\mathcal {L}} ^{-1}\left[\ \omega _{0}^{2}\ U\ \omega _{\mathrm {f} }{\frac {\frac {1}{(\omega _{\mathrm {f} }^{2}\ -\omega _{0}^{2})}}{\ s^{2}+\omega _{0}^{2}\ }}\ +{\frac {-{\frac {1}{(\omega _{\mathrm {f} }^{2}\ -\omega _{0}^{2})}}}{\ s^{2}+\omega _{\mathrm {f} }^{2}\ }}\ \right]$

Isolating the constant and adjusting for lack of numerator:

${\frac {\ \omega _{0}^{2}\ U\omega _{\mathrm {f} }\ }{\ \omega _{\mathrm {f} }^{2}-\omega _{0}^{2}\ }}\operatorname {\mathcal {L}} ^{-1}\left[\ \left({\frac {\omega _{0}}{\omega _{0}(s^{2}+\omega _{0}^{2})}}-{\frac {\omega _{0}}{\omega _{0}(s^{2}+\omega _{f}^{2})}}\right)\ \right]\,$

Performing the reverse Laplace transform on each summands:

${\frac {\ \omega _{0}^{2}\ U\omega _{\mathrm {f} }\ }{\ \omega _{\mathrm {f} }^{2}-\omega _{0}^{2}\ }}\ \left(\operatorname {\mathcal {L}} ^{-1}\left[\ {\frac {1}{\omega _{0}}}{\frac {\omega _{0}}{(s^{2}+\omega _{0}^{2})}}\right]\ -\operatorname {\mathcal {L}} ^{-1}\left[{\frac {1}{\omega _{\mathrm {f} }\ }}{\frac {\omega _{\mathrm {f} }\ }{(s^{2}+\omega _{f}^{2})}}\right]\right)\,$

${\frac {\ \omega _{0}^{2}\ U\omega _{\mathrm {f} }\ }{\ \omega _{\mathrm {f} }^{2}-\omega _{0}^{2}\ }}\ \left(\ {\frac {1}{\omega _{0}}}\operatorname {\mathcal {L}} ^{-1}\left[{\frac {\omega _{0}}{(s^{2}+\omega _{0}^{2})}}\right]\ -{\frac {1}{\omega _{\mathrm {f} }\ }}\operatorname {\mathcal {L}} ^{-1}\left[{\frac {\omega _{\mathrm {f} }\ }{(s^{2}+\omega _{f}^{2})}}\right]\right)\,$

$v_{\mathrm {in} }(t)={\frac {\ \omega _{0}^{2}\ U\ \omega _{\mathrm {f} }\ }{\omega _{\mathrm {f} }^{2}-\omega _{0}^{2}}}\ \left({\frac {1}{\omega _{0}}}\ \sin(\omega _{0}\ t)-{\frac {1}{\ \omega _{\mathrm {f} }\ }}\ \sin(\omega _{\mathrm {f} }\ t)\right)\;,$

Furthermore, there seems to be a step to simplify the expression of v(t) that has not be taken as b/b = 1 and not b should appear in the last formula in the time domain. Instead of this:

$v(t)=v_{0}\cos(\omega _{0}\ t)+{\frac {v'_{0}\ b}{\ b\ \omega _{0}\ }}\ \sin(\omega _{0}\ t)+{\frac {\omega _{0}^{2}\ U\ \omega _{\mathrm {f} }}{\ \omega _{\mathrm {f} }^{2}-\omega _{0}^{2}\ }}\left({\frac {1}{\omega _{0}}}\ \sin(\omega _{0}\ t)-{\frac {1}{\ \omega _{\mathrm {f} }\ }}\ \sin(\omega _{\mathrm {f} }\ t)\right)\;.$

Should we not have the following ?

$v(t)=v_{0}\cos(\omega _{0}\ t)+{\frac {v'_{0}}{\omega _{0}\ }}\ \sin(\omega _{0}\ t)+{\frac {\omega _{0}^{2}\ U\ \omega _{\mathrm {f} }}{\ \omega _{\mathrm {f} }^{2}-\omega _{0}^{2}\ }}\left({\frac {1}{\omega _{0}}}\ \sin(\omega _{0}\ t)-{\frac {1}{\ \omega _{\mathrm {f} }\ }}\ \sin(\omega _{\mathrm {f} }\ t)\right)\;.$

Cordially yours, Temnothorax (talk) 23:22, 24 October 2023 (UTC)

Straight wire as an open LC Circuit

His resonance frequency is given by F = 1/(2pi.square root of LC)

where L might be given by : 4/pi3 x mu0/pi2 x l x ln l/r or L = 4/pi5 x mu0 x l x ln l/r ...

and C by : pi3/4 x E0 x l / ln l/r

with l and r as the length and radius of the wire .

Example : for a 18 AWG wire dipole of ten meters long

C = 69.47 pF , L = 1.62 microH and F of resonance is 15 Mhz...

Leveugel (talk) 13:34, 22 June 2026 (UTC)Leveugel (talk) 09:36, 29 June 2026 (UTC)