Talk:LC circuit

message: It seems that I already have a section on order.  As a recent undo notes, LC filters have even order. There are RC and RL filters, starting with first order. Or, as noted in the edit summary, it comes from the order of the differential equation. The not very many, and not necessarily WP:RS that I searched, all say LC starts at 2. Gah4 (talk) 05:15, 5 April 2023 (UTC)
thread_title:  order, again 

message: The article needs more pictures of tuned circuits.  High Q tank circuits in radio transmitters should be easy for general readers to understand because they usually consist of a coil suspended in air and a variable capacitor with exposed plates.  I added a photo of a tank circuit from a transmitter, but Commons had almost none so I had to use a black & white one from a 1938 book.  I'd like to request any radio amateurs out there that would like to help, open the case of your transmitter (and/or antenna tuner) and see if you can get a good shot of the tank circuit.  Also military radio equipment or transmitters of AM radio stations would be good sources. If you are unfamiliar with the process of uploading a photo to Commons I can help. Thanks! --ChetvornoTALK 20:49, 20 June 2023 (UTC)
thread_title:  Request for photos 

message: 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)
thread_title: Laplace solution : verify correctness of full demonstration