I know this is almost a year old, but is this notebook helpful to you: https://colab.research.google.com/drive/1p5WP-twoxD5HKq9gziNWKqSi8Xo_cotg?usp=sharing -Michael
On Tuesday, June 28, 2022 at 4:52:12 PM UTC-4 [email protected] wrote: > Thank you Oscar, > onestly I don't get the same resoult of yours, because I get 3 result , > not just one like you: > one solution for for τ/R =−i/(100πRi) , one solution for τ/R =i/(100πRi) > and another is otherwise > Anyway, the otherwise seems to be possible to be simplified and become > like your solution, so thank you very much for the help. > It should be nice to avoid the 3 solution and keep just the otherwise one > and get it simplified, but I just want to thank you for your help. > > > Il giorno martedì 28 giugno 2022 alle 12:05:54 UTC+2 Oscar ha scritto: > >> On Tuesday, 28 June 2022 at 08:03:05 UTC+1 [email protected] wrote: >> >>> >>> Hi Aaron, thank you for the reply. >>> >>> The code I have in my notebook (I'm using google colab) is this now (at >>> the end of the post), but using w is not working anymore. >>> Anyway, if I move back to use the line w = 2 * sp.pi * 50 I still get a >>> unreadble result. >>> The result should be something like the nice wikipedia result I found >>> there: >>> >>> https://it.wikipedia.org/wiki/Circuito_RC#Risposta_in_frequenza_del_circuito_RC >>> Is there a way to get the a readble result? (something like the >>> following latex) >>> {\displaystyle v_{C}(t)=v_{C}(0)e^{-{\frac {t}{\tau }}}+K\cos(\omega >>> t+\theta )} >>> The wikipedia result has tau=R*C, theta is an angle displacement, and K >>> don't know what is. >>> >> >> The result that I see is similar: >> >> In [13]: diffeq >> Out[13]: >> d vo⋅cos(100⋅π⋅t) vc(t) >> ──(vc(t)) - ─────────────── + ───── >> dt C⋅R C⋅R >> >> In [12]: res >> Out[12]: >> -t >> >> ─── >> >> C⋅R >> >> 100⋅π⋅C⋅R⋅vo⋅sin(100⋅π⋅t) vo⋅cos(100⋅π⋅t) vo⋅ℯ >> >> vc(t) = ───────────────────────── + ────────────────── - >> ────────────────── >> 2 2 2 2 2 2 2 2 2 >> >> 10000⋅π ⋅C ⋅R + 1 10000⋅π ⋅C ⋅R + 1 10000⋅π ⋅C ⋅R + >> 1 >> >> The main difference is that the constant K is given explicitly here in >> terms of C, R and v0. If you want to substitute for tau and K you can do >> something like: >> >> In [19]: tau, K = symbols('tau, K') >> >> In [20]: res.subs(C, tau/R).subs(vo, K*(1 + 10000*pi**2*tau**2)) >> Out[20]: >> -t >> ─── >> τ >> vc(t) = 100⋅π⋅K⋅τ⋅sin(100⋅π⋅t) + K⋅cos(100⋅π⋅t) - K⋅ℯ >> >> You can see now something more similar. The difference is that you have >> sin and cos rather than cos with a phase shift theta but those are >> equivalent under trigonometric identities. Another difference is the >> constant K on the exponential term but the wikipedia solution is incorrect >> there: >> >> >>> In the italian wikipedia there's this piece (I try to translate) that I >>> would like to reproduce in a notebook: >>> >>> Let's see how does the RC circuit works with a sine wave. We can use >>> voltage Kirchhoff law: >>> {\displaystyle V_{0}\cos(\omega t)=R\cdot i(t)+v_{C}(t)} >>> >>> we can rewrite the equation like: >>> {\displaystyle V_{0}\cos(\omega t)=RC{\frac {dv_{C}(t)}{dt}}+v_{C}(t)} >>> >>> and then solve the differential equation with constant coefficients with >>> a known therm: >>> {\displaystyle {\frac {dv_{C}(t)}{dt}}+{\frac {1}{\tau }}v_{C}(t)={\frac >>> {V_{0}\cos(\omega t)}{\tau }}} >>> >>> where \tau =RC is still the time constant of the circuit. >>> The general solution come from the sum of the associated homogeneous >>> solution: >>> >>> >>> v_{C}(t)=v_{C}(0)e^{-{\frac {t}{\tau }}} >>> >> It is incorrect to use v_C(0) here. There should be a constant there but >> it is not generally equal to the initial condition. >> >>> and a particolar solution: >>> {\displaystyle K\cos(\omega t+\theta )\ } >>> >>> where K is a constant. So: >>> >>> >>> {\displaystyle v_{C}(t)=v_{C}(0)e^{-{\frac {t}{\tau }}}+K\cos(\omega >>> t+\theta )} >>> >> You can see that this is incorrect if you just substitute t=0. >> >> -- >> Oscar >> > -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/c10232a2-7251-4d27-b887-6ab06fd0f340n%40googlegroups.com.
