Also, here is the code - apologies for not including it before:
This notebook was prompted by the thread at:<br>
https://groups.google.com/g/sympy/c/VW2eFNBw1Cg/m/AxTlTwGUHQAJ
import sympy as sp
sp.init_printing(use_latex=True)
'''
Without the real=True part, there are some solutions with imaginary
components
'''
w, t, tau, vo, vc0, K, theta = sp.var(r'\omega, t, \tau, v_o, v_{C}(0), K,
\theta', real=True)
vc = sp.Function('v_c')
diffeq = sp.Derivative(vc(t), t) + vc(t) / tau - vo * sp.cos(w * t) / tau
diffeq
res = sp.dsolve([diffeq], [vc(t)], ics={vc(0):vc0})
res
'''
Quick note - the solution at<br>
https://it.wikipedia.org/wiki/Circuito_RC#Risposta_in_frequenza_del_circuito_RC
<br>is not quite right in that<br>
$$ {\displaystyle v_{C}(t)=v_{C}(0)e^{-{\frac {t}{\tau }}}+K\cos(\omega
t+\theta )}$$
gets the initial condition wrong; I think it should be something like:
$${\displaystyle v_{C}(t)=(v_{C}(0)-K\cos(\theta))e^{-{\frac {t}{\tau
}}}+K\cos(\omega t+\theta )}$$
'''
# Collecting vo terms
res[0].rhs.expand().collect(vo).trigsimp()
# Collecting exponential terms
res[0].rhs.expand().collect(sp.exp(-t/tau)).trigsimp()
'''
The $K$ and $\theta$ here match
https://it.wikipedia.org/wiki/Circuito_RC#Metodo_simbolico_per_la_risposta_in_frequenza
(using trig identities to get single magnitude and phase)
'''
K = vo * sp.sqrt((w*tau)**2 + 1**2) / ((w*tau)**2 + 1)
K
theta = sp.atan2(-w*tau, 1)
theta
vcneat = (vc0-K*sp.cos(theta))*sp.exp(-t/tau)+K*sp.cos(w*t+theta)
vcneat
# Are they the same?
(vcneat - res[0].rhs).trigsimp()
On Saturday, March 4, 2023 at 2:01:38 AM UTC-5 Michael Gustafson wrote:
> 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 federic...@gmail.com 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 federic...@gmail.com 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
>>>
>>
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