RC circuit current, large signal

I hear you, especially when it's already been done for you:

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Reply to
seagirt555
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It's only a sine burst if you include the case where the voltage source is turned on, ie it's not steady state, continuous. Usually when dealing with a sine wave source it's the steady state that we assume we're talking about.

Reply to
seagirt555

That's the dilemma of a differential equation solution: the sine has to be turned on.

You see the same thing in Spice.

Reply to
John Larkin

Basically the problem is that the DEQs as-written don't describe anything physical.

Either the system is steady state and you have to _assume_ a form of the solution that describes a steady-state system. The solution as given isn't that.

Or the system isn't steady-state and there must also be a step function at 0 to "switch on" the sine, in addition to the driving sine that runs for all time. The equations as written don't describe that situation, either.

Reply to
bitrex

Basically the problem is that the DEQs as-written don't describe anything physical.

Either the system is steady state and you have to _assume_ a form of the solution that describes a steady-state system. The solution as given isn't that.

Or the system isn't steady-state and there must also be a step function at 0 to "switch on" the sine, in addition to the driving sine that runs for all time. The equations as written don't describe that situation, either.

Reply to
bitrex

Am 05.09.2019 um 22:18 schrieb bitrex:

Hello,

Here is the solution from a script found in the Internet.

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T = R*C w = 2*pi*f A = 1/(1+(w*T)**2) B = w*T/(1+(w*T)**2)

ue(t) = sin(w*t)

uc(t) = B*exp(-t/T) +A*sin(w*t) -B*cos(w*t)

-------------------------------------------

Because B/T is the same as A*w,

ic(t) = w*C*(-A*exp(-t/T) +A*cos(w*t) +B*sin(w*t))

--------------------------------------------------

I have checked the result with LTspice.

Helmut

Reply to
Helmut Sennewald

The issue is that I got my EE degree before Spice existed. I had to write my own simulation code. Sometimes I still do.

Reply to
John Larkin

Reply to
bitrex

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