Hi again,
Doing a direct evaluation with the sine source is a little more involved but maybe not by much...
Starting with ljcox's current analysis for the circuit:
R*C*dVo/dt=Vi*R/r-Vo
that is the differential equation for the circuit.
One way to solve this is by the use of Laplace transform techniques,
or we can do it using an integrating factor.
To get into integrating factor form, we rearrange some terms and we get:
dVo/dt+Vo/(R*C)=Vi/(r*C)
and since Vi=sin(w*t) we substitute that and get:
dVo/dt+Vo/(C*R)=sin(w*t)/(r*C)
and now the equation is in the integrating factor form:
dy/dt+p(t)*y=g(t)
and the solution is:
(u(t)*y)'=u(t)*g(t)
where:
(u(t)*y)' is the time derivative of (u(t)*y)
y=Vo
dy/dt=dVo/dt
p(t)=1/(R*C)
g(t)=sin(w*t)/(r*C)
u(t) is the integrating factor:
u(t)=e^integrate[p(t)]dt
and now we will integrate both sides of the DE:
integrate[(u(t)*Vo)']dt=integrate[u(t)*g(t)]dt
and after substituting the values above and simplying the left side we get:
e^(t/(R*C))*Vo=integrate[e^(t/(R*C))*sin(w*t)/(r*C)]dt
and after performing the ride side integration we get:
e^(t/(R*C))*Vo=(e^(t/(R*C))*(R*sin(w*t)-w*cos(w*t)*R^2*C))/(r*w^2*C^2*R^2+r)+K
Now solving for Vo we get:
Vo=K*e^(-t/(R*C))-(w*R^2*C*cos(w*t))/(r*w^2*C^2*R^2+r)+(R*sin(w*t))/(r*w^2*C^2*R^2+r)
and lastly we would solve for K using the initial conditions.
This is now in the form i was talking about previously, but if you want to solve with a peak amplitude factor
you'll have to substitute A*sin(w*t) for sin(w*t) in g(t), then solve again, and if you want an input offset and a
peak amplitude you'll have to substitute A*sin(w*t)+B where B is the offset voltage, then solve the above again.
Now wasn't that simple?