Exciting an RC Circuit at its TRUE Pole on the Sigma (Real) Axis in the Left-Half Plane
Good question, here's my attempt. You still have to convert your input signal from the time domain to s domain, then do the math, then convert the result back to the time domain. That pole just tells you where the resonance is, the behavior of the resonance is usually &quot;growing infinitely forever&quot; - but the exact behavior needs to be found from doing the math.
Using transforms from: https://en.wikipedia.org/wiki/Laplace_transform
$$
\frac{1}{1+sRC}V_i(s)=V_o(s)\\
V_i(t) = V_{i0}e^{-t/(RC)}=V_{i0}e^{-t/\tau}\\
V_i(s) = \frac{V_{i0}}{1/\tau+s}\\
V_o(s) = \frac{1/\tau}{1/\tau+s}\frac{V_{i0}}{1/\tau+s}\\
V_o(s) = \frac{ V_{i0}/\tau}{(1/\tau+s)^2}\\
V_o(t) = (V_{i0}/\tau)\times t e^{-t/\tau}
$$
Which gives us:
$$
V_o(t)/V_i(t) = \frac{(V_{i0}/\tau)\times t e^{-t/\tau}}{V_{i0}e^{-t/\tau}}\\
V_o(t)/V_i(t) = t\times 1/\tau
$$
Predicting that, in the time domain, the ratio of output to input voltage increases linearly with t. In the s-domain, the ratio is infinity.
EDIT: This actually perfectly matches your result, even the correct slope - except there is an offset of 1 unit on the y axis. My guess is that the simulation found the DC point using \$V_{i}=1 \text{ V}\$ at \$t&lt;0\$, so changes the math. The above assumes \$V_{i}=0 \text{ V}\$ at \$t&lt;0\$. So it's pretty much the same, just a little difference because of the initial condition.
If the initial condition was V=1, you can use the superposition principle and add the solution we found (for V=0 until t=0, then decaying) to the solution if V=1 when t was negative and V=0 when t is positive
$$
V_o(t) = (V_{i0}/\tau)\times t e^{-t/\tau}+V_{i0}e^{-t/\tau}\\
V_o(t)/V_i(t) = \frac{(V_{i0}/\tau)\times t e^{-t/\tau}+V_{i0}e^{-t/\tau}}{V_{i0}e^{-t/\tau}}\\
V_o(t)/V_i(t) = t\times 1/\tau+1
$$
Now it exactly matches what you have! So the simulation does use V=1 for when t is negative, and this should clearly show what's happening. Just because something is infinity in the s domain doesn't mean it is in time.