Question: Yeah, you probably are reading this question and thinking I'm one of those engineers who hasn't learned about the s-plane. I can assure you I'm not! I know that the transfer function of an RC circuit (Vout/Vin) has a REAL pole, with no imaginary component at s= -1/RC + j0. I am curious what happens when we excite an RC circuit at its TRUE pole, which is a decaying exponential with the expression e^(-t/RC).

Of course, the output of an RC Low-pass filter will never blow up to infinity. But what about the ratio of the output to the input, which after all, is the transfer function I originally defined?

Well, let's take this to LTSpice. Below is what I've simulated:
[![enter image description here](https://i.stack.imgur.com/GChDh.png)](https://i.stack.imgur.com/GChDh.png)

I would expect Vout/Vin to be infinity everywhere, however what I see is a ramp. I suspect I'm missing something about how to properly interpret the magnitude response of my transfer function with its true pole as an input. A time-domain view such as what I have should explain something, but I can't understand why it's a ramp.

If anyone has any intuition behind this question, please let me know.

Answer: 
If the system's impulse response is \$h(t)=e^{-at}u(t)\$, \$a>0\$, where \$u(t)\$ is the unit step function, and if its input is \$x(t)=e^{-at}u(t)\$, then its response is given by

$$y(t)=te^{-at}u(t)\tag{1}$$

which is a basic result. The function \$(1)\$ is just the inverse Laplace transform of a double real-valued pole:

$$Y(s)=\frac{1}{(s+a)^2}\tag{2}$$

Clearly, the output \$y(t)\$ decays to zero. However, the ratio \$y(t)/x(t)\$ is a ramp for \$t>0\$.

Please note that the ratio \$y(t)/x(t)\$ is not the system's transfer function. The transfer function is the ratio of the Laplace transforms of the input and output:

$$H(s)=\frac{Y(s)}{X(s)}\tag{3}$$

For a first-order lowpass filter with time constant \$\tau\$, \$H(s)\$ is given by

$$H(s)=\frac{1}{1+s\tau}\tag{4}$$
It *is* ramping to infinity as you expected. The s-domain calculation is a steady-state calculation, and you have an initial transient (your input exponential was effectively a constant '1' for all t < 0). You have to wait for that to decay (i.e. become negligible) -- e.g. a long (infinite) time.

You can see that after 10 time constants, your 'gain' is already 10.