Electronic – Trouble analyzing BJT circuits when we change β from very high to 100

You can use the Shockley equation, which is another idealization, but a better one than a fixed diode drop. To apply it, you have to know, for the given diode, all the parameters:

$$I=I_\mathrm{S}\left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right)$$

I would suggest to replace the transistor with an actual diode model, from which you can pull out the parameters and calculate by hand.

In this answer to a question, I was working with the formula, in relation to the behavior of a diode model in CircuitLab. See the question's "appendix" at the bottom.

I do not believe there is any problem with the notation used by the author. He just wanted to express the following (\$x\$ input, \$y\$ output - the response to a eigenfunction \$e^{st}\$):

$$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)d\tau $$ $$ = \int_{-\infty}^\infty h(\tau)e^{s(t-\tau)}d\tau $$ $$ = e^{st} \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$

Thus, the response to \$e^{st}\$ is of the form

$$y(t) = H(s)e^{st}$$

where \$H(s)\$ is a complex constant whose value depends on \$s\$ and wich is related to the impulse response by

$$ H(s) = \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $$

There is no conflict time x frequency here.

Also, note:

If \$\lambda\$ is a pole of \$H(s)\$, the mode \$e^{{\lambda}t}\$ can be generated at the output by an initial state without the application of any input. If \$\lambda\$ is not a pole of \$H(s)\$, then this is not possible. In this case, the only way to generate \$e^{{\lambda}t}\$ at the output is to apply \$e^{{\lambda}t}\$ at the input. In other hand, if the input \$x(t)\$ is of the form \$e^{{\lambda}t}\$, where \$\lambda\$ is not a pole of \$H(s)\$, then the output due to input \$x(t) = e^{{\lambda}t}\$ is equal to \$y(t) = H(\lambda)e^{{\lambda}t}\$.