I have the following first order system
$$ G(s) = \frac{3}{1 + Ts}e^{-sL} $$
to which as an experiment I give as an input the signal represented by the green graph below (a step input). During the experiment a mistake occurred and the step response eventually rests at a saturated value of $A=85$ according to the below red graph.
I want to now figure out what the values of the parameters $T$ and $L$ are. The value for $L$ is simple enough as it's simply the time delay which I get to be $L=20$ using the graphs. It is with the time constant $T$ that I am getting stuck on. I honestly have no good idea of how to go about this, but here is what I've tried.
We have the system's transfer function $G(s)$ and we know that we have as an input a step response width a height of $70 - 30 = 40$ whose Laplace transform becomes $R(s) = 40/s$. The output becomes
$$ Y(s) = G(s)R(s) = \frac{3e^{-20s}}{1+Ts} \cdot \frac{40}{s} $$
I was thinking that this expression in the Laplace domain isn't to much use, so I made an attempt to find the inverse Laplace transform. To do this I begin by ignoring the exponential $e^{-20s}$ which is contributing to time delay and focus only on performing partial fraction decomposition on the remaining expression $ \frac{120}{s(1+Ts)} $.
$$ \frac{120}{s(1+Ts)} = \frac{A}{s} + \frac{B}{1+Ts} $$
which after solving for $A$ and $B$ yields
$$ \frac{120}{s(1+Ts)} = \frac{120}{s} - \frac{120T}{1+Ts}. $$
Now bringing back the exponential term gives
$$ G(s) = \frac{120}{s}e^{-20s} - \frac{120T}{1+Ts}e^{-20s}. $$
Using the time shift property of the Laplace transform $f(t - t_{0})u(t - t_{0}) \iff F(s)e^{-st_{0}},\ t_{0} \geq 0$ I believe the inverse Laplace transform of the left term is
$$ 120u(t-20) $$
where $u(t)$ is the step (Heaviside) function. Regarding the right term I first rewrite it by multiplying the numerator and denomiator by $\frac{1/T}{1/T}$
$$ \frac{120T}{1+Ts}e^{-20s} = \frac{120}{s + \frac{1}{T}}e^{-20s} $$
so I can now use a table to find the inverse Laplace transform of $\frac{1}{s + \frac{1}{T}}e^{-20s}$. I find that the inverse Laplace transform of $1/(s - \lambda)$ is $e^{\lambda t}u(t)$ and that in combination with the time shift property leads me to believe that the final inverse Laplace transform of $\frac{120T}{1+Ts}e^{-20s}$ is
$$ 120e^{-\frac{1}{T}(t-20)}. $$
Finally we have that the inverse Laplace transform of $G(s)$ is
$$ \mathcal{L}^{-1}(G(s)) = 120u(t - 20) - 120e^{-\frac{1}{T}(t-20)}. $$
I don't know if this is correct and even if it is, I am not sure if it's to any use. I have a gut feeling that I am going in the right direction and that I should be able to continue from here somehow.
The shown step response and the transfer function don't fit together. According to the plot for the input, the input is at $\hat{u}=30$ before the step, and the output is clearly at steady state before the step. We can now use the final value theorem to determine the value the output should have before the step: $$y(t=50)=\lim_{s\rightarrow 0}G(s)\hat{u}=3\cdot 30=90$$ However, in the plot we have that $y(50)=20$. We can thus conclude that the shown step response for sure is from a different system. We thus stop here, since any further calculations would be meaningless.