Inverse Laplace transform of the feedback system transfer function with delay I need to find a solution to the problem in the general case.
For given transfer function
$$
 W(s)=\frac{{{K}_{1}}{{e}^{-s{{T}_{1}}}}}{1+{{K}_{2}}{{e}^{-s{{T}_{2}}}}} 
$$
get the analytical formulas for the following inverse Laplace transform
$$ 
 h(t)={{\mathcal{L}}^{-1}}\left\{ \frac{W(s)}{s} \right\}
$$
Can I find a combination of hyperbolic functions expressions to apply? Is this applicable (source)?
 A: Assuming that $|K_2e^{-sT_2}|\lt 1$  we can proceed as follows.
$$
W(s)=K_1e^{-sT_1}\sum_{i=0}^n(-1)^iK_2^ie^{-i T_2s}
$$
then
$$
\mathcal{L}^{-1}\left[\frac{W(s)}{s}\right] = K_1\sum_{i=0}^n(-1)^i K_2^i \theta (t-T_1-i T_2)
$$
where $\theta(\cdot)$ is the Heaviside Theta function.
A: The inverse Laplace transform of $W(s)/s$ is the impulse response of the same system, which is the step response of the system $W(s)$. In term of signals, we have that $Y(s)=W(s)U(s)$. Going to the time-domain, we obtain the expression
$$y(t)=K_1u(t-T_1)-K_2y(t-T_2).$$
The delay $T_1$ is just going to shift the whole response by $T_1$ seconds. So, we may assume for now that $T_1=0$ for simplicity. Applying a step and assuming zero initial conditions yields for $t\in[0,T_2)$:
$$y(t)=K_1.$$
Propagating forward, we get for $t\in[T_2,2T_2)$
$$y(t)=K_1-K_2K_1=K_1(1-K_2).$$
Repeating this process over and over, we can observe that the impulse response is piecewise constant and that the values are related to the sequence
$$y_{n+1} = K_1-K_2y_n,\ y_0=K_1$$
It is a linear recurrence equation which is bounded and converging if and only if $|K_2|<1$, bounded but oscillatory if and only if $K_2=1$, and unbounded otherwise.
The general solution is given by
$$y_n = (-K_2)^nK_1+\sum_{k=0}^{n-1}(-K_2)^{n-k-1}K_1,\ n\ge0$$
and we have that $y(t)=y_i$ for all $t\in[iT_2,(i+1)T_2)$.
Adding now the offset $T_1$, we get that the impulse response $h$ of $W(s)/s$  is given by
$$h(t)=\left\{\begin{array}{rl}
0&,t\in[0,T_1)\\
y(t-T_1)&,t\ge T_1
\end{array}\right.$$
Interestingly, we can connect this result with the answer from Cesareo by noting that $$y_{n+1}-
\sum_{i=0}^ny_i=(-1)^{n+1}K_1K_2^{n+1},$$ and we recover the expression
$$h(t)=K_1\sum_{i=0}^n(-1)^iK_2^i\theta(t-T_1-iT_2),\ t<T_1+(n+1)T_2.$$
In the case where $|K_2|<1$, the limit as $n\to\infty$ exists and we have
$$h(t)=K_1\sum_{i=0}^\infty(-1)^iK_2^i\theta(t-T_1-iT_2),\ t\ge0.$$
