Inverse Laplace transform of the feedback system transfer function with delay

Question

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)?

Answer 1

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.

Answer 2

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

$$h(t)=K_1\sum_{i=0}^\infty(-1)^iK_2^i\theta(t-T_1-iT_2),\ t\ge0.$$