Finding the righthand system from the Laplace transform $G(s) = \frac{1}{1- e^{-sT}}$

Question

I want to find the impulse response $g(t) \in \mathbb{C}$, that it's two-sided Laplace transform is: $$\mathcal{L}\{ g(t)\} = G(s)=\frac{1}{1-e^{-sT}}$$

I tried to find $g(t)$, by finding the inverse impulse response. Since, if I define $f(t)$ to be the inverse impulse response. Then, as in LTI systems, the convolution $ (g\ast f)(t) = \delta(t) $. Thus, in Laplace space according to the convolution theorem: $ G(s)\cdot F(s) = 1$ $$ \Rightarrow F(s) = 1-e^{-sT} $$ Then, by choosing the right plane of the Laplace space, as the $ROC$. And matching with known transformations. $$ f(t) = \mathcal{L^{-1}}\{ F(s)\} = \delta(t) - \delta(t-T) $$ So going back to the first equation: $$ \begin{align} \delta(t) = (g\ast f)(t) &= \int_{-\infty}^{\infty}(\delta(\tau)-\delta(\tau-T))g(t-\tau)d\tau \\ &= \int_{-\infty}^{\infty}\delta(\tau)g(t-\tau)d\tau-\int_{-\infty}^{\infty}\delta(\tau-T)g(t-\tau)d\tau \\ &= g(t) - g(t-T) \end{align} $$ To conclude, I reached the dead-end with getting stuck with this functional equation: $$ g(t) - g(t-T) = \delta(t) $$ where $T$ is a constant.

Is my action/move of splitting the "integral" in the above, wrong? Since $\delta(t) - \delta(t-T)$ is a generalized function? Thanks in advance.

Answer 1

We have that $$g(t)=g(t-T)+\delta(t),\ g(s)=0,\ s<0.$$

Then, applying this formula recursively, we obtain that the impulse response is

$$g(t)=\sum_{i=0}^\infty\delta(t-iT).$$

We can then verify that this is the case since

$$g(t)-g(t-T)=\sum_{i=0}^\infty\delta(t-iT)-\sum_{i=0}^\infty\delta(t-iT-T)=\sum_{i=0}^\infty\delta(t-iT)-\sum_{i=1}^\infty\delta(t-iT)=\delta(t).$$

Answer 2

The above is the correct answer, and is a way to solve the equation. Instead of searching the inverse impulse function, or solving the equation, finding $g(t)$, in a more reasonable approach:

As we know, we have $g(t)\in\mathbb{C}$ a righthand signal, that it's two-sided Laplace transform is: $\mathcal{L}\{ g(t)\} = G(s)=\frac{1}{1-e^{-sT}}$.

As the given $ROC$ of the problem, $ROC = \{s|\Re(s)\gt 0 \}$. We get a geometric series, $$ \forall s \in ROC, \forall T \gt 0: |e^{-sT}| \lt 1\\ \Rightarrow G(s) = \frac{1}{1-e^{-sT}} = \sum^\infty_{n=0} (e^{-sT})^n = \sum^\infty_{n=0} e^{-nsT} \\\overset{\mathcal{L}^{-1}}{\underset{\text{Linearity and: }\mathcal{L}^{-1}\{e^{-sT}\} = \delta(t-T)}{\Rightarrow}} \boxed{g(t)= \sum^\infty_{n=0} \delta(t-nT)} $$ For $ T = 0$, it's undefined. For $T \lt 0 ,\exists \tau = -T \gt 0$, $$ |e^{-sT}| = |e^{s\tau}| \underset{\forall s \in ROC, \forall T\lt 0}{\gt} 1\\ \Rightarrow G(s) = \frac{1}{1-e^{-sT}} = \frac{1}{1-e^{s \tau}} = -e^{-s \tau}\frac{1}{1-e^{-s \tau}} \underset{|e^{-s \tau}|\lt 1}{=} - e^{-s \tau}\sum^\infty_{n=0} e^{-ns\tau} = - \sum^\infty_{n=0} e^{-(n+1)s\tau} \\ \underset{k=n+1}{=} - \sum^\infty_{k=1} e^{-ks\tau} \\ \underset{\text{Like above}}{\Rightarrow} g(t) = - \sum^\infty_{k=1}\delta(t-k\tau) = - \sum^\infty_{k=1}\delta(t+kT) $$ The bottom, like @Maxim said, is also a causal signal, meaning a righthand signal. All depend on the sign of T, as he said.