Prove that $\mathscr{L}^{-1}\{\frac{1}{(s+\alpha)(1-e^{-Ts})}\}=e^{-\alpha t} \left[ \frac{e^{(m+1) \alpha T}-1}{e^{ \alpha T}-1}\right]$

Question

Prove that the inverse laplace transform of : $$G(s) = \frac{1}{(s+\alpha)(1-e^{-Ts})}$$

equals $$g(t) = e^{-\alpha t} \left[ \frac{e^{(m+1) \alpha T}-1}{e^{ \alpha T}-1}\right]$$

for : $$mT<t<(m+1)T \:\:\:\: , \:\:\:\: m=0,1,2,...,$$

I can't understand what is meant by the the two conditions above.

What I could do with this problem is :

Write $\frac{1}{1-e^{-Ts}}$ as a power series : $$\frac{1}{1-e^{-Ts}} = \sum_{n=0}^{\infty}(e^{-Ts})^{n} \:\:\:\: , \:\:\:\: |e^{-Ts}|<1$$

Then : $$G(s) = \frac{1}{s+\alpha}\sum_{n=0}^{\infty}e^{-nTs} = \sum_{n=0}^{\infty}\frac{1}{s+\alpha}e^{-nTs}$$

Taking the inverse laplace transform : $$g(t) = \sum_{n=0}^{\infty}e^{-\alpha(t-nT)}u(t-nT)$$

And I could show that : $$\frac{e^{(m+1) \alpha T}-1}{e^{ \alpha T}-1} = \left[1-e^{(m+1) \alpha T}\right]\frac{1}{1-e^{ \alpha T}} \:\:\:\: , \:\:\: |e^{- \alpha T}|<1$$

$$= \left[1-e^{(m+1) \alpha T}\right]\sum_{n=0}^{\infty}(e^{\alpha T})^{n}$$ $$ = \sum_{n=0}^{\infty}e^{n \alpha T}-\sum_{n=0}^{\infty}e^{(n+m+1) \alpha T}$$ $$= \sum_{n=0}^{\infty}e^{n \alpha T}-\sum_{n=m+1}^{\infty}e^{n \alpha T} = \sum_{n=0}^{m}e^{n \alpha T}$$

And stopped here .

Answer 1

You have already done all the work:

The idea is to define $g$ as a piecewise function, i.e to simplify the expression you have found for each interval $(mT,(m+1)T)$.

If $m T < t < (m+1) T$ then: $$g(t)=\sum_{n=0}^\infty e^{-\alpha(t-nT)} u(t-nT)=\sum_{n=0}^{m} e^{-\alpha(t-nT)} u(t-nT)+\sum_{n=m+1}^\infty e^{-\alpha(t-nT)} u(t-nT)$$ But:

• If $n \leq m$ then $t-nT > mT-nT \geq 0$ so $$u(t-nT) =1$$.
• If $n \geq m+1$ then $t-nT < (m+1)T-nT \leq 0$ so $$u(t-nT) =0$$.

The expression of $g(t)$ is thus: $$g(t)=\sum_{n=0}^{m} e^{-\alpha(t-nT)} +\sum_{n=m+1}^\infty 0=\sum_{n=0}^{m} e^{-\alpha t} e^{-\alpha (-nT)}=e^{-\alpha t} \sum_{n=0}^m e^{n\alpha T}$$ and using what you have shown: $$g(t)=e^{-\alpha t} \left(\frac{e^{\alpha (m+1) T}-1}{e^{\alpha T}-1} \right)$$ for $t \in (mT,(m+1)T)$.