1
$\begingroup$

Find the inverse Laplace transform of this function (related to my question earlier):

$$f(t)=\mathcal{L}_s^{-1}\left[\frac{s}{s+\frac{1}{\tau}}\cdot\frac{A}{s}\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\sum_{n=0}^{\infty}\mathrm e^{-nTs}\right]_{(s)}$$

I'm not sure how to go.

For $t>0$ we can find that:

$$f(t)=\mathcal{L}_s^{-1}\left[\frac{s}{s+\frac{1}{\tau}}\cdot\frac{A\tanh\left(\frac{Ts}{4}\right)}{s}\right]_{(s)}=A\left(e^{-\frac{t}{\tau}}-2\tau\mathcal{L}_s^{-1}\left[\frac{1}{\left(1+e^{\frac{Ts}{2}}\right)\left(1+\tau s\right)}\right]_{(s)}\right)$$

$\endgroup$
3
$\begingroup$

From your related question we have $F(s)=H(s)\cdot G(s)=\frac{s}{s+\frac{1}{\tau}}\cdot\frac{A}{s}\tanh\left(\frac{sT}{4}\right)$ and then \begin{align} \color{blue}{F(s)}&=\frac{A}{s+\frac{1}{\tau}}\cdot\tanh\left(\frac{sT}{4}\right) =\frac{A}{s+\frac{1}{\tau}}\cdot\frac{\mathrm e^{\frac{Ts}{4}}-\mathrm e^{-\frac{Ts}{4}}}{\mathrm e^{\frac{Ts}{4}}+\mathrm e^{-\frac{Ts}{4}}}=\frac{A}{s+\frac{1}{\tau}}\cdot\frac{1-\mathrm e^{-\frac{Ts}{2}}}{1+\mathrm e^{-\frac{Ts}{2}}}\\ &\color{blue}{=\frac{A}{s+\frac{1}{\tau}}\cdot\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\cdot\frac{1}{1-\mathrm e^{-sT}}}\\ &=\frac{A}{s+\frac{1}{\tau}}\cdot\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\cdot\sum_{n=0}^{\infty} \mathrm e^{-nTs}\\ &=\frac{A}{s+\frac{1}{\tau}}\sum_{n=0}^{\infty} \left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2\mathrm e^{-nTs}\\ &=\frac{A}{s+\frac{1}{\tau}}\sum_{n=0}^{\infty} \left[\mathrm e^{-nTs}-2\mathrm e^{-\frac{2n+1}{2}Ts}+\mathrm e^{-(n+1)Ts}\right]\\ &=A\sum_{n=0}^{\infty} \frac{1}{s+\frac{1}{\tau}}\left[\mathrm e^{-nTs}-2\mathrm e^{-\frac{2n+1}{2}Ts}+\mathrm e^{-(n+1)Ts}\right]\\ \end{align} Observing that $$ \mathcal L^{-1}\left\{\frac{1}{s+a}\mathrm e^{-bs}\right\}=\mathrm e^{-a(t-b)}u(t-b)\tag{$\star$} $$ we have \begin{align} \mathcal L^{-1}\left\{F(s)\right\}=f(t)&=A\sum_{n=0}^{\infty}\left[\mathrm e^{-\frac{1}{\tau}\left(t-nT\right)} u\left(t-nT\right)- 2\mathrm e^{-\frac{1}{\tau}\left(t-\frac{2n+1}{2}T\right)} u\left(t-\tfrac{2n+1}{2}T\right)+\right.\\ &\qquad\quad\left.+\mathrm e^{-\frac{1}{\tau}\left(t-(n+1)T\right)} u\left(t-(n+1)T\right) \right] \end{align}

Another way is the following. Observe that if $$f(t)=\sum_{n=0}^{\infty}f_0(t-nT)$$ we have $$\mathcal L\{f(t)\}=\sum_{n=0}^{\infty}\mathcal L\{f_0(t-nT)\}=\sum_{n=0}^{\infty}F_0(s)\mathrm e^{-nTs}=F_0(s)\sum_{n=0}^{\infty}\mathrm e^{-nTs}=\frac{F_0(s)}{1-\mathrm e^{-Ts}}$$ So for your function we have $$ F(s)=\underbrace{\frac{A}{s+\frac{1}{\tau}}\cdot\left(1-\mathrm e^{-\frac{Ts}{2}}\right)^2}_{F_0(s)}\cdot\frac{1}{1-\mathrm e^{-sT}} $$ that is $$ F_0(s)=\frac{A}{s+\frac{1}{\tau}}\cdot\left(1-2\mathrm e^{-\frac{Ts}{2}}+\mathrm e^{-{Ts}}\right) $$ and then using $(\star)$ we have $$ f_0(t)=A\left[\mathrm e^{-\frac{t}{\tau}} u\left(t\right)- 2\mathrm e^{-\frac{1}{\tau}\left(t-\frac{T}{2}\right)} u\left(t-\tfrac{T}{2}\right)+\mathrm e^{-\frac{1}{\tau}\left(t-T\right)} u\left(t-T\right) \right] $$ and finally \begin{align} f(t)&=\sum_{n=0}^{\infty}f_0(t-nT)\\ &=A\sum_{n=0}^{\infty}\left[\mathrm e^{-\frac{1}{\tau}\left(t-nT\right)} u\left(t-nT\right)- 2\mathrm e^{-\frac{1}{\tau}\left(t-\frac{2n+1}{2}T\right)} u\left(t-\tfrac{2n+1}{2}T\right)+\right.\\ &\qquad\quad\left.+\mathrm e^{-\frac{1}{\tau}\left(t-(n+1)T\right)} u\left(t-(n+1)T\right) \right] \end{align}

A third way starting with your work $$f(t)=\mathcal{L}^{-1}\left\{k\left(\frac{1}{s+\frac{1}{\tau}}\right)\left(1-\frac{2}{\mathrm e^{\frac{T}{2}s}+1}\right)\right\} = A\,\mathrm e^{-\frac{t}{\tau}} - 2A\tau \, \mathcal{L}^{-1}\left\{\frac{1}{\left(1+\mathrm e^{\frac{T}{2}s}\right)\left(1+\tau s\right)}\right\}$$ then, observing that $$\frac{1}{1+\mathrm e^{\frac{T}{2}s}}=\sum_{n=0}^{\infty} (-1)^{n}\mathrm e^{-\frac{T}{2} (n+1) s}$$ we have \begin{align} \mathcal{L}^{-1}\left\{\frac{1}{\left(1+\mathrm e^{\frac{T}{2}s}\right)\left(1+\tau s\right)}\right\} &= \sum_{n=0}^{\infty} (-1)^{n} \, \mathcal{L}^{-1}\left\{\frac{\mathrm e^{-\frac{T}{2} (n+1) s}}{1+\tau s}\right\} \\ &= \frac{1}{\tau} \, \sum_{n=0}^{\infty} (-1)^{n} \, u\left(t - (n+1) \frac{T}{2}\right) \, \mathrm e^{- \frac{1}{\tau}\left(t - \frac{T}{2} (n+1)\right)} \end{align} which leads to $$f(t)= A\left[\mathrm e^{-\frac{t}{\tau}} - 2 \sum_{n=0}^{\infty} (-1)^{n} \, u\left(t - (n+1) \frac{T}{2}\right) \, \mathrm e^{- \frac{1}{\tau}\left(t - \frac{T}{2} (n+1)\right)}\right]$$

$\endgroup$
  • $\begingroup$ Thank you, very much! You're a great help $\endgroup$ – Jan May 14 '16 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.