Calculate the Laplace transform

Question

Help me calculate the Laplace transform of a geometric series. $$ f(t) = \sum_{n=0}^\infty(-1)^nu(t-n) $$

show that $$ \mathcal{L} \{f(t)\} = \frac{1}{s(1+\mathcal{e}^{-s})} $$

Edit: so far I know that

$$ \mathcal{L} \{f(t)\} = \frac{1}{s}\sum_{n=0}^\infty(-1)^ne^{-ns} $$

Answer 1

Here is a simpler approach.

Note that for $t>0$, we have $f(t)+f(t-1) = 1$. The Laplace transform of a time-shifted function (in this case $f(t-1) = f(t-1) u(t-1)$) is $s \mapsto e^{-s} \hat{f}$, where $\hat{f}$ is the Laplace transform of $f$. Furthermore, the Laplace transform of $1$ is just $s \mapsto \frac{1}{s}$. Hence we have $$\hat{f}(s)+ e^{-s} \hat{f}(s) = \frac{1}{s},$$ from which it follows that

$$\hat{f}(s) = \frac{1}{s(1+e^{-s})}.$$ (Your formula above is incorrect.)

Answer 2

Try to think what your sum looks like. You will see it is a periodic function, with period $2$, that we can define as.

$$f(t) = 1 ; 0<t<1$$

$$f(t) = 0 ; 1<t<2$$

In general if a function has period $P$, it's LP is given by

$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-sP}}\int_0^P e^{-st}f(t)dt$$

under appropriate conditions (i.e. the LP should exist)

It this case $f(t)$ is periodically constant, so the LP exists, and

$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\int_0^2 e^{-st}f(t)dt$$

$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\int_0^1e^{-st}dt$$

$$\mathcal L \{ f(t) \} =\frac{1}{1-e^{-2s}}\frac{1-e^{-s}}{s}$$

$$\mathcal L \{ f(t) \} =\frac{1}{1+e^{-s}}\frac{1}{s}$$