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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} $

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    @DavidMitra That is right.2012-05-08

3 Answers 3

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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.)

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    That is certainly a clever solution!2012-05-08
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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

$f(t) = 0 ; 1

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

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

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

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

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

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A two-step solution:

Fact 1. $L\big[H_n(t)]=\frac{e^{-pn}}{p};$

Fact 2. $\sum_0^{\infty}(-1)^n\frac{e^{-pn}}{p}=\frac{1}{p}\frac{1}{1+e^{-p}},\;\; Rep>0.$

Here, $H_n(t)=u(t-n).$