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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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I assume $u(t)$ is the Heavise Step function? –  Pedro Tamaroff May 8 '12 at 15:16
Hint: What is the Laplace transform of the shift $u_x(\cdot)\mapsto u(\cdot-x)$? (Use a change of variable). –  AD. May 8 '12 at 15:22
I added some extra information as to where I get stuck –  ababzy May 8 '12 at 15:38
Write the series as ${1\over s} \sum\limits_{n=0}^\infty \bigl(-e^{-s}\bigr)^n$ and apply the formula giving the sum of a convergent Geometric series. (I think the proposed answer is off...) –  David Mitra May 8 '12 at 15:38
For $\sum\limits_{n=0}^\infty (-e^{-s})^n$, the ratio is $r=-e^{-s}$, so the sum is ${1\over 1-r}={1\over 1-(-e^{-s})}={1\over 1+e^{-s}}$; or, am I off? –  David Mitra May 8 '12 at 15:52

2 Answers 2

up vote 4 down vote accepted

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! –  Pedro Tamaroff May 8 '12 at 17:50

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

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

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

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

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

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