# Laplace Transform of a Geometric Series

I need to graph the following function

$f(t) = 1 + \sum_{n=1}^\infty(-1)^n u(t-n)$

($u$ refers to the unit step function)

and find the laplace transform of this function.

The problem is similar to another one posted here Calculate the Laplace transform except for mine i need to graph a function and it is slightly different.

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If $t>0$, $u(t-0) = 1$, so the Laplace transforms will be the same. Why don't you just graph the functions $t \mapsto 1-u(t-1)$, and $t \mapsto 1-u(t-1)+u(t-2)$ and guess what the general pattern is. The graph is fairly simple. – copper.hat May 19 '12 at 4:16
me and a classmate sketched the graph, – Ozzy Gonzalez May 19 '12 at 6:03
me and a classmate sketched a graph that looked like f(1) = 1 + (-1) = 0 f(2) = 1 + (-1) +1 = 1 f(3) = 1 + (-1) +1 + (-1) = 0 f(4) = 1 + (-1) +1 + (-1) +1 = 1 not to sure if that's how you sketch piecewise functions – Ozzy Gonzalez May 19 '12 at 6:05
It should look like a square wave... – copper.hat May 19 '12 at 6:05
so from 0 to 1 f(t) = 0 and from 1 to 2 f(t) = 1 (from here it repeated itself) and we took the laplace 1/(1-e^(-sT) * integral from 0 to 2 of e^-(st)f(t)dt which is than split up into two integrals, the first from 0 to 1 (but since f(t) is 0 on this range it becomes 0) the second from 1 to 2 1/(1-e^(-2s) integral from 1 to 2 e^-(st)dt and i'm not sure if we simplified it right, but the answer came out to 1/(s(e^s+1) – Ozzy Gonzalez May 19 '12 at 6:10

Here's my picture, you should get the pattern from this...

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Maybe your can write some values in the axis! – Pedro Tamaroff May 22 '12 at 2:05
Well, the values on the y-axis are just $0$ and $1$, the values on the x-axis are just $0,1,2,...$. – copper.hat May 22 '12 at 4:11

Matlab Program

t=0:0.01:10;
G=zeros(1,length(t));
for n=1:10
G1=(-1)^n*heaviside(t-n);
G=G+G1;
end
F=1+G;
plot(t,F)
axis([0 10 -2 2])

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