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Inverse laplace transform square wave

by magnifik
Tags: inverse, laplace, square, transform, wave
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magnifik
#1
Mar3-11, 08:17 PM
P: 361
can someone please help me find the inverse laplace transform of the following:


i do not know how to solve for this due to the e^-20s shift in the denominator... is there a way to change this function to make it easier to solve?
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LCKurtz
#2
Mar4-11, 01:50 PM
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You might find some helpful information here:

http://www.intmath.com/Laplace-trans...c-function.php
magnifik
#3
Mar4-11, 05:27 PM
P: 361
Quote Quote by LCKurtz View Post
You might find some helpful information here:

http://www.intmath.com/Laplace-trans...c-function.php
from the graphs on that website i assume the inverse laplace will involve step functions. thanks for the help

magnifik
#4
Mar4-11, 06:57 PM
P: 361
Inverse laplace transform square wave

in analyzing JUST the function part of this...

f(t) = 1, when 0 < t < 10
f(t) = 0, when 10 < t < 20
F(s) = (-1/s)(e^(-10s) - 1)/(1 - e^(-20s))

is it correct to say that f(t) = U(t) - U(t -10) and that the period is 20??
vela
#5
Mar4-11, 08:05 PM
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Quote Quote by magnifik View Post
i do not know how to solve for this due to the e^-20s shift in the denominator... is there a way to change this function to make it easier to solve?
The idea is to use the fact that

[tex]\frac{1}{1-x} = 1+x+x^2+x^3+x^4+\cdots[/tex]

so that

[tex]\begin{align*}
\frac{1}{1-e^{-20s}} &= 1 + e^{-20s} + (e^{-20s})^2+ (e^{-20s})^3 + \cdots \\
&= 1 + e^{-20s} + e^{-40s} + e^{-60s} + \cdots
\end{align*}[/tex]

Hopefully, you can see why that may result in a function periodic in the time domain.
LCKurtz
#6
Mar4-11, 09:39 PM
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Quote Quote by magnifik View Post
in analyzing JUST the function part of this...

f(t) = 1, when 0 < t < 10
f(t) = 0, when 10 < t < 20
F(s) = (-1/s)(e^(-10s) - 1)/(1 - e^(-20s))

is it correct to say that f(t) = U(t) - U(t -10) and that the period is 20??
Yes. To be precise, F(s) is the Laplace Transform of the periodic extension with period 20 of your f(t).
magnifik
#7
Mar4-11, 09:57 PM
P: 361
Quote Quote by vela View Post
The idea is to use the fact that

[tex]\frac{1}{1-x} = 1+x+x^2+x^3+x^4+\cdots[/tex]

so that

[tex]\begin{align*}
\frac{1}{1-e^{-20s}} &= 1 + e^{-20s} + (e^{-20s})^2+ (e^{-20s})^3 + \cdots \\
&= 1 + e^{-20s} + e^{-40s} + e^{-60s} + \cdots
\end{align*}[/tex]

Hopefully, you can see why that may result in a function periodic in the time domain.
i understand that it can be represented as a series. however, i wouldn't know how to take the inverse laplace of the series
vela
#8
Mar4-11, 11:04 PM
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You're not taking the inverse transform of just the series. You're taking the inverse of

[tex]\begin{align*}
\frac{F(s)}{1-e^{-20s}} & = F(s)[1 + e^{-20s} + e^{-40s} + e^{-60s} + \cdots] \\
& = F(s) + e^{-20s}F(s) + e^{-40s}F(s) + e^{-60s}F(s) + \cdots
\end{align*}[/tex]
magnifik
#9
Mar5-11, 12:55 PM
P: 361
Quote Quote by vela View Post
You're not taking the inverse transform of just the series. You're taking the inverse of

[tex]\begin{align*}
\frac{F(s)}{1-e^{-20s}} & = F(s)[1 + e^{-20s} + e^{-40s} + e^{-60s} + \cdots] \\
& = F(s) + e^{-20s}F(s) + e^{-40s}F(s) + e^{-60s}F(s) + \cdots
\end{align*}[/tex]
not sure how i would do this. i know that the e^-ns represents a shift in the time domain f(t-n) but wouldn't there be infinitely many time shifts using the series expansion?
vela
#10
Mar5-11, 04:04 PM
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Yes, you need an infinite number of shifts because a periodic signal has an infinite number of cycles.

Take your f(t) in post #4 and repeatedly shift and add. What do you get?
magnifik
#11
Mar5-11, 06:54 PM
P: 361
Quote Quote by vela View Post
Yes, you need an infinite number of shifts because a periodic signal has an infinite number of cycles.

Take your f(t) in post #4 and repeatedly shift and add. What do you get?
U(t) - U(t-10) + U(t - 20) - U(t - 30) + U(t-40) - U(t - 50) + U(t - 60) + U(t - 70)...

i think i see your point.. the function will essentially be U(t) - U(t-10) repeated over and over again
LCKurtz
#12
Mar5-11, 09:18 PM
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Vela and I are leading you in two different directions for solution. I guess you learn twice as much. Say you have a periodic function f(t) with period p and define a new function f1(t) which is 1 on (0,p) and 0 elsewhere, giving you one nonzero period of your function f(t). So you can write

f1(t) = f(t)(u(t)-u(t-p))

You know that the transform of the periodic function f(t) is

[tex]F(s) = \frac{L(f_1(t))}{1-e^{-ps}}[/tex]

This tells you that when taking inverse transforms a factor of (1 - e-ps) can be suppressed and the inverse will give you one period of the original periodic function. So if you can find the inverse without that factor, just extend it periodically with period p to get the function f(t).


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