
#1
Mar311, 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? 



#2
Mar411, 01:50 PM

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#3
Mar411, 05:27 PM

P: 361





#4
Mar411, 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?? 



#5
Mar411, 08:05 PM

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[tex]\frac{1}{1x} = 1+x+x^2+x^3+x^4+\cdots[/tex] so that [tex]\begin{align*} \frac{1}{1e^{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. 



#6
Mar411, 09:39 PM

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#7
Mar411, 09:57 PM

P: 361





#8
Mar411, 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)}{1e^{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] 



#9
Mar511, 12:55 PM

P: 361





#10
Mar511, 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? 



#11
Mar511, 06:54 PM

P: 361

i think i see your point.. the function will essentially be U(t)  U(t10) repeated over and over again 



#12
Mar511, 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 f_{1}(t) which is 1 on (0,p) and 0 elsewhere, giving you one nonzero period of your function f(t). So you can write
f_{1}(t) = f(t)(u(t)u(tp)) You know that the transform of the periodic function f(t) is [tex]F(s) = \frac{L(f_1(t))}{1e^{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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