Understanding the Laplace Transform for Right-Handed Notes

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Homework Help Overview

The discussion revolves around the Laplace Transform, specifically addressing the transition of an integral involving a dummy variable from \( u \) to \( t \) in the context of right-handed notes. Participants are examining the implications of variable renaming in integrals and questioning the assumptions behind this process.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning why the integral changes from \( \int_0^T e^{-su}f(u) \, du \) to \( \int_0^T e^{-st}f(t) \, dt \) without considering \( t - nT \). There is a focus on the nature of dummy variables in integration and whether the renaming affects the interpretation of the integral.

Discussion Status

The discussion is ongoing, with participants exploring the reasoning behind the variable renaming in integrals. Some guidance has been provided regarding the nature of dummy variables, but there is still uncertainty about the implications of the variable choice in this context.

Contextual Notes

Participants note that the notation in the notes may be inconsistent, leading to confusion about the relationship between the variables \( u \) and \( t \). There is an acknowledgment of potential sloppiness in the author's presentation of the material.

hotjohn
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for the right part of the notes, why the integral of (e^-su)f(u) from 0 to T will become integral of (e^-st)f(t) from 0 to T suddenly ? why not integral of (e^-s (t-nT) )f(t-nT) from 0 to T ? as we can see, u = t +nTd given/known data

Homework Equations

The Attempt at a Solution

 

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hotjohn said:
for the right part of the notes, why the integral of (e^-su)f(u) from 0 to T will become integral of (e^-st)f(t) from 0 to T suddenly ? why not integral of (e^-s (t-nT) )f(t-nT) from 0 to T ? as we can see, u = t +nTd given/known data

Homework Equations

The Attempt at a Solution

Do you ask about this step?
laplace.jpg

If so, that's just renaming the (dummy) integration variable from u to t.
 
Samy_A said:
Do you ask about this step?
View attachment 95345
If so, that's just renaming the (dummy) integration variable from u to t.
yes , why thgere is no need to turn u into t -nT ? since it is given at the left part of the notes
 
hotjohn said:
yes , why thgere is no need to turn u into t -nT ? since it is given at the left part of the notes
It's just a name. It doesn't matter whether the integration variable is called u or t (or something else).

Example: the two following integrals are equal
##\displaystyle \int_a^b e^x dx = \int_a^b e^y dy##
 
Last edited:
hotjohn said:
yes , why thgere is no need to turn u into t-nT ? since it is given at the left part of the notes.
The author was a little sloppy. The earlier ##t## (the one related to ##u##) and the ##t## in the last integral aren't the same ##t##.
 

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