Laplace transform

  • Thread starter hotjohn
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  • #1
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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


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  • #2
Samy_A
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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

Do you ask about this step?
laplace.jpg

If so, that's just renaming the (dummy) integration variable from u to t.
 
  • #3
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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
 
  • #4
Samy_A
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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##
 
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  • #5
vela
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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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