Understanding the Laplace Transform for Right-Handed Notes

[hotjohn]

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

 

[Samy_A]

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?

If so, that's just renaming the (dummy) integration variable from u to t.

 

[hotjohn]

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

 

[Samy_A]

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
$\displaystyle \int_a^b e^x dx = \int_a^b e^y dy$

 

[vela]

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 $t$.