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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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- Thread starter hotjohn
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Samy_A

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Do you ask about this step?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

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

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yes , why thgere is no need to turnDo you ask about this step?

View attachment 95345

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

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Samy_A

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It's just a name. It doesn't matter whether the integration variable is called u or t (or something else).yes , why thgere is no need to turnu into t -nT? since it is given at the left part of the notes

Example: the two following integrals are equal

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

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vela

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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##.yes , why thgere is no need to turnu into t-nT? since it is given at the left part of the notes.

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