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How do you find the unilateral Laplace Transform of x(t)=u(t)-u(t-2)
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look it up in my engineering cheetsheet?
\[X(s)={1 \over s}- {e ^{-2s} \over s}\]
AnwarA, can you explain how you found that? did you just use a laplace transform table?
you can use the definition of laplace transform
lemme see if i can make it readable
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\[\int\limits_{0}^{\infty}e^(-s*t)f(t)dt\]
well you can easily do it using the Laplace definition or this formula for laplace transform of the unit step function: \[L[u(t-a)]={e ^{-as} \over s}\]
yea, that didn't work, well, you can ignore when the step function is zero, and integrate where t > 0, or t-2 > 0
Ah okay, thank you both.
no problem.
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