Use the unit step function to express the function http://webwork.math.ttu.edu/webwork2...18af0996c1.png and then find the Laplace Transform http://webwork.math.ttu.edu/webwork2...7ebbecd6c1.png
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Use the unit step function to express the function http://webwork.math.ttu.edu/webwork2...18af0996c1.png and then find the Laplace Transform http://webwork.math.ttu.edu/webwork2...7ebbecd6c1.png
just use the definition of th laplace transform(or you could memorize a theorem)Quote:
Originally Posted by dbhakta
for the first one
$\displaystyle \mathcal{L}(f(t))=\int_{0}^{\infty}e^{-st}f(t)dt$
$\displaystyle \int_{0}^{2}e^{-st}\cdot 0 dt +\int_{2}^{\infty}(t+1)e^{-st}dt=-\frac{1}{s}(t+1)e^{-st}+\frac{1}{s^2}e^{-st}|_{2}^{\infty}$
$\displaystyle 0-[-\frac{1}{s}(2+1)e^{-2s}+\frac{1}{s^2}e^{-2s}]=\frac{3}{s}e^{-2s}+\frac{1}{s^2}e^{-2s}$
do the same thing for the 2nd one. Good luck.
First you need to convert the information given in the form of range to heaviside:
f(t) = 0, 0 < t < 2;
→ 0.[H(t-0) - H(t-2)]
= 0
f(t) = (t+1), t > 2;
→ (t+1). H(t+2)
[(t-2)+3].H(t-2)
(t-2)H(t-2) + 3H(t-2)
then you apply laplace transform to each of the terms above:
for the first term you will get: 0
for the second term you will get(using table):
exp(-2s)/(s^2) +3 exp(-2s)/s
for the second part just change (t+1) to 1 and repeat the steps again.
and yeah you can always use the general method posted by TheEmptySet.
