Laplace transform Heaviside step function

  • Thread starter Pietair
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  • #1
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Homework Statement


What is the Laplace transform of f(t) = 0 for 0 < t < 2 and f(t) = (4-t) for 2 < t < 3 and f(t) = 1 for 3 < t < 4 and f(t) = (5-t) for 4 < t < 5 and f(t) = 0 for t > 5?

Homework Equations


7410747ec7563eab51f608f2c80a9497.png


The Attempt at a Solution


f(t) = H(t-2)(4-t) - H(t-3)(4-t) + H(t-3) - H(t-4) + H(t-4)(5-t) - H(t-5)(5-t)
f(t) = H(t-2)(4-t) + H(t-3)(2t-7) - H(t-4)(t-4) + H(t-5)(t-5)

How can I rewrite H(t-2)(4-t) + H(t-3)(2t-7) to a form suitable for Laplace transform?
 

Answers and Replies

  • #2
Tom Mattson
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For [itex]H(t-2)(4-t)[/itex], you'd like to see a [itex]t-2[/itex] in the second factor right? So just put one there, and balance it out elsewhere in the expression. Like this:

[itex]H(t-2)(4-(t-2)-2)[/itex]

See how I added and subtracted 2? This simplifies to:

[itex]H(t-2)(2-(t-2))[/itex]

Do the same for [itex]H(t-3)(2t-7)[/itex].
 
  • #3
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Thank you for your reply!

Now I am getting:
[itex]
H(t-2)(2-(t-2)) + H(t-3)(2(t-3)-1) - H(t-4)(t-4) + H(t-5)(t-5)
[/itex]

And the Laplace transform:
[itex]
((2/s)-(1/s^2))e^(-2s) + ((2/s^2)-(1/s))e^(-3s) + ((-1/s^2)e^(-4s)) + ((1/s^2)e^(-5s))
[/itex]

Is this correct?
 
  • #4
Tom Mattson
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Looks good to me.
 
  • #5
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Thanks a lot!
 

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