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Heaviside function

  • Thread starter cragar
  • Start date
  • #1
2,544
2

Homework Statement



how to i take the laplace transform of this ,

-tH(t-1)

so we need to get thr right shift so is it -(t-1 + 1 ) so do i take the laplace transform of
-(t+1) so would it be -(1/s^2 + 1/s ) *e^(-s)
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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47
You just need to calculate
[tex]- \int_0^\infty e^{-s t} t H(t - 1) \, \mathrm dt[/tex]

You can break up the integral in two parts: 0 < t < 1 and t > 1.
 
  • #3
2,544
2
we cant just force the shift , if we had t^2H(t-1)
then to shift the quadratic we would (t-2+2)^ then we would expand
(t+2)^2 to t^2+4t+4 then take the laplace transform of that .
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,302
47
What do you want to shift? H(t - 1) is zero for t < 1, so

[tex]
\int_0^\infty f(t) H(t - 1) \, \mathrm dt
=
\int_0^1 0 \cdot f(t) \, \mathrm dt + \int_1^\infty 1 \cdot f(t) \, \mathrm dt
=
\int_1^\infty f(t) \, \mathrm dt
[/tex]
or am I really stupid?
 
Last edited:
  • #5
2,544
2
um i dont know my teacher never really talked about it like that,
ur prolly right , and another thing i cant see what ever u typed in those black boxes
my computer wont let me so its hard for me to tell what ur doing .

im sure ur right , but my teacher told us to take the laplace tranform
of a function time the heaviside function
he said the function needed to have the same shift as the heaviside
function in order for the formula to work .
f(t)H(t-1) = e^(-s)*L(f(t))
 

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