Heaviside Function Homework: Laplace Transform w/ Right Shift

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Homework Help Overview

The discussion revolves around taking the Laplace transform of the expression -tH(t-1), which involves the Heaviside function and right shifts. Participants are exploring the implications of the Heaviside function on the transformation process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to calculate the integral of -tH(t-1) and consider breaking it into parts based on the Heaviside function's definition. There are questions about how to properly apply shifts and whether the function needs to match the Heaviside function's shift for the Laplace transform to be valid.

Discussion Status

The discussion is ongoing, with participants offering different interpretations of how to handle the Heaviside function in the context of the Laplace transform. Some guidance has been provided regarding the integral setup, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants note that there may be confusion stemming from differing teaching methods and interpretations of the Heaviside function's role in Laplace transforms. There is also mention of technical issues affecting the ability to view shared content, which may hinder understanding.

cragar
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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)
 
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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.
 
we can't 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 .
 
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<br /> =<br /> \int_0^1 0 \cdot f(t) \, \mathrm dt + \int_1^\infty 1 \cdot f(t) \, \mathrm dt<br /> =<br /> \int_1^\infty f(t) \, \mathrm dt[/tex]
or am I really stupid?
 
Last edited:
um i don't know my teacher never really talked about it like that,
ur prolly right , and another thing i can't see what ever u typed in those black boxes
my computer won't let me so its hard for me to tell what ur doing .

I am 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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