Laplace transform of Heaviside function

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The Laplace transform of H(-t-17) is zero for all t > 0 because H(-t-17) evaluates to zero in this range. The Heaviside function H(t) is a unit step function, and H(-t-17) represents a reflection and translation, which results in it being zero for positive t. The correct form of the shifting theorem is L(H(t-a)) = (e^-as)/s, but this does not apply here since the function is zero. Understanding the properties of the Heaviside function clarifies that the Laplace transform only considers positive time values. Thus, the Laplace transform of H(-t-17) is simply the transform of the zero function.
gravenewworld
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Homework Statement



What is the laplace transform of H(-t-17)

Homework Equations



Shifting theorem:

L(H(t-a)) = (e^-as)/s

The Attempt at a Solution



This is the only part of the problem that I can not get (this part is from a larger differential equation I'm trying to solve). I'm can't seem to figure out how the Laplace transform changes if there is a negative sign in front of the t. I'm not sure if there are some properties of the Heaviside function that will allow me to solve this that I simply don't know and that we didn't go over in class.
 
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gravenewworld said:

Homework Statement



What is the laplace transform of H(-t-17)

Homework Equations



Shifting theorem:

H(t-a) = (e^-at)/s

The Attempt at a Solution



This is the only part of the problem that I can not get (this part is from a larger differential equation I'm trying to solve). I'm can't seem to figure out how the Laplace transform changes if there is a negative sign in front of the t. I'm not sure if there are some properties of the Heaviside function that will allow me to solve this that I simply don't know and that we didn't go over in class.
It's "Heaviside" which is now fixed.

Your relevant equation is incorrect and has two errors. It should be L(H(t - a)) = (e-as)/s.

H(t) is the unit step function. H(t - a) is the translation of the unit step function by a units. H(-t - 17) = H(-(t + 17)). This involves a reflection across the vertical axis, followed by a translation to the left by 17 units.
 
gravenewworld said:

Homework Statement



What is the laplace transform of H(-t-17)

Homework Equations



Shifting theorem:

L(H(t-a)) = (e^-as)/s

The Attempt at a Solution



This is the only part of the problem that I can not get (this part is from a larger differential equation I'm trying to solve). I'm can't seem to figure out how the Laplace transform changes if there is a negative sign in front of the t. I'm not sure if there are some properties of the Heaviside function that will allow me to solve this that I simply don't know and that we didn't go over in class.

If t > 0 then -t -17 < 0 so H(-t-17) = 0 for all t > 0. Since the Laplace transform only cares about the values of a function in t > 0 the Laplace transform of H(-t-17) is equal to the Laplace transform of the zero function.
 
Thank you both. I was actually able to pull up a nice identity online for H(x), where H(x) = (x + |x|)/2x. It helped a lot as well.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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