Laplace transforms, heaviside step functions

In summary: So in summary, the original problem probably had a typo, and for the second one, sin(3π) and cos(3π) are both equal to 0, which makes the negative sign disappear and simplifies the expression.
  • #1
missavvy
82
0
Hey guys! I just have a question regarding some notes I've been looking at and trying to understand...

Homework Statement


eq0018MP.gif

Taking the transform:

eq0019MP.gif


Just wondering, can someone explain to me why the e-3s/s is positive, and not negative because it is originally -u3(t)...

Another quick one:

Write this in terms of Heaviside function:

f(t) =
{ 0, 0 <= t <= pi
{ -sin3t, t >= pi

1) So you would get f(t) = -(upit)sin3t

And somehow it ends up as f(t) = (upit)sin(3(t-pi))
Can someone explain this step as well? Just what happened to the negative in front of it, and why it became 3(t-pi)

Thanks!
 
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  • #2
missavvy said:
Hey guys! I just have a question regarding some notes I've been looking at and trying to understand...

Homework Statement


eq0018MP.gif

Taking the transform:

eq0019MP.gif


Just wondering, can someone explain to me why the e-3s/s is positive, and not negative because it is originally -u3(t)...
It's probably just a typo in the original problem.

Another quick one:

Write this in terms of Heaviside function:

f(t) =
{ 0, 0 <= t <= pi
{ -sin3t, t >= pi

1) So you would get f(t) = -(upit)sin3t

And somehow it ends up as f(t) = (upit)sin(3(t-pi))
Can someone explain this step as well? Just what happened to the negative in front of it, and why it became 3(t-pi)
Try expanding sin[3(t-pi)] using the angle addition formula for sine. The reason it was written this way is probably because it's now in the form where the time-shift property of the Laplace transform can be applied.
 

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  • #3
Thanks for the first one that makes sense.

For the second one, I attempted using the identity and get:

f(t) = -upi(t)sin(3t)
= -L{upi(t)sin3(t-pi+pi)}
= -L{upi(t)[sin3(t-pi)cos(3pi) + sin(3pi)cos3(t-pi)]}
taking the transform:
= -e-pi*s(3cos(3pi) + 3s*sin(3pi) / s2 + 9

Hmm I don't understand how it is suppose to be just upi(t)sin(3(t-pi)) = e-pi*s(3) / (s2 + 9)(s2 + 1)

What happened when you add the pi and subtract it, and what happened to the negative in front of it all?
 
  • #4
What are cos(3π) and sin(3π) equal to? Also, I didn't mean that you needed to use the time-shift property to get that result. It's just basic trig.
 

1. What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is commonly used in engineering and physics to analyze systems and their behavior over time.

2. How is a Laplace transform calculated?

A Laplace transform is calculated using an integral formula that involves the original function, a complex exponential term, and a variable of integration. This integral can be solved using techniques such as partial fraction decomposition or the Laplace transform table.

3. What is the purpose of the Heaviside step function in Laplace transforms?

The Heaviside step function, also known as the unit step function, is used in Laplace transforms to account for the time delay in a system. It represents a sudden change or "step" in a function's value at a specific time.

4. How are Laplace transforms and Fourier transforms related?

Laplace transforms and Fourier transforms are both mathematical tools used to analyze functions. However, Laplace transforms are specifically used for functions that are defined for all positive values of time, while Fourier transforms are used for functions that are defined for all values of time.

5. What are some practical applications of Laplace transforms?

Laplace transforms have a wide range of practical applications in fields such as electrical engineering, control systems, signal processing, and physics. They are used to solve differential equations, analyze systems and circuits, and model physical phenomena such as heat transfer and fluid flow.

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