Solve Laplace Transform: te^t*H(t-1)

There are lots of explanations on the web, but you have to be careful to read and understand them, not just look at formulas.
  • #1
nobodyuknow
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Homework Statement



Laplace transform: te^t*H(t-1)

Homework Equations



L{te^t} = 1/(s-1)^2
L{H(t-1)} = e-s/s

The Attempt at a Solution



Using e^(at)f(t) -> F(s-a)
with..e^(1t)H(t-1)
We can get, e^(-1s-1)/(s-1) = e^(-s-1)/(s-1) [I think?]
Thus applying tf(t) -> -dF/ds
Derive[e^(-s-1)/(s-1)] w.r.t. s, you get:
-se^(-s-1)/(s-1)^2

Which is close, but, not exact.

I know the answer is se^s-1/(s-1)^2

But how do they come to that?
I don't think the order of my working would affect the final outcome.

All help appreciated
 
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  • #2
You need to show some attempt at solving the problem on your own.
 
  • #3
nobodyuknow said:

Homework Statement



Laplace transform: te^t*H(t-1)

Homework Equations



L{te^t} = 1/(s-1)^2
L{H(t-1)} = e-s/s

The Attempt at a Solution



I know the answer is se^s-1/(s-1)^2

But how do they come to that?

How do you combine the two individual laplace transforms to calculate the final?

No. L{H(t-1)} is not e-s/s = e-1; it is e^(-s)/s. You must learn to use parentheses when writing things in plain text.

If I were doing the problem I would not bother trying to combine LTs using some "rules"; I would just write out the actual integration and go on from there.
 
  • #4
Thanks, that was a mistake sorry

So basically use the integration rule and integrate te^t and then apply the Heaviside function?
I'm sorry, I'm not too confident with the Integral method, but..

integral{te^t} = (t-1)e^(t-st)

I'm not sure how to work from there.

[Googling Heaviside Integration as of now]
 
  • #5
nobodyuknow said:
Thanks, that was a mistake sorry

So basically use the integration rule and integrate te^t and then apply the Heaviside function?
I'm sorry, I'm not too confident with the Integral method, but..

integral{te^t} = (t-1)e^(t-st)

I'm not sure how to work from there.

[Googling Heaviside Integration as of now]

What do you mean by the "integral method"? What do you think the Laplace transform of f(t) actually IS? Go back and look at the basic definitions---it is, by definition, an integral of a certain type. All the so-called "rules" you want to use are just properties of integrals, applied to some special cases.

And no, you do NOT do the integral and then apply the Heaviside function; the Heaviside function is part of the definition of f(t) in this case, and comes before any integrations.

It is not at all clear to me that you really understand what you are doing; that is why I suggest you go right back to square one and start with basic definitions.
 

FAQ: Solve Laplace Transform: te^t*H(t-1)

1. What is a Laplace Transform?

A Laplace Transform is a mathematical tool that allows us to convert a function of time into a function of complex frequency. It is typically used in engineering and physics to solve differential equations.

2. How do you solve a Laplace Transform?

To solve a Laplace Transform, you first need to take the integral of the function multiplied by e^-st, where s is the complex frequency. This will give you the Laplace Transform in terms of s. Then, you can use tables or inverse Laplace Transform techniques to find the original function in terms of time.

3. What is the significance of te^t in this Laplace Transform?

The term te^t represents the time function in the Laplace Transform. It is multiplied by a step function (H(t-1)) to ensure that the function is only evaluated after t=1. This is known as a delayed step function and is commonly used in applications such as control systems.

4. How is the Laplace Transform used in real-world applications?

The Laplace Transform has a wide range of applications in engineering and physics. It is commonly used to solve differential equations in control systems, signal processing, and circuit analysis. It is also used in solving heat transfer problems and studying fluid dynamics.

5. Are there any limitations to using Laplace Transform?

While the Laplace Transform is a powerful tool, it does have some limitations. It can only be applied to functions that have a Laplace Transform in the first place. It also assumes that the function is causal, meaning that it only depends on past values and not future values. Additionally, the inverse Laplace Transform may not exist for certain functions, making it difficult to find the original function in terms of time.

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