Laplace Transformations Step Functions

In summary, the homework statement is that the attachment is the problem. The Attempt at a Solution is that the student is trying to solve the laplace transformations but does not understand how to start with the heaviside functions for the 5t and the 30. The student is trying to find F(s) but is not sure how to treat the -3t attached to the U9(t).
  • #1
Mikesgto
18
0

Homework Statement



The attachment is the problem.

Homework Equations


The Attempt at a Solution



I understand how to go about solving the laplace transformations but I have no idea how to start with the Heaviside functions for the 5t and the 30. What I got was 5t+30U6(t) but it turned out to be wrong. Any help would be greatly appreciated.
 

Attachments

  • problem.png
    problem.png
    1.6 KB · Views: 393
Physics news on Phys.org
  • #2
That's close. If t=7, the forcing function should be equal to 30, but your expression would give 5(7)+30U6(7) = 35+30 = 65. Can you see why it's wrong and how to fix it?
 
  • #3
5t-5U6(t)(t-6)

I got this one after a ton of work, but I am failing to grasp the concept going behind this.
 
  • #4
You can look at it this way. You only want the 5t term to contribute between t=0 and t=6, so after t=6, you have to subtract out 5t to cancel the other 5t term:

F(t) = 5t U0(t) + 30 U6(t) - 5t U6(t)

(Often, it's understood that the input is 0 for t<0, so you don't write the U0(t) in explicitly.) When t>6, the first and last terms cancel, and you're just left with F(t)=30.

You can also look at it like this, which is the way I prefer. If you rearrange the terms slightly, you can write

F(t) = 5t (U0(t) - U6(t)) + 30 U6(t)

The factor U0(t)-U6(t) is equal to 1 between t=0 and t=6 and is equal to 0 everywhere else, so it turns the 5t term on between t=0 and t=6 and disables it for the other times.
 
  • #5
ok, so the next problem i have is attached and the solution i came up with is
3t-3tU9(t)

Am I on the right path there?
 

Attachments

  • problem11.png
    problem11.png
    1.4 KB · Views: 348
  • #6
Yup, that's right.
 
  • #7
Thanks so much for all the help. The next part of my solutions are finding out F(s). Do I treat the -3t attached to the U9(t) as my f(t)?
 
  • #8
I'm not sure what you're asking. It might help if you show us your calculations.
 

Related to Laplace Transformations Step Functions

1. What is a Laplace Transformation Step Function?

A Laplace Transformation Step Function is a mathematical tool used to convert a function of time into a function of frequency. It is often used in engineering and physics to analyze systems and signals that change over time.

2. How does a Laplace Transformation Step Function work?

A Laplace Transformation Step Function involves taking the integral of a function of time multiplied by a step function. The step function ensures that the integral only considers values after a certain point in time, making it a useful tool for analyzing systems that have a sudden change or "step" in their behavior.

3. What is the purpose of using a Laplace Transformation Step Function?

The main purpose of using a Laplace Transformation Step Function is to simplify the analysis of systems that change over time. By converting a function of time into a function of frequency, it becomes easier to solve differential equations and evaluate the stability and behavior of the system.

4. What are some real-world applications of Laplace Transformation Step Functions?

Laplace Transformation Step Functions are commonly used in electrical engineering to analyze circuits and signals, in control systems to design and optimize controllers, and in physics to study dynamic systems such as vibrations and oscillations.

5. Are there any limitations to using Laplace Transformation Step Functions?

One limitation of using Laplace Transformation Step Functions is that they assume that the system being analyzed is linear and time-invariant. This means that the system's behavior does not change over time and its response is proportional to its input. Additionally, they may not be suitable for analyzing systems with discontinuous or singular inputs.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
157
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
823
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
Back
Top