Laplace transformation problem

In summary, the conversation is about finding the Laplace transform of 4u(t-ர)cos2t and using the unit step function to do so. The attempted solution involves using the formula F(s)=s/s^2+4 and integrating using the Laplace transform. However, there is some confusion about where the factor of 4e^-ர(cosர + sinர) comes from, and the expert suggests recognizing the periodic nature of trig functions in order to simplify the solution. The final answer should not include the variable a.
  • #1
jamshaid
6
0

Homework Statement



L{4u(t-ர)cos2t}

Homework Equations



I have used UNIT STEP FUNCTION but could not get the result

The Attempt at a Solution

 
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  • #2
What did you get? Show your work, not just your final result from your attempt.
 
  • #3
L{4u(t-ர)cos2t
f(t) = cos2t
F(s) = s/ s^2+4
therefore

4s e^-ர(cosர + sinர)/s^2 +4 Ans
________________________________
that is my work
 
  • #4
jamshaid said:
L{4u(t-ர)cos2t
f(t) = cos2t
F(s) = s/ s^2+4
therefore

4s e^-ர(cosர + sinர)/s^2 +4 Ans
________________________________
that is my work

Where is the factor the of [itex]4e^{-\tau}(\sin\tau+\cos\tau)[/itex] coming from? What rule are you trying to apply here?
 
  • #5
here is my solve, if I have done correctly.
laplace.gif
 
  • #6
Please expert help me.
 
  • #7
Unfortunately, I find your handwriting a bit hard to read. Could you type out your final answer?
 
  • #8
jamshaid said:
here is my solve, if I have done correctly.
View attachment 28107

It looks good to me (although you have a very strange way of writing [itex]\pi[/itex]). On a side note, since trig functions are [itex]2\pi[/itex] periodic, you should have immediately recognized that [itex]\cos(2t+2\pi)=\cos(2t)[/itex] without appealing to a trig identity.
 
  • #9
I think the exponential term in the front incorrectly has an a in it rather than s.
 
  • #10
here is my final Answer
4e^(-as) s/(s^(2) + 4)

what do you think...
 
  • #11
The variable a shouldn't be in the final answer.
 
  • #12
vela said:
The variable a shouldn't be in the final answer.

can you giving me some hints?
 

FAQ: Laplace transformation problem

1. What is a Laplace transformation?

A Laplace transformation is a mathematical technique used to solve differential equations. It transforms a function of time into a function of frequency, making it easier to solve complex problems in physics and engineering.

2. What types of problems can be solved using Laplace transformation?

Laplace transformation can be used to solve problems in fields such as electrical engineering, mechanical engineering, and physics. It is commonly used to analyze systems with differential equations, such as circuits, control systems, and mechanical vibrations.

3. How is Laplace transformation different from Fourier transformation?

Laplace transformation is similar to Fourier transformation in that it also transforms a function from the time domain to the frequency domain. However, Laplace transformation can be used to solve more complex problems, as it includes a complex variable that allows for the analysis of transient behavior.

4. What are the advantages of using Laplace transformation?

Laplace transformation provides a powerful tool for solving differential equations, as it simplifies complex problems into algebraic equations that are easier to solve. It also allows for the analysis of transient behavior, making it useful in understanding the response of systems to sudden changes.

5. Are there any limitations to using Laplace transformation?

While Laplace transformation is a powerful tool, it does have some limitations. It can only be used for linear systems, and the function being transformed must be well-behaved (i.e. it must have a finite number of discontinuities).

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