Why don't my two approaches to finding c(t) match in Laplace transform problem?

In summary: Your input has helped me a lot.In summary, the first method gave me c(t)=60e-4tu(t) while the second method yielded c(t)=15(1-e-4t)c(t)=10e-4tr(t). I don't understand why the second method gives me the correct answer, but the first method didn't.
  • #1
ViolentCorpse
190
1

Homework Statement


I'm given a transfer function
C(s)=10R(s)/(s+4)

And I have to find c(t) for r(t)=6u(t)

The Attempt at a Solution


First I did this problem by taking inverse laplace of the transfer function, and inserting the value of r(t) in it.

Next I did the same problem by first taking laplace of r(t), inserting it in the transfer function equation and then taking the inverse laplace to get c(t).

The answers I got from the two approaches don't match. I got c(t)=60e-4t by the first method and c(t)=15(1-e-4t) by the second method.

I don't understand why they don't match. I know that the second approach is giving me the right answer, but I can't figure out what's wrong with the former approach?

Thank you for your time!
 
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  • #2
Did you perhaps transform the product in transformed space to a product in normal space?
 
  • #3
ViolentCorpse said:
First I did this problem by taking inverse laplace of the transfer function, and inserting the value of r(t) in it.
What do you mean you inserted the value of r(t) in it? Please show your actual work.
 
  • #4
C(s)=10R(s)/(s+4)

1st method:
c(t)=10e-4tr(t) (Is this what you meant by transforming a product to normal space, Orodruin?)
Since r(t)=6u(t)
So, c(t)=60e-4tu(t)

2nd method:
r(t)=6u(t)
R(s)=6/s
So
C(s)=60/s(s+4)
Applying partial fractions to get
C(s)=15/s - 15/(s+4)
c(t)=15(1-e-4t)
 
Last edited:
  • #5
ViolentCorpse said:
c(t)=10e-4tr(t) (Is this what you meant by transforming a product to normal space, Orodruin?)

Yes, the product of two Laplace transforms is not the Laplace transform of the product, it is the Laplace transform of the convolution of the functions. You therefore cannot do what you did in this step.
 
  • #6
Orodruin said:
Yes, the product of two Laplace transforms is not the Laplace transform of the product, it is the Laplace transform of the convolution of the functions. You therefore cannot do what you did in this step.
I vaguely recalled something like that the moment you mentioned it before. I feel so stupid.

Thanks so much, Orodruin!
 

Related to Why don't my two approaches to finding c(t) match in Laplace transform problem?

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 often used in engineering and physics to solve problems involving differential equations.

2. What is a problem in Laplace transform?

A problem in Laplace transform refers to any difficulty or challenge encountered while using the Laplace transform to solve a particular mathematical or engineering problem. This could include issues such as convergence, improper integrals, or incorrect assumptions about the initial conditions.

3. How do you solve a problem in Laplace transform?

To solve a problem in Laplace transform, it is important to understand the underlying principles and techniques involved. This includes knowing how to apply the Laplace transform to a function, how to manipulate complex expressions, and how to use properties of the Laplace transform to simplify the problem. It may also be helpful to consult with other experts or references for guidance.

4. What are some common mistakes made when using Laplace transform?

Some common mistakes made when using Laplace transform include forgetting to account for initial conditions, using the wrong transformation formula, and not checking for convergence before applying the transform. It is also important to be careful when manipulating complex expressions and to double check all calculations for accuracy.

5. How is Laplace transform used in real-world applications?

Laplace transform has many real-world applications, particularly in engineering and physics. It can be used to solve problems involving electrical circuits, mechanical systems, and heat transfer. It is also useful in signal processing, control systems, and quantum mechanics. Overall, the Laplace transform is a powerful tool for solving complex mathematical problems in various fields of science and engineering.

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