# Homework Help: Problem in Laplace transform

1. Dec 2, 2014

### ViolentCorpse

1. The problem statement, all variables and given/known data
I'm given a transfer function
C(s)=10R(s)/(s+4)

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

3. 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?

2. Dec 2, 2014

### Orodruin

Staff Emeritus
Did you perhaps transform the product in transformed space to a product in normal space?

3. Dec 2, 2014

### vela

Staff Emeritus
What do you mean you inserted the value of r(t) in it? Please show your actual work.

4. Dec 3, 2014

### ViolentCorpse

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: Dec 3, 2014
5. Dec 3, 2014

### Orodruin

Staff Emeritus
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. Dec 3, 2014

### ViolentCorpse

I vaguely recalled something like that the moment you mentioned it before. I feel so stupid.

Thanks so much, Orodruin!