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

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The discussion centers on a discrepancy in two methods for finding c(t) from a given transfer function C(s)=10R(s)/(s+4) with r(t)=6u(t). The first method involves taking the inverse Laplace transform directly, resulting in c(t)=60e-4t, while the second method, which applies the Laplace transform to r(t) first, yields c(t)=15(1-e-4t). The key issue identified is that the product of Laplace transforms does not equate to the Laplace transform of the product of the original functions; instead, it represents the convolution of those functions. This misunderstanding led to the incorrect result in the first method. The second approach is confirmed to be correct, clarifying the importance of proper application of Laplace transform properties.
ViolentCorpse
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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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Did you perhaps transform the product in transformed space to a product in normal space?
 
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.
 
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:
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.
 
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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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