Laplace Transform of Product of Two Functions

But the formula for the Laplace transform of a product is different, so I just needed to apply that formula. In summary, to find the Laplace transform of u(t-∏/2)et, you can write it as e^{(t-\frac \pi 2)}e^{\frac \pi 2} and use the formula \mathcal L(f(t-a)u(t-a) = e^{-as}\mathcal Lf(t) to get the correct answer.
  • #1
trust
12
0

Homework Statement



Laplace Transform of u(t-∏/2)et

(u is unit step function)

Homework Equations



Laplace Transform Table (any)

The Attempt at a Solution



I tried using the Laplace transform for the unit step function and the exponential function.

L{u(t-∏/2)} = e-(∏s)/2

L{et} = 1/(s-1)

L{u(t-∏/2)et} = (e-(∏s)/2)(1/(s-1))

When I check my answer this is all correct except I'm missing an e∏/2 term. (The correct answer is (e-(∏s)/2)(1/(s-1))(e∏/2). Does anyone know where this extra term comes from?
 
Physics news on Phys.org
  • #2
trust said:

Homework Statement



Laplace Transform of u(t-∏/2)et

(u is unit step function)

Homework Equations



Laplace Transform Table (any)

The Attempt at a Solution



I tried using the Laplace transform for the unit step function and the exponential function.

L{u(t-∏/2)} = e-(∏s)/2

L{et} = 1/(s-1)

L{u(t-∏/2)et} = (e-(∏s)/2)(1/(s-1))

When I check my answer this is all correct except I'm missing an e∏/2 term. (The correct answer is (e-(∏s)/2)(1/(s-1))(e∏/2). Does anyone know where this extra term comes from?

I assume you have the formula[tex]
\mathcal L(f(t-a)u(t-a) = e^{-as}\mathcal Lf(t)[/tex]Write your function as[tex]
e^{(t-\frac \pi 2)}e^{\frac \pi 2}[/tex]and use that.
 
  • #3
trust said:

Homework Statement



Laplace Transform of u(t-∏/2)et

(u is unit step function)

Homework Equations



Laplace Transform Table (any)

The Attempt at a Solution



I tried using the Laplace transform for the unit step function and the exponential function.

L{u(t-∏/2)} = e-(∏s)/2

L{et} = 1/(s-1)

L{u(t-∏/2)et} = (e-(∏s)/2)(1/(s-1))

When I check my answer this is all correct except I'm missing an e∏/2 term. (The correct answer is (e-(∏s)/2)(1/(s-1))(e∏/2). Does anyone know where this extra term comes from?

The Laplace transform of a product is NOT the product of the transforms. Why don't you just do the problem directly: u(t - π/2) is 0 for t < π/2 and 1 if t > π/2, so you just have a simple integral of an exponential.

RGV
 
  • #4
Thanks, I got it now. I was thinking that the transform of the product was the product of the transforms.
 
Last edited:

1. What is the Laplace transform of the product of two functions?

The Laplace transform of the product of two functions f(t) and g(t) is equal to the convolution of their individual Laplace transforms, which is denoted as F(s)*G(s).

2. How is the Laplace transform of a product useful in solving differential equations?

The Laplace transform of a product can be used to solve differential equations by reducing them to algebraic equations, which are often easier to solve. This method is particularly useful for solving initial value problems.

3. Can the Laplace transform of a product be used to find the inverse Laplace transform?

Yes, the inverse Laplace transform of a product can be found by using the convolution theorem, which states that the inverse Laplace transform of a convolution is equal to the product of the individual inverse Laplace transforms.

4. Are there any special cases for the Laplace transform of a product?

Yes, if one of the functions is a unit step function, the Laplace transform of the product can be simplified to a single term. Also, if one of the functions is a Dirac delta function, the Laplace transform of the product will be equal to the product of the other function and its derivative at t=0.

5. Is there a way to simplify the Laplace transform of a product of two exponential functions?

Yes, the Laplace transform of a product of two exponential functions can be simplified using the properties of the Laplace transform. Specifically, the product of two exponential functions can be written as a single exponential function with a modified coefficient.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
130
  • Calculus and Beyond Homework Help
Replies
7
Views
768
  • Calculus and Beyond Homework Help
Replies
1
Views
616
  • Calculus and Beyond Homework Help
Replies
3
Views
789
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
886
Back
Top