Laplace Transform of Product of Two Functions

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 9K views
trust
Messages
12
Reaction score
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
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.
 
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
 
Thanks, I got it now. I was thinking that the transform of the product was the product of the transforms.
 
Last edited: