Laplace Transform of Product of Two Functions

[trust]

Homework Statement

Laplace Transform of u(t-∏/2)e^t

(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{e^t} = 1/(s-1)

L{u(t-∏/2)e^t} = (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?

 

[LCKurtz]

Homework Statement

Laplace Transform of u(t-∏/2)e^t

(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{e^t} = 1/(s-1)

L{u(t-∏/2)e^t} = (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<br /> \mathcal L(f(t-a)u(t-a) = e^{-as}\mathcal Lf(t)Write your function as<br /> e^{(t-\frac \pi 2)}e^{\frac \pi 2}and use that.

 

[Ray Vickson]

Homework Statement

Laplace Transform of u(t-∏/2)e^t

(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{e^t} = 1/(s-1)

L{u(t-∏/2)e^t} = (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

 

[trust]

Thanks, I got it now. I was thinking that the transform of the product was the product of the transforms.