Laplace Transform for a circuit

[Pete_01]

Homework Statement

It's been a while since I've done laplace transforms, but I need some help with one!
I am supposed to find the laplace tranform for the following signal. This is for my circuits class btw.

x_a(t) = 2u(t-1)

Homework Equations

Laplace tables. http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

The Attempt at a Solution

I am guessing that the u(t-a) part is a unit step function correct? From there, I am not exactly sure how to go. I don't see any transforms that fit exactly.

 

[Mark44]

#25 is u(t - c). Yes, this is a unit step function.

And L{kf(t)} = kL{f(t)} because of the linearity of the Laplace transform.

 

[Pete_01]

Thanks for the reply mark!
So using 25, I would have 2*e^-s/s ?

#25 is u(t - c). Yes, this is a unit step function.

And L{kf(t)} = kL{f(t)} because of the linearity of the Laplace transform.

 

[Mark44]

Yes.

 

[DeeNos]

Hello, thank you for reaching out for help with the Laplace transform for your circuit homework. You are correct, u(t-a) is a unit step function which represents a sudden change in the signal at time t=a. To find the Laplace transform of this signal, we can use the property of the Laplace transform that states:

L{u(t-a)f(t-a)} = e^(-as)F(s)

Using this property, we can rewrite the signal xa(t) as:

xa(t) = 2u(t-1) = 2u(t-1)1(t)

Where 1(t) is the identity function. Now, we can apply the property mentioned above to find the Laplace transform:

L{xa(t)} = L{2u(t-1)1(t)} = e^(-1s) * L{2} = 2e^(-s)

Therefore, the Laplace transform for the given signal is 2e^(-s). I hope this helps you with your homework. If you have any further questions, please feel free to ask. Good luck with your studies.