# Unit step function and Laplace

• Jan 10th 2008, 07:24 PM
sixstringartist
Unit step function and Laplace
I usually try to use Latex and make everything look pretty and easy to read, but Im in a bit of a rush so please excuse my ugly equations.

Im doing one-sided Laplace transforms involving unit step functions and Im not sure how to deal with one of the form f(x) * u(t) * u(c-t)

Explicitly my problem is:

f(t) = [2t^2 - 10t] * u(t) * u(2-t)

I know what the graph looks like, but Im not sure how to find F(s) in this situation.

• Jan 10th 2008, 08:32 PM
mr fantastic
Quote:

Originally Posted by sixstringartist
I usually try to use Latex and make everything look pretty and easy to read, but Im in a bit of a rush so please excuse my ugly equations.

Im doing one-sided Laplace transforms involving unit step functions and Im not sure how to deal with one of the form f(x) * u(t) * u(c-t)

Explicitly my problem is:

f(t) = [2t^2 - 10t] * u(t) * u(2-t)

I know what the graph looks like, but Im not sure how to find F(s) in this situation.

Does * represent multiplication or convolution?
• Jan 10th 2008, 08:32 PM
CaptainBlack
Quote:

Originally Posted by sixstringartist
I usually try to use Latex and make everything look pretty and easy to read, but Im in a bit of a rush so please excuse my ugly equations.

Im doing one-sided Laplace transforms involving unit step functions and Im not sure how to deal with one of the form f(x) * u(t) * u(c-t)

Explicitly my problem is:

f(t) = [2t^2 - 10t] * u(t) * u(2-t)

I know what the graph looks like, but Im not sure how to find F(s) in this situation.

u(t)u(2-t)=u(2-t)

(that is for the product to be non zero both u(t) and u(2-t) must both be non-zero
and when they are they are both 1, ..)

RonL
• Jan 11th 2008, 05:00 AM
sixstringartist
they represent multiplication,

but I dont see how u(t)u(2-t) = u(2-t) because that would include negative values. If anything it can be written as u(t)u(2-t) = u(t) - u(t-2) is that how I am supposed to evaluate this?
• Jan 11th 2008, 06:49 AM
sixstringartist
Ive found what I was looking for:

$
Au(t)u(T-t) = A[u(t) - u(t-T)]
$

Thanks for the help
• Jan 11th 2008, 05:59 PM
mr fantastic
Quote:

Originally Posted by sixstringartist
they represent multiplication,

but I dont see how u(t)u(2-t) = u(2-t) because that would include negative values. If anything it can be written as u(t)u(2-t) = u(t) - u(t-2) is that how I am supposed to evaluate this?

CaptainBlack made a boo boo .... u(t)u(2-t) = u(2-t) is not correct. Note however that u(t)u(t-2) = u(t-2) is correct .....The good captain made a simple and easy-to-make slip-up.

Well spotted and good for you in finding the correct expression.
• Jan 12th 2008, 02:01 AM
CaptainBlack
Quote:

Originally Posted by mr fantastic
CaptainBlack made a boo boo .... u(t)u(2-t) = u(2-t) is not correct. Note however that u(t)u(t-2) = u(t-2) is correct .....The good captain made a simple and easy-to-make slip-up.

Well spotted and good for you in finding the correct expression.

Opps, and so:

$u(t)u(2-t) = \left\{ \begin{array} {ll} 1,& u \in [0,2]\\ 0, & \mbox{otherwise} \end{array} \right.$

(OP double check the end points yourself to see if the interval over which this is 1 is open or closed)

RonL
• Feb 29th 2008, 11:49 AM
shenky365
Attachment 5208

HI

I am new here and i am having a little trouble finding the constants A, B, C and t in this question. Please i need help.
the question is :-

f(t) = AH(t) + (-8t + B) H(t - 5) - (-8t - C) H(t - 22) where H(t) is the unit step.
thanks

Attachment 5209