Step Function Help

1. Nov 24, 2009

jumbogala

1. The problem statement, all variables and given/known data
The piecewise continuous function

f(x) = { -8 for x ≤ 8
-x – 1 for x > 8}

Rewrite it in step functions.

2. Relevant equations

3. The attempt at a solution
I think it's 1-(-8)(h(x-8)) + (-x-1)(h(x-8))

Where h(x-a) is the Heaviside function (x-8) meaning it "jumps up" to 1 at x = a.

Is that right? It seems wrong to me.

2. Nov 24, 2009

Staff: Mentor

Why would you need a step function for x≤8?

Oh, and the function doesn't look continuous at x=8, unless I'm missing something. Is there a typo in the definition?

3. Nov 24, 2009

jumbogala

It's not continuous at x = 8. You have to write the entire function using step functions so that it's in the form

f(x) = ___________ + u8(x)* __________

"Fill in the blanks", basically. It will be two step functions added together.

4. Nov 24, 2009

Staff: Mentor

But in your original post (OP), you said

5. Nov 24, 2009

jumbogala

That's what the question says =/ Each piece of the function is continuous in pieces. The goal of this is to find the Laplace transform of f(x), if that helps.

6. Nov 24, 2009

Staff: Mentor

Well, I'm no expert, but being continuous means no discontinuities:

http://en.wikipedia.org/wiki/Piecewise_continuous

But whatever. It seems like if you want to use a step funtion for x≤8, then that one should run backwards in x, no?

7. Nov 24, 2009

jumbogala

Oh well I suppose it's not discontinuous at x = 8. The definition of the function just changes at x = 8, according to Wikipedia.

Yes, I think so. The step function I got was -8 + u8(t)(7-t). (Still not sure if that's entirely correct...)

So that's what I need to find the Laplace transform of. Laplace of -8 is -8/s

Laplace transform of u8(t)(7-t) is what's troubling me.

8. Nov 24, 2009

HallsofIvy

Staff Emeritus
No, jumbogala, you were right here and, unfortunately, berkeman was wrong (I'll never let him live it down!). "Piecewise continuous" means "continuous in pieces"- that is that it is continuous everywhere except possibly at a finite number of point, separating the continuous "pieces".

And, berkeman, he does NOT "want to use a step funtion for x≤8", he wants to use a step function for the entire function. That is, write a formula, including a step function so that the entire function is given by that step function.

Let's see you suggested f(x)= 1-(-8)(h(x-8)) + (-x-1)(h(x-8)). Let's check to see if that is correct. h(x) is 0 if x< 0 so h(x-8) is 0 if x is less than 8. If x< 8 then f(x)= 1-(-8)(0)+ (-x-1)(0)= 1. No, that isn't what you want- you want -8. Also h(x-8) is equal to 1 if x is larger than or equal to 8. If x is larger than or equal to 8, f(x)= 1- (-8)(1)+(-x-1)(1)= 1+8- x-1= 8- x. No, that isn't it either.

Think about this: If f(x)= u(x) for x< a and f(x)= v(x) for $x\ge a$, and we want to write it as f(x)= p(x)h(x-a)+ q(x), what must p and q equal? Well, if x< a, h(x-a)= 0 so f(x)= p(x). Obviously, we must have p(x)= u(x). If $x\ge a$, h(x-a)= 1 so f(x)= u(x)+ q(x)(1)= v(x) so q(x)= v(x)- u(x).

In this problem u(x)= -8 and v(x)= -x-1

9. Nov 25, 2009

Staff: Mentor

Thanks, Halls. Like I said, I'm no expert in this. Still, piecewise continuous used to mean something different to me. Guess I need to review some, eh?

10. Nov 25, 2009