# Heaviside (Unit step function) question

1. Jul 11, 2008

### math_04

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

The function f(t) is defined for t>=0 by

f(t) = 1 for 0<= t <= 1 , t-2 for 1 <=t <= 2 and 0 for t >2

Express f(t) in terms of the Heaviside function and hence or otherwise find L(f(t)), the Laplace transform of f(t)

2. Relevant equations

3. The attempt at a solution

So i figured out the Heaviside function is

f(t) = u (t) - (t-2)u(t-2) + u(t-2)

Is the above function right? One more thing, how do u figure out the signs like how do u know it is u(t-2) instead of -u(t-2)?

So to find L(f(t) how do u do do that. I know that u(t) has to be changed to something else but thing is I have no idea how to manipulate these kind of functions. Please do offer help.

Thanks.

2. Jul 11, 2008

### CompuChip

Did you check it? Suppose 0 <= t <= 1. Then f(t) = 1.
What does the expression you gave evaluate to? What is the step function when the argument is negative and when it is positive?

So let's try it this way: can you construct a function in terms of theta functions that is constant 1 on the interval [a, b] and zero outside that?

3. Jul 11, 2008

### HallsofIvy

Staff Emeritus
That is exactly the same as u(t)- (t-3)u(t-2). Is that what you intended to write or is one of those supposed to be u(t-1)?

Why didn't you just try evaluating it for various t to see? In particular, if t is between 1 and 2, assuming that you meant f(t)= u(t)- (t-2)u(t-2)+ u(t-2), u(t) and u(t-1) would be 1 while u(t-2) would be 0. You would have f(t)= 1- (t-2)+ 0= 3- t. No, that's not what you want. Remember that once t> 0, u(t) stays 1. You can't just ignore it when t becomes greater than 1. Yes, since f(t)= 1 for t between 0 and 1, You want to start with u(t). But if you were to write f(t)= u(t)+ a(t)u(t-1), when t is between 1 and 2, you have f(t)= 1+ a(t)= t-2. What is a(t)? Finally, if you write f(t)= u(t)+ a(t)u(t-1)+ b(t)u(t-2), for t > 2, you have f(t)= 1+ a(t)+ b(t)= 0. Since you already know what a(t) is, it is easy to solve for b(t).

u(t-a) is defined to be 0 if t< a, 1 if $t\ge a$. So f(t)u(t-a) is 0 if t< a, f(t) if $t\ge a$. Since the laplace transform of f(t) is defined to be
$$\int_0^\infty f(t)e^{-st}dt$$
the Laplace transform of f(t)u(t-a) is
$$\int_a^\infty f(t)e^{-st}dt$$
because the function value for t< a is just 0.

Last edited: Jul 11, 2008
4. Jul 11, 2008

### Defennder

In general, if

$$f(t) = \begin{array}{c}g(t) \ \mbox{for} \ a\leq t \leq b \\ 0 \ \mbox{everywhere else} \ \end{array}$$

then $$f(t) = g(t)[u(t-b)-u(t-a)]$$.