# Derivative of unit step function

1. May 3, 2003

### Will

[SOLVED] Derivative of unit step function

How does one do this, for example x= e^(-3t)u(t-4); how do you get x' ??

2. May 4, 2003

### enigma

Staff Emeritus
Do laplace transforms on it.

3. May 4, 2003

### Hurkyl

Staff Emeritus
Write out the definition of the unit step function and it might be easier to see.

4. May 4, 2003

### Will

I think I got it now. I used the property L{f'}(s) = sL{f}(s) - f(0)
Is that correct?

5. May 4, 2003

### Tyger

The Fourier Transform

can also be used. It can be used for many unbounded functions.

6. May 4, 2003

### Hurkyl

Staff Emeritus
You could just differentiate it directly.

x(t) = e^(-3t)u(t-4)

is equivalent to:

Code (Text):

x(t) =  e^(-3t)   (for t > 4)
0      (for t < 4)

with x(4) depending on the precise definition of u.

Differentiating on each piece gives:

Code (Text):

x'(t) = (-3) e^(-3t)   (for t > 4)
0            (for t < 4)

And x'(4) does not exist because x(t) is discontinuous at t = 4

IOW:

x'(t) = (-3) e^(-3t) u(t - 4) for t [x=] 4

Last edited: May 4, 2003
7. May 6, 2003

### ahrkron

Staff Emeritus
Sometimes you can safely assume the derivative of a step to be a delta function (for instance, when you integrate a delta, you get a step).

They need to be used as distributions, and there may be some requirements on the functions you use along with them (integrability, continuity,...).

I'm sorry I don't remember much about it.