# The derivative of a unit step function is the impulse functions

Discussion in 'Math' started by Joe24, Jan 23, 2009.

1. ### Joe24 Thread Starter Active Member

May 18, 2007
52
0
Hello all,

By definition, we are taught that the derivative of the unit step function is the impulse function (or delta function, which is another name).

u(t) = 1 for t>0
= 0 otherwise

So when t is equal to some infinitesimal point to the right of 0, then u(t) shoots up to equal to a constant 1. From that point on, u(t) = 1 for all time (to positive infinity).

So the derivative of a constant we know is equal to 0. And also the derivative of an infinite slope is not defined. The infinite slope occurs when
u(t) goes from magnitude 0 to 1.

So how is the derivative defined to be equal to 1.

Thanks

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Where did you find the explanation about the derivative?

3. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Functions belong to different classes based on continuity and differentiability. The generalized functions such as the unit step and the unit impulse do not belong to the class of continuous functions, let alone those that are differentiable.

They are called generalized functions and treated as if certain properties were true. The unit ramp belongs to C0, the class of continuous functions. The unit parabola belongs to C1, the class of continuous functions with a continuous 1st derivative.

Continuous functions
http://en.wikipedia.org/wiki/Continuous_function
Generalized functions
http://en.wikipedia.org/wiki/Generalized_function
Weak derivatives
http://en.wikipedia.org/wiki/Weak_derivative
Lebesgue Spaces
http://en.wikipedia.org/wiki/Lebesgue_space

4. ### Ratch New Member

Mar 20, 2007
1,068
3
Joe24,

It's not. You are referring to the unit step function, right? It is easily proven that the derivative of the unit step function is the impulse function. Because the area under the impulse function is indefinite, it was defined to be 1 by Paul Dirac who proposed it. So the function value is not considered to be 1, but the area under the impulse function is. Ratch

Last edited: Jan 24, 2009
5. ### Joe24 Thread Starter Active Member

May 18, 2007
52
0
MIK3:
The relationship between the two I have know for a long time. Just that recently I was in a class where the subject came up again. But as I thought about it for a little while, it just didn't make sense to me. The explanation comes from the way I think about this particular relationship in my own head.

Ratch:
Yes, I was referring to the unit step function. And I see where I might have confused the readers with my last question. What I should have asked was at time t = 0, we know that the impulse function is equal to 1. So taking the derivative of u(t) at t = 0, should equal to 1. But at time t = 0, from what I have read on the web and in some texts, the u(0) = to either 0, or not defined, depending on which reference I am reading. Which I don't understand why some define this unit step at this particular point in time different from others.

In either case, the derivative is NOT equal to 1...

Papabravo:
What would be the assumptions here?

Thanks

6. ### Ratch New Member

Mar 20, 2007
1,068
3
Joe24,

No we don't. The impulse function is indefinite. It can be defined many ways. The area under the function is defined to be 1. There is a difference between the value of the function (which is indefinite) and the area it encompasses (which is defined to be 1).

That's right. The derivative of a unit step function is indefinite, not 1. The AREA of the unit function derivative (impulse function) is 1, however.

Ratch

Last edited: Jan 24, 2009
7. ### Joe24 Thread Starter Active Member

May 18, 2007
52
0
Thanks alot for all the input. I appreciate it much.