How Does Integration of Delta Functions Translate to Heaviside Functions?

[Ted123]

Homework Statement

Compute the integral

$$F(x) = \int^x_{-\infty} f(t) \;dt$$

of the linear combination of Dirac delta-functions

$$f(t) = -2\delta (t) + \delta (t-1) + \delta (t-2)$$.

Express the result analytically (piecewise on a set of intervals) and draw a sketch of the function $$F(x)$$.

The Attempt at a Solution

Does $$F(x) = -2H(x) + H(x-1) + H (x-2)$$ where H is the Heaviside function?

I know how to express the Heaviside/Delta functions in terms of 'jumps' in a graph but the actual values could be anything couldn't they? For instance:

$$\begin{displaymath} F(x) = \left\{ \begin{array}{lr} 0, & \;x \leq 0\\ -2, & \;0 < x \leq 1\\ -1, & \;1<x\leq 2\\ 0, & \;x > 2 \end{array} \right.$$

and

$$\begin{displaymath} F(x) = \left\{ \begin{array}{lr} 1, & \;x \leq 0\\ -1, & \;0 < x \leq 1\\ 0, & \;1<x\leq 2\\ 1, & \;x > 2 \end{array} \right. \end{displaymath}$$

both respresent that linear combination of Heaviside functions don't they?

 

[L-x]

Homework Statement

Compute the integral

$$F(x) = \int^x_{-\infty} f(t) \;dt$$

of the linear combination of Dirac delta-functions

$$f(t) = -2\delta (t) + \delta (t-1) + \delta (t-2)$$.

Express the result analytically (piecewise on a set of intervals) and draw a sketch of the function $F(x)$.

The Attempt at a Solution

Does $$F(x) = -2H(x) + H(x-1) + H (x-2)$$ where H is the Heaviside function?

I know how to express the Heaviside/Delta functions in terms of 'jumps' in a graph but the actual values could be anything couldn't they? For instance:

$$\begin{displaymath} F(x) = \left\{ \begin{array}{lr} 0, & \;x \leq 0\\ -2, & \;0 < x \leq 1\\ -1, & \;1<x\leq 2\\ 0, & \;x > 2 \end{array} \right.$$

and

$$\begin{displaymath} F(x) = \left\{ \begin{array}{lr} 1, & \;x \leq 0\\ -1, & \;0 < x \leq 1\\ 0, & \;1<x\leq 2\\ 1, & \;x > 2 \end{array} \right. \end{displaymath}$$

both respresent that linear combination of Heaviside functions don't they?

should make it a bit easier to read, can't get your array to work though.

 

[Ted123]

should make it a bit easier to read, can't get your array to work though.

This should make it easier!

[PLAIN]http://img824.imageshack.us/img824/9868/heaviside.png

 

[L-x]

You have the correct shape, however you may define a value at x= 0, 1, 2 for the heavside step function. Personally I think 1/2 is the most sensible, because it means that the function is odd and the approximation H(x)={1+tanh(kx)}/2 holds exactly in the limit k->infinity.

0, 1 and 0.5 are all valid choices to use, and will often depend on what exactly you are using H for.

http://www.wolframalpha.com/input/?i=0.5(1+tanh+(x))

 

[Ted123]

You have the correct shape, however you may define a value at x= 0, 1, 2 for the heavside step function. Personally I think 1/2 is the most sensible, because it means that the function is odd and the approximation H(x)={1+tanh(kx)}/2 holds exactly in the limit k->infinity.

0, 1 and 0.5 are all valid choices to use, and will often depend on what exactly you are using H for.

http://www.wolframalpha.com/input/?i=0.5(1+tanh+(x))

In my definition of H, H(0) is undefined.