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} <br /> 0, &amp; \;x \leq 0\\ <br /> -2, &amp; \;0 &lt; x \leq 1\\<br /> -1, &amp; \;1&lt;x\leq 2\\<br /> 0, &amp; \;x &gt; 2<br /> \end{array} <br /> \right.

and

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

and

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