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The impulse signal

The impulse function, also known as Dirac delta function or Dirac impulse, is defined by

\begin{displaymath}
\delta(\mathrm{x})= \left\{
\begin{array}{ll}
\infty & \textrm{if $x=0$}\\
0 & \textrm{else}\\
\end{array} \right.
\end{displaymath} (3.3)

Technically, this is not even a function at all. However, it has some interesting and important properties. It is the differentiation of the step function, or, in other words, the integral of the impulse signal is the step signal, and its value (i.e., the area under the function) is one. So the impulse signal is a function which is zero everywhere except at zero, where it has an infinitely narrow spike with infinite height but it integrates to a value of one. The latter can quite easily be seen by (taking as granted that its integral is a step)


\begin{displaymath}
\int_{-\infty}^{\infty}\delta(x) \mathrm{dx} =
\mathrm{step}...
...}(c_1) - \lim_{c_2 \to -\infty} \mathrm{step}(c_2) =
1 - 0 = 1
\end{displaymath} (3.4)

An intuitive way of looking at this is to let w in the definition of the ramp go to zero and look at the differentiation, the box function (Fig. 3.1). The smaller w gets, the higher gets the box function but the area under the box is always one. In the limit the box has infinite height and infinite small width, but the area is still one. So we can find the derivative of the step signal without getting into formal troubles, it is the impulse signal $\delta(x)$.

This means also, that when we shift the impulse to some value t, multiply it with some function f and integrate the result, we get the value of the function at t, mathematically

\begin{displaymath}
\int_{-\infty}^{\infty}\mathrm{f}(x)\delta(x-t) \mathrm{dx} = f(t)
\end{displaymath} (3.5)


next up previous contents
Next: The impulse train Up: Important functions in signal Previous: The ramp and step   Contents

1999-12-29