Reading the signal literature I often come across the expression $k \delta (t)$ where $k$ is a constant. I presume this is a notation to suggest we are referring to the area or strength of the Dirac function since the multiplication of a constant, $k$, by $\delta (t)$ seems to make no sense.

My question is what does $k \delta(t)$ mean? Am I correct in assuming that on its own it doesn't represent $k$ multiplied by $\delta(t)$?


1 Answer 1


You are correct. In your example, $k$ would refer to the "area underneath" the impulse. Mathematically speaking, the Dirac delta isn't a function in the typical sense of the term. Instead, it is more of a distribution, characterized by the fact that when integrated across any interval that contains $t=0$, the result is unity. That is:

$$ \int_{-\epsilon}^{\epsilon} \delta(t) dt = 1 \ \forall\ \epsilon > 0 $$

The distribution is typically defined as follows:

$$ \delta(t) = \begin{cases} \infty,\ t = 0 \\ 0, \text{ otherwise}\end{cases} $$

Multiplication by a constant $k$ obviously wouldn't change this definition at all. So instead, the scaling merely changes the area underneath the distribution:

$$ \int_{-\epsilon}^{\epsilon} k\delta(t) dt = k \ \forall\ \epsilon > 0 $$

Extending this concept of multiplication by a constant $k$ to multiplication by a function $f(t)$ yields the following:

$$ \int_{-\epsilon}^{\epsilon} f(t) \delta(t) dt = f(0) \ \forall\ \epsilon > 0 $$

or more generally:

$$ \int_{T-\epsilon}^{T+\epsilon} f(t) \delta(t-T) dt = f(T) \ \forall\ \epsilon > 0 $$

which is known as the sifting property of the Dirac delta and is used extensively in linear systems theory.

  • $\begingroup$ Remark that the Dirac delta is typically defined as the limit of a sequence of functions with unit area and support that approaches to 0. $\endgroup$
    – thang
    Commented Jan 22, 2013 at 19:42

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