2

We have the result $(x * \delta)(t) = x(t)$, where $*$ denotes convolution. Here let $x(t)$ is a real valued function with respect to time and $\delta(t)$ is the unit impulse function.

$$ (x * \delta) (t) = \int_{-\infty}^\infty x(\tau) \delta(t-\tau)\ \mathrm d\tau \tag1 $$

This equation means that we multiply the signal $x(\tau)$ with shifted versions of delta function by sweeping it right from $-\infty$ to $\infty$, right?

I have a confusion here. Actually unit impulse signal is a signal with area unity, right? And suppose $x(t)$ is rectangular pulse with unit area like shown:

And $x(t)$ is comprised of infinite impulses from $t=0$ to $t=1$. From equation $(1)$ it means there are infinite impulses in the single rectangular pulse. Which means the area of single rectangle pulse is infinite because of infinite impulses in that region and surely it is a contradiction. Where has my understanding gone wrong?

dexterdev
  • 782

1 Answers1

1

Wikipedia: "The Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite, $$ \delta(t)=\begin{cases} \infty,\ &t=0\\ 0,\ &t\neq0 \end{cases} $$ and which is also constrained to satisfy the identity $$ \int_\mathbb{R}\delta(t)dt = 1. $$

"From a purely mathematical viewpoint, the Dirac delta is not strictly a function, because any extended-real function that is equal to zero everywhere but a single point must have total integral zero. The delta function only makes sense as a mathematical object when it appears inside an integral. While from this perspective the Dirac delta can usually be manipulated as though it were a function, formally it must be defined as a distribution that is also a measure. In many applications, the Dirac delta is regarded as a kind of limit (a weak limit) of a sequence of functions having a tall spike at the origin. The approximating functions of the sequence are thus "approximate" or "nascent" delta functions."

So, according to the above, your definition ("unit impulse signal is a signal with area unity") to this "function" is wrong. Accordingly, the claim that a rectangular function can be functionally decomposed into an infinite sequence of such delta "functions", is also wrong.

user91011
  • 1,113