0
$\begingroup$

The unit impulse function, in a few textbooks that I have referred, has a value of 0 at t≠0 , and an area of unity (1). The height of the impulse function also tends to infinity at t=0. But since it's height cannot be it's magnitude (as we generally use it's unit area in calculations), what would be the value/amplitude/magnitude of the unit impulse function at t=0?

Before you say that my question has answered itself, and that it's value is one(1) at t=0, note that this is not defined as such in any standard textbooks. This question has been bothering me for quite a while now, so any attempt at elaborating a solution will be highly appreciated.

$\endgroup$
4
$\begingroup$

The "impulse function" is not actually a function but a distribution, and its value at $t=0$ is not defined.

The impulse only really makes sense when multiplied by a signal inside an integral, where it is used to "select" the value of the signal at specific time instants.

Personally, I try to avoid impulses whenever possible. Look for Lapidoth's book on digital communications (available for free on his website) for a complete treatment of Fourier analysis (as relevant for communications) without a single delta. Not many people share my squimishness with deltas, though, and if you're taking a signals course they are pretty much unavoidable.

Note also that, once you go to the discrete time domain, everything becomes nicer: instead of the Dirac delta, you use the Kronecker delta, which is a regular function and has a value of 1 at $n=0$.

$\endgroup$
1
$\begingroup$

everything MBaz says is true, but we engineers like to treat $\delta(t)$ as a function with the properties of

$$ \delta(t) = 0 \qquad \forall t \ne 0 $$

and

$$ \int\limits_{-a}^{+a} \delta(t) \, \mathrm{d}t = 1 \qquad \text{for any } a>0 $$

but the math guys tell us we can't do that. they say that if

$$ f(x) = g(x) $$

almost everywhere in $E$, then

$$ \int_E f(x) \, dx \ = \ \int_E g(x) \, dx $$

i asked these math guys about it here.

like concepts of infinity, the dirac delta function is sorta a female canine.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ rhymes with "rich". $\endgroup$ – robert bristow-johnson Sep 27 '18 at 20:34
  • 1
    $\begingroup$ I never can decide... $\endgroup$ – Fat32 Sep 27 '18 at 20:49
  • 1
    $\begingroup$ @Fat32, i added the definitive cover image. $\endgroup$ – robert bristow-johnson Sep 27 '18 at 21:56
  • 1
    $\begingroup$ I asked Riemann and he refused (to integrate that $\delta(x)$) :-) $\endgroup$ – Fat32 Sep 27 '18 at 23:37
  • 1
    $\begingroup$ (Riemann told me a different story) and it turns out that the integral of that $\delta(x)$ is zero. $\endgroup$ – robert bristow-johnson Sep 28 '18 at 19:04
0
$\begingroup$

In addition to the excellent answers others have given, it should be noted that, at least intuitively, the impulse can be thought of as the limiting case of an infinitely narrow pulse with unit area, i.e. $$ \delta(t) = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \operatorname{rect} \left( \frac{t}{\epsilon} \right). $$ (And yes, I'm aware a kitten has died somewhere because of this mathematical sin.)

You can substitute just about any zero-mean probability density function for the above relation and take the limit as the variance goes to zero.

This relation shows that the amplitude approaches infinity as the width approaches zero. For this reason it isn't a "true" function, but is, like others have said, a "distribution" or a "generalized function". Generally, the Dirac delta function only has meaning inside an integral, such as a Fourier integral or a probability integral.

$\endgroup$
  • 2
    $\begingroup$ yeah, but us indecent, kitten-killing EEs often have naked Dirac deltas hanging out. like with the Dirac comb. $\endgroup$ – robert bristow-johnson Sep 27 '18 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.