2
$\begingroup$

The definition of Dirac delta function states that it gives a value of $\infty$ at t=0 and 0 elsewhere.

But, the definition of unit impulse function states that it gives a value of 1 at t=0 and 0 elsewhere.

Many textbooks state that Dirac delta and unit impulse are the same function.

So What is the difference between Dirac delta and unit impulse function?

$\endgroup$
  • 3
    $\begingroup$ The “definition” of a Dirac delta says nothing of the kind of things you say it does. If you find a textbook or paper that gives what you state above as the definition of a Dirac delta, throw the book away. If your professor wrote it on the blackboard, find a different professor or course. $\endgroup$ – Dilip Sarwate Oct 1 '18 at 10:27
2
$\begingroup$

The Dirac delta $\delta(x)$ is a continuous argument (generalized) function where $x$ is a continuous variable and can be time, space, frequency etc. Unfortunately, a simple definition like

$$\delta(x) = \begin{cases} \infty ~~~&,~~~ x=0 \\ 0 ~~~&,~~~ x \neq 0 \end{cases} $$

does or can not properly describe its behaviour. You need to define its properties under the integral sign, and that requires advanced mathematical stuff for a formal verification. But a simplified Riemann definition of $\delta(x)$ is given as:

$$\int_{-\infty}^{\infty}f(x)\delta(x-a) dx = f(a)$$ for a sufficiently smooth test function $f(x)$. All properties of $\delta(x)$ follows from this Riemann (seeming) integral operation.

On the other hand, the discrete counterpart, the unit-impulse or the Kronecker delta is a sequence of integer index $m$ with very well defined behaviour as:

$$\delta[m] = \begin{cases} 1 ~~~&,~~~ m=0 \\ 0 ~~~&,~~~ m \neq 0 \end{cases} $$ These two functions serve the same purpose for discrete and continuous systems, hence share the same name impulse. Yet they are two different, distinct entities; the Dirac impulse is a very problematic concept, whereas the discrete impulse is a mere simplicity.

$\endgroup$
  • $\begingroup$ excuse me, what does "does not properly describe it's behaviour" mean? $\endgroup$ – Anonymous Oct 1 '18 at 10:08
  • 2
    $\begingroup$ Too see why that definition cannot describe Dirac impulse fully, you should better read a chapter of a signal processing book. In summary, Dirac impulse is defined with respect to its effects under integral sign: for a smooth (continuous) test function (at t=0) of $\phi(t)$ the integral $$\int_{-\infty}^{\infty} \phi(t) \delta(t) dt = \phi(0)$$, and properties of $\delta(x)$ follow from this defintion. For example $x(t)\delta(t-a) = x(a)\delta(t-a)$ or $x(t) \star \delta(t-d) = x(t-d)$ or $\delta(at) = \frac{1}{|a|} \delta(t)$... $\endgroup$ – Fat32 Oct 1 '18 at 10:34
  • 1
    $\begingroup$ Do not worry, I upvoted your answer. A separate question and thread is probably wiser. I suppose Lebesque is too much, looking for something between him and Riemann $\endgroup$ – Laurent Duval Oct 1 '18 at 21:02
  • 1
    $\begingroup$ @LaurentDuval There is a book from Lighthill which is about ...distributions...; (introduction to Fourier Analysis and Generalised Functions - 1959) afaik it's one of the most referenced books while defering the formal discussion of $\delta(x)$, so may be you can have a look at it for an easy introduction to formal treatment of distributions... $\endgroup$ – Fat32 Oct 1 '18 at 21:08
  • 1
    $\begingroup$ @LaurentDuval Very nice effort. A good read for spare time enthusiasm. Ineed, imho mathematical topology, set theory and measure theory are the fundamental obstackles before distributions... $\endgroup$ – Fat32 Oct 13 '18 at 21:29
2
$\begingroup$

"Unit impulse" is a generic term that, depending on context, refers to one of the following distributions:

They are both referred to as "unit impulse", because they both represent the response of a system to a short "impulsive" input, and have similar properties that translate between continuous and discrete domains, such as the sifting property: $$ f(t_0) = \int_{-\infty}^\infty \delta(t - t_0) f(t) dt $$ for Dirac delta and $$ f[k_0] = \sum_{k = -\infty}^\infty \delta[k - k_0] f_k $$ for the Kronecker delta.

$\endgroup$
1
$\begingroup$

In

https://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

gives as a loose definitiniton

$$ \delta(x)=\begin{cases} +\infty & x=0 \\ 0 & x\ne0 \end{cases} $$ so perhaps biting your head off is a bit severe. I'm not about to throw Wikipedia away.

The rest of the article along with https://en.wikipedia.org/wiki/Kronecker_delta answers most of your question

$\endgroup$
  • 1
    $\begingroup$ that loose definitiniton is crappy. someone should fix that in Wikipedia. $\endgroup$ – robert bristow-johnson Oct 1 '18 at 18:05
  • 1
    $\begingroup$ just calling it the way i see it $\endgroup$ – Stanley Pawlukiewicz Oct 1 '18 at 18:17
  • $\begingroup$ yer fine, Stan. i also checked the Wikipedia page and it says that. but, while i am not the Dirac distribution Nazi (i am generally fine treating it as a function that is a limit of functions all having an area of 1) that definition from Wikipedia is crappy and should be fixed. e.g. suppose we define it like this: $$ f_a(x) \triangleq \begin{cases} \frac{1}{a^2} \qquad & \text{for } x=0 \\ 0 &\text{for } x\ne 0 \\ \end{cases}$$ then define $$ \delta(x) \triangleq \lim_{a\to 0} f_a(x) $$ what's the integral of $\delta(x)$ in the limit? $\endgroup$ – robert bristow-johnson Oct 1 '18 at 18:25
  • $\begingroup$ The loose definition is confusing,how can we replace infinity(at x=0) with the integral of the function I.e 1 (at x=0) ? $\endgroup$ – Anonymous Oct 2 '18 at 15:08
  • 1
    $\begingroup$ look up the radius of an electron on the web. The sites I looked at say $<10^{-18}$ m and that’s an upper bound. Lower bounds seem hard to find. I think the loose definition appeals to ones intuition rather well but I find myself advocating here for something I loosely hold. If you think the loose definition stinks from a mathematical perspective, that is fine. Physics and Math intersect but don’t always harmonize $\endgroup$ – Stanley Pawlukiewicz Oct 2 '18 at 22:08
0
$\begingroup$

Dirac Delta function and unit impulse function are the same. But many texts define that $\delta(t) = \left \lbrace \begin{array}{lr} \infty, \hspace{0.5cm} t=0 \\ 0, \hspace{0.5cm} else \ \end{array} \right . $ But I am of the opinion that this definition is not perfect. Because $\infty$ is something that is not defined. How can one insert something that is not defined as a quintessential part of a definition? So I think what Simon Haykins defined it, is perfect:

$\delta(t) = 0 \hspace{0.5cm} for \hspace{0.5cm} t \neq 0 $ and $\int_ \limits{-\infty}^{+\infty}\delta (t)dt = 1$

$\endgroup$
-1
$\begingroup$

Dirac Delta function(better distribution) and unit impulse are exactly same but we interpret the definition in a different way.

When we refer to this function as Dirac Delta we consider the magnitude of the distribution at every point. Hence this gives value infinite at t=0 and zero other wise.

When we refer to signal as unit impulse the values we refer are the areas of integration around that particular point with limits t- to t+. This gives value 1 to the unit impulse representation of the function.

To clarify further they are in CT domain ie continues time domain but there DT version is kroneker delta, this is defined on only integral values of n, whereas you see the unit impulse or Dirac Delta function as you may know is defined for all real values of t.

Source:Signals and Systems by Oppenheim (Professor at MIT)

$\endgroup$

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.