# Unit impulse vs Kronecker delta vs Dirac delta?

Apparently these 3 terms are similar, but what is exactly difference between them and what is exact value of each of them at time zero,1 or infinity?

UPDATE: After seeing MattL's comment below, I am shortening the answer and tried to provide a more accurate description of Unit Impulse and Dirac Delta function.

Kronecker Delta is defined for the discrete time domain. It is defined as a function of 2 indices. $$\delta_{ij} = 0 \text{ if }i \ne j\\ = 1 \text{ if } i = j$$.

Unit Impulse function is defined for both discrete and continuous time values as independent variable. For discrete time, it is defined as $$\delta[n]$$, $$\delta[n] = 1 \text{ for } n = 0\\ = 0 \text{ for } n \ne 0.$$

Now, Unit impulse and Dirac Delta are defined for continuous time as independent variable and are used interchangeably sometimes. They are not technically a function but are defined as limiting values of function as the width around $$t=0$$ reduces.

Area under integral under $$-\infty \lt t \lt +\infty$$ is 1. ie. $$\int_{-\infty}^{+\infty}\delta(t) = 1$$

As an example, consider a rectangle function from $$-T/2$$ to $$+T/2$$ with a height of $$1/T$$ and $$0$$ everywhere else. The area of this rectangle $$\int_{-\infty}^{+\infty}x_T(t) = 1$$. As the $$lim_{T \rightarrow 0}$$ height goes to $$\infty$$ as $$T \rightarrow 0$$. $$lim_{T\rightarrow 0}\int_{-\infty}^{+\infty}x_T(t) = 1$$

Similarly, we can define Dirac Delta as limiting case of Gaussian, Sinc, Exponential functions etc. where the limit of the parameter which defines the width is taken such that the width goes to 0.

• What do you mean by unit impulse is defined for continuous time? It is very well defined for discrete time – Dsp guy sam Apr 12 at 14:44
• Ok, what I meant is unit impulse and dirac delta are used interchangeably in continuous time. I missed to add that unit impulse is defined for discrete time also. That is $\delta[n-k]$. I was only thinking about continuous time signals when I answered this question. – jithin Apr 12 at 14:47
• The sentence "For unit impulse and dirac delta, at $t=0$ [...]" is inaccurate. Why is $x(t)=0$ (just) for $t=1$ or $t=\pm\infty$? What about all other values of $t$? Also, it is the limit of infinitely many functions, not only the Gaussian that you mention. Note that the Dirac delta impulse can also be written as the limit of the sinc function. I think the answer would be better if it were reduced in length, and if you remove all inaccuracies. – Matt L. Apr 12 at 15:52
• @MattL.that was because OP asked "what is exact value of each of them at time zero,1 or infinity?" so Iwrote the values for those $t$ :-). I will correct the inaccuracies. – jithin Apr 12 at 16:11
• OK, but the point is - and that's where the OP's understanding is wrong - the Dirac delta impulse doesn't have any value anywhere, because it's no conventional function. It only makes sense under and integral. – Matt L. Apr 12 at 16:14

Kronecker Delta used mainly in context of discrete time signals is defined as $$\delta_{ij}$$ is 1 only when $$i=j$$ or $$i-j=0$$, $$0$$ otherwise. So alternatively it is also written as $$\delta[i-j]$$. This value is exactly 1 and finite (refer below for the distinction with Dirac Delta)

The unit impulse is just the kronecker Delta with $$j=0$$ hence we only refer to unit impulse with one parameter $$\delta_i$$. Since $$j=0$$, this is alternatively written as $$\delta[i ]$$.Hence is 1 at $$i=0$$, $$0$$ otherwise.

The continuous time unit impulse can be thought of derivative of continuous time unit step (modified to have a slightly less steep than the sudden step at $$t=0$$). Refer to Oppenheim's book on signals and systems Chapter 1.

The Dirac Delta is used in context of continuous time signals to define impulses. It is defined as having an infinitely small width and hence a large magnitude, however the area under the curve integrates to 1. Refer to Oppenheim's book on signal and systems for various shapes that a Dirac Delta function could take, in all cases the area being 1.

• unit impulse is also written like $\delta[n-k]$ i.e., with two parameters... – Fat32 Apr 12 at 14:36
• Ofcousre if one wants to "shift" the impulse then it would be written in that form. That's basic signal shift operation on signals. – Dsp guy sam Apr 12 at 14:37
• In your last para"hence large magnitude" ? Means infinity? – Man Apr 12 at 15:50
• It would depend on the value of the width of the Dirac Delta, in limit of the with approaching to zero (as is the definition of Dirac Delta), yes, the magnitide is approaching infinity – Dsp guy sam Apr 12 at 15:58
• It would depend on the value of the width of the Dirac Delta, in limit of the width approaching to zero (as is the definition of Dirac Delta), yes, the magnitide is approaching infinity – Dsp guy sam Apr 12 at 15:58