What is the value of a delta function?

Question

So my understanding is that a delta function $\delta(t)$ has the value of $\infty$ at $t = 0$ and $\delta(t) = 0$ everywhere else. Essentially, it's just one big spike on a graph right?

In my signal processing lecture, I came across the following:

Here, we arrive at $X_1[k] = \frac{1}{2}\delta [k - 1] + \frac{1}{2}\delta [k + 1]$, which means $X_1[k]$ only has non-zero value at $k - 1$ and $k + 1$. However, using my definition of a delta function, shouldn't $X_1[k = \pm 1] = \infty$?

This isn't the case when you look at the graph. It shows $X_1[k = \pm 1] = \frac{1}{2}$, so what's wrong? Is there some other delta function I just don't know?

Answer 1

Two answers here.

1. First, the answer from the classical point of view.

Delta function is defined as something which has the following property:

$$\int_{-\infty}^\infty \delta(x) f(x)dx=f(0)$$

It is not a function but some special symbol.

Now, $a \delta(x)$ is defined as heaving the following property:

$$\int_{-\infty}^\infty a\delta(x) f(x)dx=a f(0)$$

That's it.

2. Actually one can consider delta function as a function that takes values from the set of divergent integrals rather than real numbers.

By definition via Fourier transform,

$$\delta (x-\alpha )={\frac {1}{2\pi }}\int _{-\infty }^{\infty } \cos(px-p\alpha ) dp$$

This gives formally $\delta(0)=\frac1\pi \int_0^\infty dx$ You can consider this divergent integral the value of Dirac Delta at zero. It is not a real number though.

This way, $a\delta(0)$ would be $\frac1\pi \int_0^\infty a dx$, also a divergent integral.

Answer 2

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Building on what others have already said, the two functions are just that—two distinct functions. One is the real-valued Kronecker Delta function $\kdelta:\Z\to\R$ defined as $$\kdelta(n)\eqd\brbl{\begin{array}{cl} 1 & \text{if $n=0$}\\ 0 & \text{otherwise} \end{array}}$$ and the other is the Dirac Delta $\delta:\R\to\Rx$, which obviously is not a real-valued function at all. This is not so much a problem in and of itself (functions can be mappings to real numbers, complex numbers, colors, faces of a die, ...) but in the case of $\delta$, it maps not into a well behaved set like the real numbers $\R$, but rather into the set of extended real numbers $\Rx\eqd\R\setu\setn{-\infty,+\infty}$. The set $\Rx$ is an Alice-in-Wonderland type world and when working therein requires great care and proper safety equipment. In $\Rx$ crazy things can happen like $\infty+\infty=\infty$.

Because $\delta(x)$ maps to $\Rx$, many would not call it a function at all, but rather an extended function, or a distribution (see for example Mallat Appendix A.7). A distribution may be something of a loose cannon when left to its own devices, but when constrained to the inside of an integral (or more generally inside an inner product) it becomes quite well-behaved, predictable, and useful.

Building on what @Anixx and @Thomas-Andrews have already pointed out, the Dirac Delta distribution can be defined, together with $\int\dx$, as an operator (a mapping from one linear space to another/same linear space) with the following definition: $$\inprod{\ff(x)}{\delta(x)} \eqd \int_{\R}\ff(x)\delta(x)\dx \eqd \ff(0)$$

So the Dirac Delta distribution $\delta(x)$ and the Kronecker delta function are distinctly different. But that being said (that they are very different in form), they are very similar in function. In particular, note the following:

(1) Both "sum" to 1: \begin{align*} \boxed{\int_{\R} \delta(x)\dx} &= \int_{\R} 1\cdot\delta(x)\dx \\&\eqd \boxed{1} && \text{by definition of $\delta$} \\ \\ \boxed{\sum_{n\in\Z}\kdelta(n)} &= \kdelta(0) && \text{by definition of $\kdelta$} \\&= \boxed{1} \end{align*}

(2) Both have a similar time-shift property: \begin{align*} \boxed{\int_{\R} \delta\brp{x-a}\ff(x)\dx} &= \int_\R \delta\brp{u}\ff(u+a)\du && \text{where $u=x-a$ $\implies$ $\dx=\du$} \\&= \boxed{\ff(a)} \\ \\ \boxed{\sum_{n\in\Z}\kdelta(n-k)y(n)} &= \sum_{n\in\Z}\kdelta(m)y(m+k) && \text{where $m=n-k$ $\implies$ $n=m+k$} \\&= \kdelta(0)y(0+k) && \text{by definition of $\kdelta$} \\&= \boxed{y(k)} \end{align*}

(3) Both induce a projection operator. A projection operator $P$ is a linear operator such that $P^2=P$: \begin{align*} \boxed{P^2}\ff(x) &= PP\ff(x) \\&\eqd P\int_\R \delta(x)\ff(x) \dx && \text{where here $P$ is the Dirac Delta operator} \\&\eqd P\ff(0) && \text{by definition of $\delta$} \\&\eqd \int_\R \delta(x)\ff(0)\dx && \text{where here $P$ is the Dirac Delta operator} \\&= \ff(0) \int_\R \delta(x)\dx && \text{by linearity of the integral operator} \\&= \ff(0) && \text{because $\int_{\R} \delta(x)\dx=1$} \\&= \int_\R\delta(x)\ff(x)\dx && \text{by definition of $\delta$} \\&= \boxed{P}\ff(x) && \text{where here $P$ is the Dirac Delta operator} \\ \\ \boxed{P^2}\fy(n) &= PP\fy(n) \\&\eqd P\sum_{n\in\Z} \kdelta(n)\fy(n) && \text{where here $P$ is the Kronecker Delta operator} \\&\eqd P\fy(0) && \text{by definition of $\kdelta$} \\&\eqd \sum_{n\in\Z} \kdelta(n)\fy(0) && \text{where here $P$ is the Kronecker Delta operator} \\&= \fy(0) \sum_{n\in\Z} \kdelta(n) && \text{by linearity of the summation operator} \\&= \fy(0) && \text{because $\sum_{n\in\Z} \kdelta(n)=1$} \\&= \sum_{n\in\Z}\kdelta(n)\fy(n) && \text{by definition of $\kdelta$} \\&= \boxed{P}\fy(n) && \text{where here $P$ is the Kronecker Delta operator} \end{align*}

(4) Both can be used for sampling. The Dirac can be used—inside an integral—as a projection operator to map a function $\ff(t)$ to a single point $\ff(a)$ for some value $t=a$. That is, it can be used to sample $\ff(t)$ at a given time $t=a$. In the field of Digital Signal Processing (DSP) a continuous time function $\ff(t)$ can be transformed (mapped) to a
sequence (a function with domain $\Z$) $\seqn{\ldots, x_{n-1}, x_n, x_{n+1}, \ldots}$, where each element $x_k$ of this sequence is $$x_k \eqd \int_\R \ff(t)\delta(t-kT)\dt = \ff(kT)$$ Likewise, the Kronecker delta can be used to sample (OK, maybe a bit of a stretch here?) a sequence $\seqn{\ldots, x_{n-1}, x_n, x_{n+1}, \ldots}$ in the sense $$y(n)= \cdots + x_{-2}\kdelta(n+2) + x_{-1}\kdelta(n+1) + x_0\kdelta(n) + x_1\kdelta(n-1) + x_2\kdelta(n-2) + \cdots$$