# Confusion on unit impulse function $\delta(t)$

$\delta(t)$ is a singular function and when I'm learning Signals and Systems I learned that $\delta(t)$ is an even function, and all of its odd order derivatives are odd function. Then we have $\delta^{'}(t)=-\delta^{'}(-t)$, let t=0 then $\delta^{'}(0)=0$. This is obviously wrong! But Where?

And if it's wrong,then $\delta^{'}(0)$=?

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I'm not sure if the derivative is well defined at $0$. – F M Jun 4 '11 at 2:19
It's not really a function either. – Unreasonable Sin Jun 4 '11 at 2:35
FYI "special funtions" (which this was tagged with before) refers to the study of particular important named functions, e.g. Riemann zeta, the Bessel functions, the gamma and digamma functions etc. Understanding distributions or "generalised functions" such as the delta function (an unfortunate name, I know) belongs to the subject of functional analysis. – kahen Jun 4 '11 at 8:32

Short answer: $\delta$ isn't a function, it is a "generalized function" supported at $0$, which can be defined as a "weak derivative" of the step function. Because it isn't a function, you can't evaluate it at points, and you can't evaluate the derivatives at points, so it doesn't really make sense to say they are "even" or "odd" functions.

Long answer: When first introduced, or when seen in the context of an engineering class, people often gloss over the fact that $\delta(t)$ is not quite a function. It is the "derivative" of the step function, or it is the "limit" of a bunch of rectangular looking functions that get taller and thinner. And for what you tend to want to do with it, these work as definition and intuition. But they are not quite correct, and so some things (like what you tried to do) don't quite work.

So what is $\delta(t)$? The best way to define it is as a "generalized function" or "distribution," namely a linear functional which takes in a function and spits out a value. In this case, $\delta(f)=f(0)$.

But if the delta function isn't really a function, what does it mean to take its derivative? First, we need to look at a wide class of functionals, which explains why they are called generalized functions.

Let $f$ smooth and bounded function. Define an operator $\Lambda_f$ which takes a smooth compactly supported function $g$ and returns $$\displaystyle \Lambda_f(g)=\int_{\mathbb R} fg dx.$$

We want to define the derivative of a functional such that $D\Lambda_f = \Lambda_{Df}$. We would therefore like to find a way to define $\Lambda_{Df}$ without using $Df$. (Note, by $Df$ I just mean $f'$, but the prime is hard to read in the subscript) We can do this using integration by parts. Because $g$ is compactly supported, we won't get any boundary terms, and so we get $$\int_{\mathbb R} f'g dx = \int_{\mathbb R} f(-g') dx$$

Thus, given an arbitrary functional $\Lambda$, we can define $D\Lambda$ by the formula $D\Lambda(g)=-\Lambda(g')$. This "derivative" is called the weak derivative. If $f$ is differentiable, the weak derivative yields the derivative, but for functions that aren't differentiable (like the step function), this gives a new notion of derivative. In this case, we get "generalized functions" which are NOT of the form $\Lambda_f$ for some function $f$.

A quick word of warning: we equip some sort of norm on our space of functions, and when we look at functionals, we only want to think about the ones which are bounded operators. Thus, while we can define the weak derivative of any functional, the weak derivative of a bounded functional might no longer be bounded, and hence the weak derivative doesn't always "exist" where we want it to.

In any case, $\delta$ is the weak derivative of the step function, and it satisfies $\delta(f)=f(0)$. We can take the derivative to get $D\delta(g)=-\delta(g')=-g'(0)$, and similarly $D^n\delta(g)=(-1)^ng^{(n)}(0)$. There isn't really a good sense in which you could call these odd or even "functions", as you don't really evaluate them at points, but rather at functions, although the even derivatives do vanish when you plug in an odd function and vice versa.

Generalized functions and their weak derivatives enjoy a lot of nice properties (all the nice properties of derivatives, and more!), and are quite useful for expressing ideas that are intuitive but hard to make rigorous otherwise. How much you really want to worry about the details depends almost entirely on whether you plan on being an analyst.

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You could think of $\delta'$ being odd as being induced by the Heaviside function, which is odd. – user12014 Jun 8 '12 at 6:06

This is obviously wrong! But Where?

The derivative of $\delta(t)$ is not defined at $t=0$ (think of $f(x)=|x|$ as another example).

But, wait, there's more: actually even the function $\delta(t)$ itself is not defined at $t=0$.

But, wait, there's more: actually $\delta(t)$ is not a function.

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actually impulse signal is not an ordinary function because any extended-real function that is equal to zero everywhere except at a single point must have total integral zero.it is called as generalized functions or distribution because it can be represented by many signals. so when it is not an ordinary function we can not talk about even and odd function.this signal is said to be even distribution (not an ordinary function)because del(t)=del(-t)

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