How can a unit step function be differentiable??

Question

Recently, I am taking a Signal & System course at my college. In all of the signal & system textbooks I have read, we see that it is written " When we differentiate a Unit Step Function, we get an Impulse function. " But as far as I have read, a unit step function is a piece-wise linear function as well as it is a continuous function but it is non differentiable. My question is that how can a non differentiable function be differentiated to obtain the impulse function?? Where is the contradiction.....

I am confused...would be happy if anybody helps...

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Answer 1

The derivative of the unit step function (or Heaviside function) is the Dirac delta, which is a generalized function (or a distribution). This wikipedia page on the Dirac delta function is quite informative on the matter.

One way to define the Dirac delta function is as a measure $\delta$ on $\mathbb{R}$ defined by $$ \delta(A) = \begin{cases} 0 &: \text{ if } 0 \notin A \\ 1 &: \text{ if } 0 \in A \end{cases} $$ Then one can write down precisely what is meant by the expression $$ \int fd \delta = f(0) $$

Answer 2

In a "practical sense", it comes from $$ \Theta\left(x\right) = \int_{-\infty}^{x}\delta\left(x'\right)\,{\rm d}x' $$

It's is quite useful but you have to handle carefully as, for example, @Prahlad Vaidyanathan already pointed out.