Derivative of t per unit step - meaning of $t\delta (t)$

Question

I derived the function ($u(t)$ is the unit step):

$$x(t)=t \space u(t)$$

$$dx/dt=1 \space u(t)+t \space \delta(t)$$

What is the graphical meaning of the second term?

Thank you very much.

Answer 1

The function $x(t)$ is a ramp and is not differentiable at $t=0$ (the left and right derivatives differ). The term $t\,\delta(t)$ accounts for that non-differentiability (the Heaviside step takes a finite value at $t=0$).

Graphically speaking, you can picture it as two infinitely close Dirac deltas with opposite signs (which implements a differentiation operator by convolution).

Answer 2

$\delta (t)$ is the Dirac delta function (well... distribution). This "function" assumes the value $\infty$ at zero and is zero everywhere else. The funny thing is

$$\int_{\mathbb{R}} \delta (t) \ dt = 1 $$

I suggest consulting the Wiki page. The Dirac delta is like a point pulse.

Answer 3

If you want some graphical meaning here, you should compare $\frac{dx}{dt}$ as a function that returns the slope of the graph of $x(t)$ with $u(t)$. The difference between these two graphs is exactly $t \delta(t)$. Try arguing graphically that $t \delta(t) = 0$, then convince yourself of this using the definition of $\delta(t)$ as a limit of proper functions.