# Adding Dirac Delta with Unit Step

What does the graph of the function $x(t)=-\delta(t)+u(t)$ look like? $$\delta(t)\ldots\text{ Dirac delta impulse}\\ u(t)\ldots\text{unit step function}$$

Will the impulse at the origin start from 0 downwards or from 1? I'm having trouble because the unit step is discontinuous at t=0, so I don't know how it will shift the impulse. I am talking about continuous time.

• Since the (positive) "impulse" is of "infinite" height, it does not matter in the least whether it "starts" at $0$ or at $1$ (or at $\frac 12$ for mealy-mouthed straddlers unable to reach a decision) on the vertical axis as it "zooms" downwards towards $-\infty$. Dec 29, 2014 at 20:24

Note that $\delta(t)$ is no function but a distribution. This means that it doesn't have function values but it is only defined by its integral. Consequently, the (generalized) function $x(t)=-\delta(t)+u(t)$ has no value at $t=0$ (the only useful value would be $-\infty$ but that's kind of hard to sketch). For sketching the function graph, you can define function values for $t\neq 0$ in the usual way. At $t=0$ you simply sketch a negative $\delta$-impulse, for $t<0$ the function is zero, and for $t>0$ you assign it the value $1$.
• this is a good example of how electrical engineers play fast and loose with $\delta(t)$. i just like to pretend it's a "regular" function that has width of 1 Planck time. Dec 30, 2014 at 21:08