For my homework I was given the signal by the illustration:

I was asked to draw the illustration for $$x_1(t) = 3x(-1+0.5t) - \delta(-t+3)$$

I have figured out how to draw the $3x(-1+0.5t)$ part, but I'm not sure how to deal with the $\delta(-t+3)$ part. What does it mean to subtract a signal by a unit impulse?

  • $\begingroup$ In this case the answer is simple, but the general case is indeed interesting... $\endgroup$ May 16, 2023 at 13:25

1 Answer 1


I have figured out how to draw the $3x(-1+0.5t)$ part

Then you have figured out

which, purely from transformations point of view, is a horizontal shift right by 1, horizontal stretch by a factor of 2, and vertical stretch by a factor of 3. And you must know that the

$-\delta(-t + 3)$ part

except as $\delta(t)$, is

so $3x(-1 + 0.5t) - \delta(t)$ is

which should let you figure out $3x(-1 + 0.5t) - \delta(-t + 3)$.

For scalings other than unity, e.g. $2\delta(t)$, one can either draw it with height of 2, or height of 1 and the value labeled next to it. The sign should be preserved in the plot though, with $-\delta(t)$ pointing down.

Note that $\delta(t)$ isn't a usual function, but a distribution, and it's not quite right to interpret it as having a value "at" anywhere (it instead must be enclosed with an integral). Graphically there's no issue in that it'll correspond to the intended symbolic expression.

General case

Let $x_2(t) = 3x(-1 + 0.5t) + \delta(t + 1)$. Then, the delta overlaps a non-zero value. I'm unsure whether there's a convention here; there's two proposals, left being mine and right @MattL's:

There's arguments both ways, that I'd summarize as, left behaving more function-like, while right emphasizing the non-functional nature of $\delta(t)$, and each being more visually convenient in different scenarios. Discussed here.

  • $\begingroup$ It would probably be instructive to explain what the little circle at $-1$ actually means. For a student it might appear to be the "height" or value of the impulse. $\endgroup$
    – Matt L.
    May 16, 2023 at 15:32
  • $\begingroup$ @MattL. I actually forgot that it should be an arrow. I do recall (not too clearly) from college days that the height does denote value, though that doesn't scale well - what would be the alternative? Label explicitly? $\endgroup$ May 16, 2023 at 15:45
  • $\begingroup$ Well, whatever makes it clear that it's not a value neither a height, but it's the integral/"area" of the Dirac impulse. There's no real way to sketch it properly other than by some symbol the meaning of which must be made clear. Remember that a Dirac impulse doesn't have a value at zero argument, it's a distribution (generalized function), which only makes sense under an integral. $\endgroup$
    – Matt L.
    May 16, 2023 at 15:55
  • $\begingroup$ @MattL. Is there a convention for overlaps, i.e. if last plot instead has $\delta(t + 1)$? If yes and it's to place the tail at the value of the function, I'll edit it in - if yes and it's not that, I can open a question - if no, guess I'll propose this convention now. And sure, I'll make a note on interpreting it. $\endgroup$ May 16, 2023 at 15:59
  • $\begingroup$ Usually a Dirac impulse is represented by an arrow of some arbitrary length with its "area" noted next to it. It doesn't interfere at all with any ordinary function. So for the given example, if you had $\delta(t+1)$ you'd just add an arrow at $t=-1$ pointing up or down, depending on the sign of the Dirac impulse. $\endgroup$
    – Matt L.
    May 16, 2023 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.