# How to sketch the following discrete time signal $\sum_{k=0}^{n} \delta[n-k]$?

I tried to solve it like in the picture but It feels so wrong though.

• What exactly is your question? Surely if you just want to sketch the signal, you just sketch $\delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3] + \ldots$ without invoking anything else?
– Peter K.
Feb 22 '18 at 16:07
• @PeterK. How can you be so sure that the upper sum limit will be a positive integer? It might be -4 for example. Feb 22 '18 at 18:03
• Good point! Are negatives allowed? Usually the summation goes up from the lower limit. My solution would work, but you’d have to take account of the direction of the summation (up for positive $n$ and down for negative).
– Peter K.
Feb 22 '18 at 18:09
• More thought required! I’ll try to give a full answer after lunch.
– Peter K.
Feb 22 '18 at 18:11
• The answer to this question depends entirely on the meaning associated with a sum in which the lower limit exceeds the upper limit. One standard interpretation is that the sum is empty (and has value $0$) if the lower limit is larger that the upper limit. With this convention, $\sum_{k=0}^n f[n]$ has value $0$ for any $f$ if $n < 0$ and so that sum is just $u[n]$. Another interpretation is analogous to what we use in integrals, $$\int_a^b = -\int_b^a$$ making the sum be $u[n]$ for $n > 0$ and $-u[-n]$ for $n<0$ with a nice ambiguity for the value at $n=0$. Feb 22 '18 at 21:49

Your signal is: $$x[n] = \sum_{k=0}^n \delta[n - k]$$ and you're saying you want to sketch it.

Based on your comment, I'm going to assume that $$x[-1] = \left . x[n] \right |_{n=-1} = \left . (\delta[n] + \delta[n+1]) \right |_{n=-1} = 0 + 1 = 1$$ and that $$x[+1] = \left . x[n] \right |_{n=+1} = \left . \left(\delta[n] + \delta[n-1]\right) \right |_{n=+1} = 0 + 1 = 1$$

I believe that means $$x[n] = 1, ~~ \forall n \in \mathbb{Z}$$ because $$x[n] = \cdots + \delta[n - (n-2)] + \delta[n - (n-1)] + \delta[n - n] + \cdots$$ if summing up or $$x[n] = \cdots + \delta[n - (n+2)] + \delta[n - (n+1)] + \delta[n - n] + \cdots$$ if summing down.

It's a little odd having the upper limit of the summation depend on the current time and, as I said in the comment, having the summation go down to the upper limit if the upper limit is below the lower one.

• So this means that the signal is equal to 1 and it would be sketched as u[n]+u[-n-1]. As for the sum depending on the n, i don’t think it’s unsual, we get it all the time, especially when applying convolution. Thank you for your answer. Feb 22 '18 at 19:35
• @user3140379 Yes, but I think $x[n] = 1$ is simpler than $x[n] = u[n] + u[-n-1]$. YMMV
– Peter K.
Feb 22 '18 at 20:46
• Yeah but the former helps more in terms of sketching :) Feb 22 '18 at 20:56
• @user3140379 Hence "YMMV" (Your Mileage May Vary). :-)
– Peter K.
Feb 22 '18 at 21:00
• I didn’t get that the first time :) Cheers! Feb 22 '18 at 21:10