I am trying to follow some notes on "Analytical Computation of the Convolution Sum (Graphical Method)" but am getting stuck on what I am doing, or if what I am doing is correct.

There are 5 given steps:

  1. Plot both the sequence $x[n]$ and $h[n]$ as functions of $k$ and flip one of them (either one) about $k = 0$ to obtain $x[-k]$ or $h[-k]$. Normally we chose the “simpler” signal to flip (often the one that begins at time $k = 0$ if possible).
  2. Label the index value in $x[-k]$ or $h[-k]$ that corresponds to $k = 0$ as the index $n$ to obtain $x[n-k]$ or $h[n-k]$. Write the equation for $x[n-k]$ or $h[n-k]$ by substituting $n = k$ into the original equation for $x[n]$ or $h[n]$.
  3. If $x[n-k]$ or $h[n-k]$ is a finite-length signal of length $L$, label the last non-zero value as the index $n-L+1$. Doing so will help us define the indices for the summation ranges in the convolution equation for each piece-wise case.
  4. We next identify all piece-wise cases involved in the convolution, i.e., non-overlapping and various over-lapping cases, that span the range of $n \in (-∞, ∞)$.
  5. For each identified case we evaluate $y[n]$ via the convolution sum. This involves first determining the indices of the convolution sum equation. We then evaluate the sum and simplify as much as possible.

I am given: $$x[n] = 2(u[n] - u[n-5])$$ $$h[n] = (n+1) (u[n] - u[n-10])$$

I decided to flip $x[n]$ and in doing so came up with:

Case 1: $n < 0$, no overlap $y[n] = 0$

Case 2: $0 \le n < 4$, partial overlap

Case 3: $4 \le n \le 9$, full overlap

Case 4: $9 < n \le 13$, partial overlap

Case 5: $13 < n$, no overlap $y[n] = 0$

I know that for discrete convolution, $$\sum_{k = -\infty}^{\infty} h[k] x[n-k]$$ but I am not sure how how to apply this to my cases. In addition I am not sure if my cases are even correct. I understand what the convolution needs to do graphically but how to generalize the summation for the cases is not clear to me.

  • $\begingroup$ Additional literature or examples for this topic would be appreciated if anyone has any too. $\endgroup$
    – MeljahU
    Feb 4 at 9:53

1 Answer 1


Your cases are correct.

Your two sequences should look like (Note: the graph below is slightly wrong and will be fixed shortly, see comments):

enter image description here

Here I've graphed the starting position, i.e. $n=0$.
Now for each $n$, shift the $x[n-k]$ sequence to the correct position, multiply the two sequences $h[k]$ and $x[n-k]$ pointwise and sum. That's convolution.

A few example: \begin{align} &\tt{n} &\quad \quad &\tt{y[n]}\\\\ &0 &\quad \quad &1\cdot 2 = 2\\ &1 &\quad \quad &1\cdot 2 + 2 \cdot 2 = 6\\ &\cdots\\ &7 &\quad \quad &4.2 + 5⋅2+6⋅2+7⋅2+8⋅2=60\\ \end{align}

  • $\begingroup$ Shouldn't $x[n-k]$ go to $n-4$, since at $u[0] = 1$? $\endgroup$
    – MeljahU
    Feb 4 at 21:40
  • $\begingroup$ Yep you’re right. I’ll edit! The picture edit will be done a little later. Good catch! The process is the same though, hopefully my answer helped in that regard $\endgroup$
    – Jdip
    Feb 4 at 23:09
  • $\begingroup$ I worked out that case 2 is $y[n] = \sum_{0}^{3} (k+1)(2) = (n+1)(n+2)$ and indeed it works for $n \in \{ 0, 1, 2, 3\} $. For case 3 though I am unsure how to set the summation up so that it works with respect to the unit step (I understand graphically what needs to happen). I tried, $\sum_{k=4-(5-1)}^{9} (k+1)(2(u[n-k] - u[n-k-5]))$ which I can get a sort of rough closed form of, $(n+1)(n+2)u[n-k] - (n+1)(n+2)u[n-k-5]$ but since it has $k$ still in the unit step it is nonsense. $\endgroup$
    – MeljahU
    Feb 5 at 1:27

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