# How to perform an analytical computation of the convolution sum for a discrete-time system?

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.

• Additional literature or examples for this topic would be appreciated if anyone has any too. Feb 4 at 9:53

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

• Shouldn't $x[n-k]$ go to $n-4$, since at $u = 1$? Feb 4 at 21:40
• 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
– Jdip
Feb 4 at 23:09
• 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. Feb 5 at 1:27