# Intervals of convolution product

Given are two Signals $x_1[n]$ and $x_2[n]$. $x_1[n]$ is in Intervall $[0,2]$ different than null and $x_2[n]$ is in Intervall $[0,3]$. The convolution product is Null outside intervals: $[0,6]$ and $[-2,10]$. Can someone explain me why? I understand for 0,6 it's simply multiplication but why $[-2,10]$?

• Neither answer is correct. It's [0 5]. The length of a convolution is equal to the sum of the individual lengths minus1 (3+4-1 = 6). The smallest non zero index is the sum of the first two indices 0+0 = 0 and the largest non zero index is the sum of the last two indices 2+3 = 5 – Hilmar Oct 25 '17 at 11:24

## 1 Answer

An easy way to deduce the nonzero range of discrete-time convolutions is the following:

Consider convolution of $$x[n]$$ with $$y[n]$$: $$z[n] = x[n] \star y[n] .$$ Let the shorther sequence $$x[n]$$ has a nonzero range of $$M_1 \leq n \leq M_2$$ , while the nonzero range of longer $$y[n]$$ be $$N_1 \leq n \leq N_2$$ .

1- Expand $$x[n]$$ as weighted impulses: $$x[n] = \sum_{k=M_1}^{M_2} x[k] \delta[n-k]$$ .

2- Distribute the convolution over expanded $$x[n]$$: $$(\delta[n-M_1]+...+\delta[n-M_2]) \star y[n] = \delta[n-M_1] \star y[n] + ... + \delta[n-M_2] \star y[n]$$

3- Consider the first impulse $$\delta[n-M_1]$$, and the last impulse $$\delta[n-M_2]$$.

4- Apply shifting property for those two impulses : $$\delta[n-M_1] \star y[n] = y[n-M_1] ~~~,~~~ \delta[n-M_1] \star y[n] = y[n-M_2] .$$

5- Determine the nonzero ranges of $$y[n-M_1]$$ and $$y[n-M_2]$$:

$$y[n]$$ is nonzero over $$N_1 \leq n \leq N_2$$, then $$y[n-M_1]$$ will be nonzero for $$N_1 \leq n-M_1 \leq N_2 ~~~~~~\Rightarrow ~~~~~~ N_1 + M_1 \leq n \leq N_2 + M_1 .$$

Similarly for $$y[n-M_2]$$ the nonzero range will be: $$N_1 \leq n-M_2 \leq N_2 ~~~~~~\Rightarrow ~~~~~~ N_1 + M_2 \leq n \leq N_2 + M_2 .$$

One can see that the output $$z[n]$$ has a nonzero range that begins at the leftmost sample of $$y[n-M_1]$$, and ends at the rightmost sample of $$y[n-M_2]$$. Yielding: $$N_1+M_1 \leq n \leq N_2 + M_2 .$$

Length of $$z[n]$$ can be found as: $$L_z = L_x + L_y -1 = N_2+M_2 - N_1-M_1 + 1 .$$

Applying this to yourcase, $$N_1 = 0$$, $$M_1 = 0$$, $$N_2 = 3$$ and $$M_2=2$$ which yields the nonzero range to be $$0 + 0 = 0 \leq n \leq 5 = 2 + 3 .$$