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 $[-2,10]$?
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2$\begingroup$ 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 $\endgroup$ – Hilmar Oct 25 '17 at 11:24
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The best way to deduce the nonzero range of the discrete convolutions is the following method.
Consider the convolution of $x[n]$ with $y[n]$ as $$z[n] = x[n] \star y[n] .$$ Let the shorther sequence be $x[n]$ with nonzero range of $M_1 \leq n \leq M_2$ and the nonzero range of $y[n]$ be $N_1 \leq n \leq N_2$ .
1- Decompose the shorter sequence $x[n]$ into discrete impulses: $x[n] = \sum_{k=M_1}^{M_2} x[k] \delta[n-k]$ .
2- Consider the first and last impulses $\delta[n-M_1]$ and $\delta[n-M_2]$.
3- Consider distribution of convolution over addition. $$ (\delta[n-M_1]+...+\delta[n-M_2]) \star y[n] = \delta[n-M_1] \star y[n] + ... + \delta[n-M_2] \star y[n] $$
4- Consider shifting property of the impulses to locate the partial convolutions; i.e. $$ \delta[n-M_1] \star y[n] = y[n-M_1] ~~~,~~~ \delta[n-M_1] \star y[n] = y[n-M_2] .$$
5- Consider the nonzero ranges of those partial convolutions.
Since $y[n]$ was nonzero over $N_1 \leq n \leq N_2$ then $y[n-M_1]$ will be nonzero for $$ N_1 \leq n-M_1 \leq N_2 ~~~~~~\Rightarrow ~~~~~~ N_1 + M_1 \leq n \leq N_2 + M_1 .$$
And similarly for the last impulse whose partial output is $y[n-M_2]$ the nonzero range will be $$ N_1 \leq n-M_2 \leq N_2 ~~~~~~\Rightarrow ~~~~~~ N_1 + M_2 \leq n \leq N_2 + M_2 .$$
Based on these we can conclude that the output $z[n]$ will have a nonzero range which begins at the left member of the first convolution and ends at the right member of the last impulse convolution. This yields $z[n]$ to be nonzero for $$N_1+M_1 \leq n \leq N_2 + M_2 .$$
The length of $z[n]$ is founs to be $$L_z = L_x + L_y -1 = N_2+M_2 - N_1-M_1 + 1 .$$
So in your case, $N_1 = 0$, $M_1 = 0$, $N_2 = 3$ and $M_2=2$ which yields the nonzero range to be $$ 0 + 0 = 0 \leq n \leq 5 = 2 + 3 .$$
