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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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    $\begingroup$ 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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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 $$ 0 + 0 = 0 \leq n \leq 5 = 2 + 3 .$$

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