How to index output of convolution product Unfortunately indexing the output of convolution product confuses me. Suppose $x[n]$ is of length $15$ (i.e. $0 &lt;= n &lt;= 14$) and $y[n]$ has the length of $9$ ($0 &lt;= n &lt;= 8$). if $z[n]$ be the output of convolution product of $x[n]$ and $y[n]$, that is $$z[n] = (x * y)[n] = \sum_{k=-\infty}^{+\infty} x[k]y[n-k]$$
As a result, $z[n]$ would has the length of $23$, right?
Now the question is what is the range of $n$ for $z[n]$? Is it $0 &lt;= n &lt;= 22$ or $-4 &lt;= n &lt;= 18$?
From formula for convolution I think the latter is true. But the problem arises when I work with seismic signals.
Suppose $acc[t_i]$ is seismic signal of Kobe earthquake where $0 &lt;= t_i &lt;= 41.99 sec$ with sampling period of $T_s = 0.01 sec$ (length of signal $4200$) as depicted below:

Now it is desired to denoise $acc[t_i]$ using moving average filter with impulse response $h[t_i]$:
$$ h[t_i] =
\begin{cases}
\frac{1}{30},  &amp; \text{if $0 &lt;= t_i &lt;= 0.29$} \\
0, &amp; \text{if $0.3 &lt;= t_i &lt;= 41.99$}
\end{cases}$$
where $T_s = 0.01 sec$. After applying $h[t_i]$ to $acc[t_i]$ by convolution, the output is the sequence $ACC[t_i]$ of length 8399, as shown in following figure.

Obviously the results of filtering process is the interval marked by red bullets. However according to convolution formula, I think I should label time axis between approximately $-21sec$ to $63sec$.
Is that right or something is missing?
If $x[n]$ and $y[n]$ are both causal and starting at index $0$, then the result of convolution will also be causal and it will start at index $0$. Just plug in $n=-4$ in the expression for $z[n]$, you will find that it will be $0$.
$$z[n] = \sum^{\infty}_{k=-\infty}x[k]y[n-k]$$
First non-zero term in above expression is at $n=0$. Because for $k&lt;0$, $x[k]$ will be $0$, you can re-write the above expression as:
$$z[n] = \sum^{\infty}_{k=0}x[k]y[n-k]$$
In the above expression, put $n&lt;0$, you will see that $y[n-k]$ will always be 0.
There will be a transient delay of $N-1$ samples, when you are using Moving-Average FIR Filter of length $N$, not an advance. The delay is due to the fact that you need $N-1$ previous samples of the input before you can produce an average of $N$ samples. You cannot have an output before you feed input to a causal FIR filter. Expecting an output at $-21sec$, is expecting to look into future which is not correct here.
The figures you have posted is showing that you have 8399 samples of output starting at $n=0$. As you can see, approximately at $5sec$, the seismic activity starts, Moving average just smoothens-out the original input signal with a transience of $0.3sec$.