Vector length output of discrete time convolution Suppose that the impulse response of a discrete time filter is $h[i]$ where $i=0,1,2,...,N-1$ and the input sequence to the filter is $x[i]$ for $i=0,1,2,.., M-1,$ what would be the length of output vector length? 

I have concluded it is $N$  because  according to discrete time convolution

\begin{align}
y[0] &amp;= x[0]h[0]\\
y[1] &amp;= x[0]h[1]+x[1]h[0]\\
y[2] &amp;= x[0]h[2]+x[1]h[1]+x[2]h[0]\\
\vdots &amp;=\vdots\\
y[N] &amp;= x[1]h[N-1]+x[2]h[N-2]+...+x[N]h[0]
\end{align}

Is this correct?
The output $y$ is convolution of $x[n]$ and $h[n]$:

$$y[n]=\sum_{l=-\infty}^{l=+\infty} x[l]h[n-l]$$
As you said, let's assume only $h[0],\cdots,h[N-1]$ and $x[0],\cdots,x[M-1]$ are non-zero.


When $0\le l\le N-1$, then we have a non-zero $x[l]$.
Now consider $h[n-l]$. As long as $0 \le n-l \le M-1$, then we have a non-zero $h[n-l]$.


add up the sides of the two inequalities to get

$$0\le n\le N+M-2$$

as the set of indices where  $y[n]$ is nonzero. There are $N+M-2+1=N+M-1$ terms.