Convolution and DFT size Assuming that filtered signal $y(n)$ is given  as

$$
y(n)=x(n)\star h(n)=\sum_{m=0}^{2N-1}x(m)h(n-m), \quad n \in[0,1,\ldots, 2N]
$$

where $\star$ is convolution parameter. $x(n)$ is discrete signal of lenght $2N$ and $h(n)$ is filter of the same lenght as $x(n)$. If I want to take for example $N$-point FFT of $y(n)$ how it would look like

\begin{align}
DFT_{N}(y(n))=\sum_{n=0}^{N-1}(x(n)\star h(n))&amp;e^{-j\frac{2\pi nk}{N}}=\sum_{n=0}^{N-1}\sum_{m=0}^{\mathbf{2N-1}}(x(m)h(n-m))e^{-j\frac{2\pi nk}{N}}\\
&amp;\textrm{or}\\
DFT_{N}(y(n))=\sum_{n=0}^{N-1}(x(n)\star h(n))&amp;e^{-j\frac{2\pi nk}{N}}=\sum_{n=0}^{N-1}\sum_{m=0}^{\mathbf{N-1}}(x(m)h(n-m))e^{-j\frac{2\pi nk}{N}}\quad ?
\end{align}
Note that if $x[n]$ and $y[n]$ are non-zero in the interval $n\in [0,2N-1]$ (and, for convenience, they are defined as being zero outside that interval), the (linear) convolution sum can be written as

$$y[n]=(x\star h)[n]=\sum_{m=0}^nx[m]h[n-m],\qquad 0\le n&lt;4N-1\tag{1}$$

Note that the upper limit is $n$ because the argument of the term $h[n-m]$ becomes negative for $m&gt;n$.

So if, for whatever reason, you're only interested in $y[n]$ in the interval $n\in [0,N-1]$, you can set the upper limit in the convolution sum $(1)$ to $N-1$ (assuming that $x[n]$ and $h[n]$ are defined as zero outside the interval $n\in [0,2N-1]$). However, the result wouldn't change if you chose the upper summation interval as $2N-1$, because you would just add zeros to the result.

In short, both formulas in your question give the same result, under the given assumptions on the definitions of $x[n]$ and $y[n]$.