the sub-range of circular and linear convolution
Given a finite length discrete-time sequence $x[n]$ nonzero for the range $ 0 \leq n &lt; L_x$, and another finite length sequence $y[n]$, nonzero for the range $0 \leq n &lt; L_y$. Their linear convolution will be 
$$ z[n] = x[n] \star y[n] $$

which is nonzero for the range $0 \leq n &lt; L_x + L_y - 1$. In other words, the convolution will have a length of $$L_z = L_x + L_y - 1$$

The $N$-point circular convolution of $x[n]$ and $y[n]$ (usually implemented through $N$-point DFTs) will be $w[n]$ 

$$w[n]   = x[n] \circledast_N y[n] $$ 

Since most typically the circular convolution is used to implement the linear convolution, then we are interested in the relation between $z[n]$ and $w[n]$; i.e., what's the range of $n$ for which they are the same ?

The answer depends on a relation between $L_z$ and $N$ and stated as follows:


if $~~ L_z \leq N ~~~ $ then $w[n] = \begin{cases} { z[n]  ~~~,~~~ 0 \leq n &lt; L_z \\ ~0~ ~~~~~,~~~~ L_z \leq n &lt; N }\end{cases} $


$$\\$$


if $~~ N &lt; L_z ~~~ $ then $w[n] = \begin{cases} { \text{aliased}  ~~~,~~~ 0 \leq n &lt; L_z-N \\ z[n]~ ~~~,~~~~ L_z-N \leq n &lt; N }\end{cases}$


Note that as a consequence of the second case, if $L_z - N &gt; N$ or $ 2N \leq L_z$ there will be no clean samples.

The following Matlab stem-plot shows the overlap between a linear and a circular convolution of two sequences of lengths $L_x = 32$ , $L_y=10$ and DFT length $N=25$ and the linear convolution of length $L_z = 41$.



Since $N &lt; L_z$, the first $L_z-N = 16$ samples of the circular convolution in the range $0 \leq n &lt; 16$ will be aliased and only the remaining $N-(L_z-N) = 2N-L_z = 9$ samples in the range $ 16 \leq n &lt; 25$ will be clean; i.e. equal to the linear convolution samples. 

Note that the initial $16$ aliased samples of the circular convolution for $0 \leq n &lt; 16$ is set to zero for clarification, and plotted them on together with the linear convolution which has a length of $41$ samples. The circular convolution just has $N=25$ samples only. Hence the last plotted $16$ zeros of the circular convolution actually belong to the first $16$ samples of the next period of the periodic result of the circular convolution.