Use MATLAB to solve following example for $N=50$ and

$($a$)M=12,($b$)M=5,$ and $($c$)M=20.$

Building a Square Wave from DTFS Coefficients The contribution of each term to the square wave may be illustrated by defining the partial$-$sum approximation to $x[n]$ in Eq$.$

hat$(x{)}_J[n]={\displaystyle \underset{k=0}{\overset{J}{}}}B[k]cos(k{}_{o}n),$

where $J\frac{N}{2}.$ This approximation contains the first $2J+1$ terms centered on $k=0$ in Eq$.(3\mathrm{.}10).$ Assume a square wave has period $N=50$ and $M=12.$ Evaluate one period of the $J$ th term in Eq$.$ and the $2J+1$ term approximation hat$(x{)}_J[n]$ for $J=1,3,5,23,$ and $25.$

Solution: Figure $3\mathrm{.}14$ depicts the $J$ th term in the sum, $B[J]cos(J{}_{o}n),$ and one period of hat$(x{)}_j[n]$ for the specified values of $J.$ Only odd values for $J$ are considered, because the evenindexed coefficients $B[k]$ are zero when $N=25$ and $M=12.$ Note that the approximation improves as $J$ increases, with $x[n]$ represented exactly when $J=\frac{N}{2}=25.$ In general, the coefficients $B[k]$ associated with values of $k$ near zero represent the low$-$frequency or slowly varying features in the signal, while the coefficients associated with values of $k$ near $+-\frac{N}{2}$ represent the high$-$frequency or rapidly varying features in the signal.