Let $\mathrm{N}=4$

We make the length of $\mathrm{x}(\mathrm{n})$ and $\mathrm{h}(\mathrm{n})$ equal to 4 $\mathrm{by}$ zero padding.

$\therefore \mathrm{x}(\mathrm{n})=\{1,2,0,0\}$

By Definition, $D F T\left[x_{1}(n)\right]=X_{1}(k)=W \times x_{1}(n)$

where $\mathrm{W}$ is the Twiddle Factor Matrix for $\mathrm{N}=4$

$\therefore X(k)=\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {-j} & {-1} & {j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {j} & {-1} & {-j}\end{array}\right] \times\left[\begin{array}{l}{1} \\ {2} \\ {0} \\ {0}\end{array}\right]$

$=\left[\begin{array}{c}{1+2+0+0} \\ {1-2 j+0+0} \\ {1-2+0+0} \\ {1+2 j+0+0}\end{array}\right]$

$=\left[\begin{array}{c}{3} \\ {1-2 j} \\ {-1} \\ {1+2 j}\end{array}\right]$

Hence, $X(k)=\{3,1-2 j,-1,1+2 j\}$

Similarly, $h(n)=\{2,2,1,1\}$

$\therefore H(k)=\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {-j} & {-1} & {j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {j} & {-1} & {-j}\end{array}\right] \times\left[\begin{array}{l}{2} \\ {2} \\ {1} \\ {1}\end{array}\right]$

$\begin{aligned} &\left[\begin{array}{l}{2+2+1+1} \\ {2-2 j-1+j} \\ {2-2+1-1} \\ {2+2 j-1-j}\end{array}\right] \\=&\left[\begin{array}{c}{6} \\ {1-j} \\ {0} \\ {1+j}\end{array}\right] \end{aligned}$

Hence, $\mathrm{H}(\mathrm{k})=\{6,1-j, 0,1+j\}$

Let, $y(n)=x(n) \otimes h(n)$

By Circular Convolution Property of DFT,

$D F T \ [y(n)]=D F T[x(n) \otimes h(n)]$

$\therefore Y(k)=X(k) H(k)$

For $\mathrm{k}=0, \mathrm{Y}(0)=\mathrm{X}(0) \mathrm{H}(0)=3 \times 6=18$

For $\mathrm{k}=1, \mathrm{Y}(1)=\mathrm{X}(1) \mathrm{H}(1)=(1-2 j)(1-j)=-1-3 j$

For $\mathrm{k}=2, \mathrm{Y}(2)=\mathrm{X}(2) \mathrm{H}(2)=(-1) \times 0=0$

For $\mathrm{k}=3, \mathrm{Y}(3)=\mathrm{X}(3) \mathrm{H}(3)=(1+2 j)(1+j)=-1+3 j$

Hence, $Y(k)=\{18,-1-3j, 0,-1+3 j\}$

By Definition, $I D F T \ [Y(k)]=y(n)=\frac{1}{N} W^{*} \times Y(k)$

$\therefore y(n)=\frac{1}{4}\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {j} & {-1} & {-j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {-j} & {-1} & {j}\end{array}\right] \times\left[\begin{array}{c}{18} \\ {-1-3 j} \\ {0} \\ {-1+3 j}\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{c}{18+(-1-3 j)+0+(-1+3 j)} \\ {18+j(-1-3 j)-0-j(-1+3 j)} \\ {18-(-1-3 j)+0-(-1+3 j)} \\ {18-j(-1-3 j)-0+j(-1+3 j)}\end{array}\right]$

$\begin{aligned} &=\frac{1}{4}\left[\begin{array}{c}{16} \\ {24} \\ {20} \\ {12}\end{array}\right] \\=&\left[\begin{array}{c}{4} \\ {5} \\ {3}\end{array}\right] \end{aligned}$

Hence, the circular convolution of the sequences $x(n)$ and $h(n)$ is $y(n)=\{4,6,5,3\}$