Solution:

Given: the circular convolution of the following sequences,

$ x_1(n)=\{1,1,2,1\} \text { and } x_2(n)=\{1,2,3,4\}\\ $

To find DFT of x1(k):

$ X_I(k)=\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 2 \\ 1 \end{array}\right]=\left[\begin{array}{c} 5 \\ -1 \\ 1 \\ -1 \end{array}\right] $

To find DFT of x2(k):

$ \begin{gathered} X_2(k)=\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & -j & -1 & j \\ 1 & -1 & 1 & -1 \\ 1 & j & -1 & -j \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 3 \\ 4 \end{array}\right]=\left[\begin{array}{c} 10 \\ -2+j 2 \\ -2 \\ -2-j 2 \end{array}\right] \\ X_3(k)=X_1(k)^* X_2(k)=\{5,-1,1,-1\} *\{10,-2+j 2,-2,-2-j 2\}=\{50,2-j 2,-2,2+j 2\} \end{gathered} $

To find IDFT of $X 3(n)$ :

$ \begin{aligned} X_3(n) & =\frac{1}{N}\left[\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & j & -1 & -j \\ 1 & -1 & 1 & -1 \\ 1 & -j & -1 & j \end{array}\right]\left[\begin{array}{c} 10 \\ 2-j 2 \\ -2 \\ 2+j 2 \end{array}\right] \\ & =\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]\left[\begin{array}{c} 10 \\ 2-j 2 \\ -2 \\ 2+j 2 \end{array}\right] \\ & =\left[\begin{array}{c} 13 \\ 14 \\ 11 \\ 12 \end{array}\right] \end{aligned} $