For $\mathrm{N}=4,$ Twiddle factor is $W_{4}^{n k}=e^{-j 2 \pi k n / 4}$

Twiddle factor Matrix

$W=\left[\begin{array}{cccc}{W_{4}^{0}} & {W_{4}^{0}} & {W_{4}^{0}} & {W_{4}^{0}} \\ {W_{4}^{0}} & {W_{4}^{1}} & {W_{4}^{2}} & {W_{4}^{3}} \\ {W_{4}^{0}} & {W_{4}^{2}} & {W_{4}^{4}} & {W_{4}^{6}} \\ {W_{4}^{0}} & {W_{4}^{3}} & {W_{4}^{6}} & {W_{4}^{9}}\end{array}\right]$

Consider, $W_{4}^{0}=e^{-0 / 4}=1$

$W_{4}^{1}=e^{-j \pi 2 / 4}=\cos \frac{\pi}{2}-j \sin \frac{\pi}{2}=0-j \times 1=-j$

$W_{4}^{2}=e^{-j \pi 4 / 4}=\cos \pi-j \sin \pi=-1-0=-1$

$W_{4}^{3}=e^{-j 6 \pi / 4}=\cos \frac{3 \pi}{2}-j \sin \frac{3 \pi}{2}=0-j \times-1=j$

$W_{4}^{4}=e^{-j 8 \pi / 4}=\cos 2 \pi-j \sin 2 \pi=1-0=1$

$W_{4}^{6}=e^{-j 12 \pi / 4}=\cos 3 \pi-j \sin 3 \pi=-1-0=-1$

$W_{4}^{9}=e^{-j 18 \pi / 4}=\cos \frac{9 \pi}{2}-j \sin \frac{9 \pi}{2}=0-j \times 1=-j$

$\therefore W=\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {-j} & {-1} & {j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {j} & {-1} & {-j}\end{array}\right]$

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

$\therefore D F T[x(n)]=X(k)=W \times x(n)$

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

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

$=\left[\begin{array}{c}{4} \\ {-2} \\ {0} \\ {-2}\end{array}\right]$

Hence, $X(k)=\{4,-2,0,-2\}$