Parseval's relation of DFT:

Statement: If $D F T\left[x_{1}(n)\right]=X_{1}(k) ;$ and $\operatorname{DFT}\left[x_{2}(n)\right]=X_{2}(k)$ then $\sum_{n=0}^{N-1} x_{1}(n) x_{2}^*(n)=\frac{1}{N} \sum_{k=0}^{N-1} X_{1}(k) X_{2} ^*(k),$ where $x_{1}(n)$ and $x_{2}(n)$ are N-point sequences.

Explanation: If $x_{1}(n)=x_{2}(n)$ then $x_{1}(n) x_{1}^{*}(n)$ $=\left|x_{1}(n)\right|^{2},$ which is the energy content of the signal in time domain. Similarly, $X_{1}(k) X_{1}^{*}(k)=\left|X_{1}(k)\right|^{2}$ represents the energy content of the signal at the given frequency.

For 4-point sequence, $\mathrm{N}=4$

Let, $X[n]=\{1,2,1,0\} \cdots(1)$

By definition, $\quad D F T[x(n)]=X(k)=W \times x(n)$

where W is Twiddle factor Matrix

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

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

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

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

Energy of the signal $=\sum_{n=0}^{N-1}\left|x_{1}(n)\right|^{2}$

$=|x(0)|^{2}+|x(1)|^{2}+|x(2)|^{2}+|x(3)|^{2}$

$=|1|^{2}+|2|^{2}+|1|^{2}+|0|^{2}(\text { From } 1)$

$=6 \cdots(3)$

Similarly,

$\frac{1}{N} \sum_{k=0}^{N-1}\left|X_{1}(k)\right|^{2}$

$=\frac{1}{4}\left\{|X(0)|^{2}+|X(1)|^{2}+|X(2)|^{2}+|X(3)|^{2}\right\}$

$=\frac{1}{4}\left\{|4|^{2}+|-2 j|^{2}+|0|^{2}+|2 j|^{2}\right\}(\text { From } 2)$

$=\frac{1}{4}\left\{16+[\sqrt{0^{2}+(-2)^{2}}]^{2}+0+[\sqrt{0^{2}+2^{2}}]^{2}\right\}$

$=\frac{1}{4}\{16+4+0+4\}$

$=6 \cdots(4)$

From $(1)$ and $(2), \sum_{n=0}^{N-1}\left|x_{1}(n)\right|^{2}=\frac{1}{N} \sum_{k=0}^{N-1}\left|X_{1}(k)\right|^{2}$

Parseval's theorem is proved for the sequence $x(n)=$ $\{1,2,1,0\}$