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Prove the Parseval's theorem for the sequence $x(n)=\{2,4,2,4\}$
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For 4 point sequence, $N=4$

By definition, $W_{N}^{n k}=e^{-j 2 \pi k n /N}=e^{-j 2 \pi k n / 4}$

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

Similarly, $W_{4}^{0}=1 ; W_{4}^{2}=-1 ; W_{4}^{3}=j ; W_{4}^{4}=1$

$W_{4}^{6}=-1 ; W_{4}^{9}=-j$

$\therefore$ Twiddle factor Matrix

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

Let, $X(k)=\{2_{\uparrow},4,2,4\} \dots(1)$

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

$\therefore X(k)=\left[\begin{array}{cccc}{1} & {1} & {1} & {1} \\ {1} & {-j} & {-1} & {j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {j} & {-1} & {-j}\end{array}\right] \times\left[\begin{array}{l}{2} \\ {4} \\ {2} \\ {4}\end{array}\right]$

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

$=\left[\begin{array}{c}{12} \\ {0} \\ {-4} \\ {0}\end{array}\right]$

Hence, $X(k)=\{12,0,-4,0\} \dots(2)$

Parseval's theorem for DFT :

If $\operatorname{DFT}[x(n)]=X(k)$ then

$\sum_{n=0}^{N-1}\left|x_{1}(n)\right|^{2}=\frac{1}{N} \sum_{k=0}^{N-1}\left|X_{1}(k)\right|^{2},$ where $x(n)$ is a N-point sequences.

Now, $\sum_{n=0}^{N-1}|x(n)|^{2}=|x(0)|^{2}+|x(1)|^{2}+|x(2)|^{2}+|x(3)|^{2}$

$=|2|^{2}+|4|^{2}+|2|^{2}+|4|^{2}$ (From Equation 1)

$=40 \cdots (3)$

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

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

$\left.=\frac{1}{4}\left\{|12|^{2}+|0|^{2}+|-4|^{2}+|0|^{2}\right\}\right)$ (From Equation 2)

$=\frac{1}{4}\{144+0+16+0\}$

$=40 \cdots (4)$

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

Hence, Parseval's theorem is proved for the sequence

$\mathrm{x}(\mathrm{n})=\{2,4,2,4\}$

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