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Find DFT of $x(n)=\{1234\}$ Using these results and not otherwise find DFT.
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$\begin{aligned} x_{1}(n) &=\{4123\} \\ & x_{2}(n)=\{2341\} \\ x_{3}(n) &=\{6464\} \end{aligned}$

Solution:

Given:

$x(n)=\{1,2,3,4\}$

By definition of DFT.

$x(k)=\sum_{n=0}^{N-1} x(n) w^{nk}_{N}$ $\mathrm{K}=0,1, \cdots, N-1$

$\therefore x(k)=\left[\begin{array}{rrrr}{1} & {1} & {1} & {1} \\ {1} & {-j} & {-1} & {j} \\ {1} & {-1} & {1} & {-1} \\ {1} & {j} & {-1} & {-j}\end{array}\right]\left[\begin{array}{l}{1} \\ {2} \\ {3} \\ {4}\end{array}\right]$

$x(k)=\left[\begin{array}{c}{10} \\ {-2+2 i} \\ {-2} \\ {-2-2 j}\end{array}\right]$

$x_{1}(n)=\{4,1,2,3\}$

$x_{1}(n)=x(n-1)$

By using time shift

$\therefore x_{1}(k)=e^{-j \frac{2 \pi l k}{N}} x(k)$

Here L = 1

$\therefore x_{1}(k)=e^{-j \frac{2 \pi R}{4}} x(k)$

$x_{1}(k)=e^{-j \pi R / 2} x(k)$

$\therefore x_{1}(0)=x(0)=10$

$x_{1}(1)=e^{-j \pi / 2} x(2)=-j(-2+2 j)=2+2 j$

$x_{2}(2)={e}^{-j 2 \pi / 2} x(2)=-1(-2)=2$

$x_{1}(3)=e^{-j 3 \pi / 2} x(3)=j(-2-2 j)=2-2 j$

$\therefore x_{1}(k)=\left[\begin{array}{c}{10} \\ {2+2 j} \\ {2} \\ {2-2 j}\end{array}\right]$

$x_{2}(n)=\{2341\}$

$x_{2}(n)=x(n+1)$

By time shift property

$X_{2}(k)=e^{j \frac{2 \pi k}{4}} x(r)$

$x_{2}(b)=e^{j \frac{\pi k}{2}} x(R)$

$x_{2}(0)=x(0)=10$

$x_{2}(1)=e^{j \pi / 2} x(1)=+j(-2+2 j)=-2-2 j$

$x_{2}(2)=e^{j \pi} x(2)=-1(-2)=2$

$x_{2}(3)=e^{j 3 \pi / 2} x(3)=-j(-2-2 j)=-2+2j$

$x_{3}(n)=\{6,4,6,4\}$

$x_{3}(n)=x(n+1)+x(n-1)$

By linearity and Time shift prop.

$X_{3}(k)=e^{j \frac{2 \pi R}{4} x(k)+e^{-\frac{-j \pi k}{4}} x(R)}$

$=e^\frac{\pi R} {4} x(R)+e^{-j \frac{\pi k}{2}} x(k)$

$=\left[\begin{array}{c}{10} \\ {+2+2j} \\ {2} \\ {2-2 j}\end{array}\right]+\left[\begin{array}{c}{10} \\ {-2-2j} \\ {2} \\ {-2+2 j}\end{array}\right]$

$x_{3}(x)=\left[\begin{array}{r}{20} \\ {0} \\ {4} \\ {0}\end{array}\right]$

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