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In an LTI system the input sequence $x(n)=\{1,1,1\}$ and the impulse response $\mathrm{h}(\mathrm{n})=\{-1,1\}$.Find the response of the LTI system by using DFT $-$ IDFT method.
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Solution:

\begin{aligned}\\ \dot{y}(n) &=x(n) * h(n) \\\\ x(n) &=\{1,1,1,0\} \\\\ h(n) &=\{-1,-1,0,0\} \\\\ g(n) &=x(n)+j h(n) \\\\ &=\{1-j, 1-j, 1,0\}\\ \end{aligned}\\

\begin{aligned}\\ G(k) &=\sum_{n=0}^{N-1} g(n) e^{-j 2 \pi / N n k} \\\\ &=\sum_{n=0}^3 g(n) e^{-j \pi / 2 n k} \\\\ G(0) &=\sum_{n=0}^3 g(n) e^0=1-j+1-j+1=3-2 j\\ \end{aligned}\\

\begin{aligned}\\ &G(1)=\sum_{n=0}^3 g(n) e^{-j \pi / 2 n}=\underline{-1-2 j} \\\\ &G(2)=\sum_{n=0}^3 g(n) e^{-j \pi n}=1 \\\\ &G(3)=\sum_{n=0} g(n) e^{-j 3 \pi / 2 n}=1 \\\\ &G(k)=\{3-2 j,-1-2 j, 1,1\} \\\\ &G *(k)=\{3+2 j,-1+2 j, 1,1\} \\\\ &G^*(N-k)=\{3+2 j, 1,1,-1+2 j\} \\\\ &X(k)=1 / 2[G(k)+G *(N-k)]\\ \end{aligned}\\

\begin{aligned}\\ &=\{3,-j, 1, j\} \\\\ H(k) &=\frac{1}{2 j}\left[G(k)-G^*(N-k)\right] \\\\ &=\{-2,-1+j, 0,-1-j\} \\\\ Y(K) &=X(k) \cdot H(k) \\\\ &=\{-6,1+j, 0,1-j\}\\ \end{aligned}\\

\begin{aligned}\\ y(n) &=I D F T[Y(k)] \\\\ &=1 / 4 \sum_{k=0}^3 Y(k) e^{j \pi / 2 n k} \\\\ y(0) &=1 / 4[-6+1+j-j+1]=-1 \\\\ y(1) &=1 / 4 \sum_{k=0}^3 Y(k) e^{j \pi / 2 k}=-2\\ \end{aligned}\\

\begin{aligned}\\ &y(2)=1 / 4 \sum_{k=0}^3 Y(k) e^{j \pi}=-2 \\\\ &y(3)=1 / 4 \sum_{k=0}^3 Y(k) e^{j 3 \pi / 2 k}=-1\\ \\ &y(n)=\{-1,-2,-2,-1\}\\ \end{aligned}\\