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Find the output y(n) of a filter whose impulse response is h(n)=(1,1,1) and input signal x(n)=(3,-1,0,1,3,2,0,1,2,1) using over lab save method.
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Solution:

Overlap save method:

\begin{aligned} & \mathrm{x}_1(\mathrm{n})=\{0,0,3,-1,0\} \\\\ & \mathrm{M}-1=2 \text { zeros } \mathrm{L}=3 \text { data points } \\\\ & \mathrm{x}_2(\mathrm{n})=\{-1,0,1,3,2\}\\ \end{aligned}

2 data from previous 3 new data $x_3(n)=\{3,2,0,1,2\}$ and $x_4(n)=\{1,2,1,0,0\}$

Given $\mathrm{h}(\mathrm{n})=\{1,1,1\}$

Increase the length by adding zeros( L+M-1=5)

i.e. $h(n)=\{1,1,1,0,0\}$

$\left[\begin{array}{lllll}1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1\end{array}\right]\left[\begin{array}{c}0 \\ 0 \\ 3 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 0 \\ 3 \\ 2 \\ 2\end{array}\right]$

$y_2(n)=x_2(n) @ h(n)=\{4,1,0,4,6\}$

\begin{aligned} & {\left[\begin{array}{lllll} 1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \end{array}\right]\left[\begin{array}{c} -1 \\ 0 \\ 1 \\ 3 \\ 2 \end{array}\right]=\left[\begin{array}{l} 4 \\ 1 \\ 0 \\ 4 \\ 6 \end{array}\right]} \\ & \mathrm{y}_3(\mathrm{n})=\mathrm{x}_3(\mathrm{n}) @ \mathrm{h}(\mathrm{n})=\{6,7,5,3,3\} \\ & {\left[\begin{array}{lllll} 1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ 0 \\ 1 \\ 2 \end{array}\right]=\left[\begin{array}{l} 6 \\ 7 \\ 5 \\ 3 \\ 3 \end{array}\right]} \\ & \mathrm{y}_4(1)=\mathrm{x}_4(\mathrm{n}) @ \mathrm{h}(\mathrm{n})=\{1,3,4,3,1\} \\ & {\left[\begin{array}{lllll} 1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 1 \\ 0 \\ 0 \end{array}\right]=\left[\begin{array}{l} 1 \\ 3 \\ 4 \\ 3 \\ 1 \end{array}\right]} \\ & \end{aligned}

$Y(n)=\{3,2,2,0,4,6,5,3,3,4,3,1\}$