written 5.3 years ago by |
Let period of $x_{1}(n)$ and $x_{2}(n)$ be 5
$x_{1}(n)=\{1,-1,-2, 3,-1\} \ and\ \quad x_{2}(n)=\{1,2,30,0\}$
In matrix form, $[y(n)]_{5 \times 1}=\left[x_{1}(n)\right]_{5 \times 5}\left[x_{2}(n)\right]_{5 \times 1}$
$\left[\begin{array}{c}{y(0)} \\ {y(1)} \\ {y(2)} \\ {y(3)} \\ {y(4)}\end{array}\right]=\left[\begin{array}{cccccc}{x_{1}(0)} & {x_{1}(4)} & {x_{1}(3)} & {x_{1}(2)} & {x_{1}(1)} \\ {x_{1}(1)} & {x_{1}(0)} & {x_{1}(4)} & {x_{1}(3)} & {x_{1}(2)} \\ {x_{1}(2)} & {x_{1}(1)} & {x_{1}(0)} & {x_{1}(4)} & {x_{1}(3)} \\ {x_{1}(3)} & {x_{1}(2)} & {x_{1}(1)} & {x_{1}(0)} & {x_{1}(4)} \\ {x_{1}(4)} & {x_{1}(3)} & {x_{1}(2)} & {x_{1}(1)} & {x_{1}(0)}\end{array}\right] \times\left[\begin{array}{c}{x_{2}(0)} \\ {x_{2}(1)} \\ {x_{2}(2)} \\ {x_{2}(3)} \\ {x_{2}(4)}\end{array}\right]$
$=\left[\begin{array}{ccccc}{1} & {-1} & {3} & {-2} & {-1} \\ {-1} & {1} & {-1} & {3} & {-2} \\ {-2} & {-1} & {1} & {-1} & {3} \\ {3} & {-2} & {-1} & {1} & {-1} \\ {-1} & {3} & {-2} & {-1} & {1}\end{array}\right] \times\left[\begin{array}{c}{1} \\ {2} \\ {3} \\ {0} \\ {0}\end{array}\right]$
$=\left[\begin{array}{c}{1-2+9+0+0} \\ {-1+2-3+0+0} \\ {-2-2+3+0+0} \\ {3-4-3+0+0} \\ {-1+6-6+0+0}\end{array}\right]$
$\therefore\left[\begin{array}{l}{y(0)} \\ {y(1)} \\ {y(2)} \\ {y(3)} \\ {y(4)}\end{array}\right]=\left[\begin{array}{c}{8} \\ {-2} \\ {-1} \\ {-4} \\ {-1}\end{array}\right]$