Question

A second order discrete time system is characterized by the difference equation,
$y\left(n\right)-3y(n-1)+2y(n-2)=x\left(n\right)-2x(n-1)$
The value of y(2) when $n\ge 0$ and x(n) = u(n) and the initial condition are given as y(-1) = y(-2) = 1 is_____

• A. 4

Answer 1

The correct option is A 4
For $x\left(n\right)=u\left(n\right)$
We have, $X\left(z\right)=\frac{1}{1-z^{-1}}$

Now for difference equation,
$y\left(n\right)-3y(n-1)+2y(n-2)=x\left(n\right)-2x(n-1)$
Taking unilateral z-transform we get,

$Y\left(z\right)-3\left[z^{-1}Y\right(z)+y(-1\left)\right]+2\left[z^{-2}Y\right(z)+z^{-1}y(-1)+y(-2\left)\right]=\frac{1}{1-z^{-1}}-\frac{2z^{-1}}{1-z^{-1}}$

$y\left(z\right)[1-3z^{-1}+2z^{-2}]+2z^{-1}-1=\frac{1-2z^{-1}}{1-z^{-1}}$

$y\left(z\right)\left[\right(1-z^{-1}\left)\right(1-2z^{-1}\left)\right]-[1-2z^{-1}]=\left(\frac{1-2z^{-1}}{1-z^{-1}}\right)$

$[1-2z^{-1}]\left[y\right(z\left)\right[1-z^{-1}]-1]=\frac{1-2z^{-1}}{1-z^{-1}}$

$y\left(z\right)[1-z^{-1}]-1=\frac{1}{1-z^{-1}}$

$y\left(z\right)=\frac{1}{(1-z^{-1}{)}^2}+\frac{1}{(1-z^{-1})}$

Taking inverse z-transform,

$y\left(n\right)=(n+1)u\left(n\right)+u\left(n\right)=(n+2)u\left(n\right)$

$y\left(2\right)=[2+2]u\left(2\right)=4$