Example 1

Find the response of the system $s(n+2)-3s(n+1)+2s(n) = delta (n)$, when all the initial conditions are zero.

Solution − Taking Z-transform on both the sides of the above equation, we get

$$S(z)Z^2-3S(z)Z^1+2S(z) = 1$$

$Rightarrow S(z)lbrace Z^2-3Z+2rbrace = 1$

$Rightarrow S(z) = frac{1}{lbrace z^2-3z+2rbrace}=frac{1}{(z-2)(z-1)} = frac{alpha _1}{z-2}+frac{alpha _2}{z-1}$

$Rightarrow S(z) = frac{1}{z-2}-frac{1}{z-1}$

Taking the inverse Z-transform of the above equation, we get

$S(n) = Z^{-1}[frac{1}{Z-2}]-Z^{-1}[frac{1}{Z-1}]$

$= 2^{n-1}-1^{n-1} = -1+2^{n-1}$

Example 2

Find the system function H(z) and unit sample response h(n) of the system whose difference equation is described as under

$y(n) = frac{1}{2}y(n-1)+2x(n)$

where, y(n) and x(n) are the output and input of the system, respectively.

Solution − Taking the Z-transform of the above difference equation, we get

$y(z) = frac{1}{2}Z^{-1}Y(Z)+2X(z)$

$= Y(Z)[1-frac{1}{2}Z^{-1}] = 2X(Z)$

$= H(Z) = frac{Y(Z)}{X(Z)} = frac{2}{[1-frac{1}{2}Z^{-1}]}$

This system has a pole at $Z = frac{1}{2}$ and $Z = 0$ and $H(Z) = frac{2}{[1-frac{1}{2}Z^{-1}]}$

Hence, taking the inverse Z-transform of the above, we get

$h(n) = 2(frac{1}{2})^nU(n)$

Example 3

Determine Y(z),n≥0 in the following case −

$y(n)+frac{1}{2}y(n-1)-frac{1}{4}y(n-2) = 0quad givenquad y(-1) = y(-2) = 1$

Solution − Applying the Z-transform to the above equation, we get

$Y(Z)+frac{1}{2}[Z^{-1}Y(Z)+Y(-1)]-frac{1}{4}[Z^{-2}Y(Z)+Z^{-1}Y(-1)+4(-2)] = 0$

$Rightarrow Y(Z)+frac{1}{2Z}Y(Z)+frac{1}{2}-frac{1}{4Z^2}Y(Z)-frac{1}{4Z}-frac{1}{4} = 0$

$Rightarrow Y(Z)[1+frac{1}{2Z}-frac{1}{4Z^2}] =frac{1}{4Z}-frac{1}{2}$

$Rightarrow Y(Z)[frac{4Z^2+2Z-1}{4Z^2}] = frac{1-2Z}{4Z}$

$Rightarrow Y(Z) = frac{Z(1-2Z)}{4Z^2+2Z-1}$