n=0时 y(2)+0.6y(1)+0.08y(0)=1 y(2)+0.6+0=1 ∴y(2)=4 n=1时 y(3)+0.6y(2)+0.08y(1)=1 y(3)+0.6*0.4+0.08=1 ∴ y(3)=0.68 对差分方程y(n+2)+0.6y(n+1)+0.08y(n)=1两边取Z变换，得 Z2 {Y(z)-[y(0)+y（1）Z-1]}+0.6Z{Y(z)- y(0)}+0.08Y(z)=U(z) Y(z)( Z2+ 0.6Z + 0.08 ) = U（z）+Z U（z）+Z = Z/(Z-1)+Z = (Z/(Z-1))*Z = ZU(z) Y(z)( Z2+ 0.6Z + 0.08 ) = ZU(z) ∴ H（z）=Y(z)/U(z) = z/(z^2+0.6z+0.08) 特征方程 Z2+0.6Z+0.08=0 Z1 = - 0.4 Z2 = - 0.2 ∣Zi∣＜1 所以系统稳定