Topic: Networks, Signals and Systems
A system is defined by the difference equation $y(n)=1.8y(n-1)-0.72y(n-2)+x(n)+0.5x(n-1)$ then the system is
- A. stable
- B. unstable
- C. marginally stable
- D. none of the above
Answer: B. unstable
Explanation: Applying the Z-transform to the difference equation yields the transfer function $H(z) = \frac{Y(z)}{X(z)} = \frac{1 + 0.5z^{-1}}{1 - 1.8z^{-1} + 0.72z^{-2}}$.
Multiply numerator and denominator by $z^2$: $H(z) = \frac{z(z + 0.5)}{z^2 - 1.8z + 0.72}$.
The characteristic equation is $z^2 - 1.8z + 0.72 = 0$. Factoring this yields roots at $(z - 1.2)(z - 0.6) = 0$.
The poles are at $z = 1.2$ and $z = 0.6$. Because one pole ($z=1.2$) has a magnitude greater than 1, it lies outside the unit circle, making the causal system unstable.