The transfer function of a discrete time LTI system is given by $$H\left( z \right) = {{2 – {3 \over 4}{z^{ – 1}}} \over {1 – {3 \over 4}{z^{ – 1}} + {1 \over 8}{z^{ – 2}}}}$$ Consider the following statements: S1 : The system is stable and causal for $$ROC:\left| z \right| > {1 \over 2}$$ S2 : The system is stable but not causal for $$ROC:\left| z \right| < {1 \over 4}$$ S3 : The system is neither stable nor causal for $$ROC:{1 \over 4} < \left| z \right| < {1 \over 2}$$ Which one of the following statements is valid?

The correct answer is: C. Both S1 and S3 are true.

The system is stable if all poles of the transfer function lie inside the unit circle. The poles of the transfer function are at $z=-1$ and $z=2$. Since both poles lie inside the unit circle, the system is stable.

The system is causal if the impulse response of the system is zero for all time $t<0$. The impulse response of the system is given by

$$h(t) = {2 \over 4} \delta(t) – {3 \over 8} \delta(t-1) + {1 \over 8} \delta(t-2)$$

Since the impulse response of the system is not zero for all time $t<0$, the system is not causal.

The region of convergence (ROC) of a transfer function is the set of all values of $z$ for which the transfer function is defined and converges. The ROC of the transfer function is given by

$$\left| z \right| > {1 \over 2}$$

Since the ROC of the transfer function includes the unit circle, the system is stable. However, since the ROC of the transfer function does not include the origin, the system is not causal.

Therefore, both S1 and S3 are true.