Question

A linear time-invariant system is characterized by \[ H(z)=\frac{3-4z^{-1}} {1-3.5z^{-1}+1.5z^{-2}}. \] Which of the following is not correct?

• A. System is stable in \(\frac12<|z|<3\)
• B. System is unstable in \(|z|>3\)
• C. System is causal, its ROC is \(|z|>3\) and system is stable
• D. System is anti-causal, its ROC is \(|z|<\frac12\) and system is unstable

Hint:

A stable system must have an ROC containing the unit circle. Always check pole locations first.

Answer 1

The Correct Option is C

Concept: The stability and causality of a system depend upon the pole locations and ROC.

Step 1: Determine poles.
Factorizing, \[ 1-3.5z^{-1}+1.5z^{-2} = (1-3z^{-1})(1-0.5z^{-1}). \] Thus poles are \[ z=3, \qquad z=\frac12. \]

Step 2: Check stable ROC.
For stability, ROC must include the unit circle. Hence \[ \frac12<|z|<3. \] Therefore option (A) is correct.

Step 3: Check causal ROC.
For causality, \[ ROC: |z|>3. \] But this ROC does not include the unit circle. Hence the system is unstable. Therefore the statement \[ \text{causal and stable} \] is incorrect. \[ \boxed{\text{Option (C) is not correct}} \]