To solve this problem, we need to analyze the given digital signal processing system and determine whether it is an FIR or IIR filter and whether it is stable or unstable.
- FIR (Finite Impulse Response) Filter: An FIR filter has a finite number of terms in its impulse response. The output depends only on the current and previous inputs. It does not involve any feedback (i.e., no terms of the form \( y(n-k) \), where \( k \geq 1 \)).
- IIR (Infinite Impulse Response) Filter: An IIR filter has an infinite impulse response, meaning it involves feedback, and its output depends on previous outputs (i.e., terms like \( y(n-k) \) for \( k \geq 1 \) are present). An IIR filter can be unstable depending on the coefficients and feedback structure.
- Stability: The stability of a system is determined by the location of the poles of its transfer function in the Z-plane. A system is stable if all poles lie inside the unit circle, and unstable if any pole lies outside the unit circle.
The given system is described by the equation:
\[ y(n) = 2x(n) + x(n-1) + 2y(n-1) \]
We can see that the output \( y(n) \) depends not only on the current and previous input values \( x(n) \) and \( x(n-1) \) but also on the previous output \( y(n-1) \), which is characteristic of an IIR filter.
The feedback term \( 2y(n-1) \) indicates that this is an IIR filter. If the coefficients in the feedback loop are not chosen properly, it can lead to an unstable system, especially since there is a feedback term in the system's equation.
Since the system has feedback and involves previous output terms, it is an IIR filter. Moreover, the presence of a feedback loop without further analysis of the system’s poles indicates the possibility of instability.
The system is An unstable IIR filter.