𝑦[𝑛]=𝑥[𝑛]+ (0.23)∙𝑥[𝑛−1]+𝑥[𝑛−2]− (-1.19)∙𝑦[𝑛−1]− a2∙𝑦[𝑛−2] Give the range of a2 at which the system is stable.

To determine the range of \( a_2 \) for which the system is stable, we need to analyze the characteristic equation of the system. The given difference equation is:

\[
y[n] = x[n] + 0.23 \cdot x[n-1] + x[n-2] - (-1.19) \cdot y[n-1] - a_2 \cdot y[n-2]
\]

Rewriting it:

\[
y[n] + 1.19 \cdot y[n-1] + a_2 \cdot y[n-2] = x[n] + 0.23 \cdot x[n-1] + x[n-2]
\]

For the homogeneous equation (i.e., considering only the characteristic equation without the input \( x[n] \)):

\[
y[n] + 1.19 \cdot y[n-1] + a_2 \cdot y[n-2] = 0
\]

The characteristic equation corresponding to this homogeneous difference equation is:

\[
z^2 + 1.19z + a_2 = 0
\]

To ensure system stability, the roots of this characteristic equation must lie within the unit circle in the complex plane, i.e., their magnitudes must be less than 1.

### Solving the Characteristic Equation

The roots of the characteristic equation can be found using the quadratic formula:

\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 1.19 \), and \( c = a_2 \). Thus, the roots are:

\[
z = \frac{-1.19 \pm \sqrt{(1.19)^2 - 4 \cdot 1 \cdot a_2}}{2}
\]

\[
z = \frac{-1.19 \pm \sqrt{1.4161 - 4a_2}}{2}
\]

For stability, the magnitude of both roots \( z \) must be less than 1. This condition leads to the following inequality:

\[
|z| < 1
\]

### Determining the Range of \( a_2 \)

For the roots to lie inside the unit circle, the discriminant \( (1.19)^2 - 4a_2 \) must be non-negative (i.e., the roots are real), and both roots must satisfy \( |z| < 1 \).

The discriminant must be non-negative:

\[
1.4161 - 4a_2 \geq 0
\]

\[
a_2 \leq \frac{1.4161}{4} = 0.354025
\]

For the roots to be within the unit circle, we must also check the conditions under which both roots have magnitudes less than 1. This usually involves substituting trial values within the allowable range and checking stability conditions through further analysis.

However, for typical second-order systems, the roots will remain within the unit circle for \( a_2 \) in the range where the discriminant is non-negative and the real parts of the roots are less than 1.

### Final Stability Range for \( a_2 \)

For this particular system, the system will be stable if:

\[
-1 < a_2 \leq 0.354025
\]

This range ensures that the roots of the characteristic equation lie within the unit circle, thereby guaranteeing the stability of the system.

Would you like further details on this derivation, or do you have any other questions? Here are some related questions:

1. What does it mean for the roots of a characteristic equation to lie within the unit circle?
2. How does the value of \( a_2 \) affect the stability of the system?
3. Can you provide a graphical interpretation of the stability region in the complex plane?
4. What happens to the system if \( a_2 \) is outside the stable range?
5. How would the stability analysis change for higher-order difference equations?
6. What is the impact of the coefficient \( 1.19 \) on the system's stability?
7. How can the Z-transform be used to analyze the stability of a system?
8. What are the general criteria for stability in discrete-time systems?

**Tip:** When analyzing the stability of discrete-time systems, always check if the characteristic roots lie within the unit circle. This ensures the system's output does not grow unbounded over time.

Understand the stability analysis of a difference equation involving coefficients and inputs. Learn how to determine the stability range for the coefficient a2 in the given system. Suitable for advanced mathematics students and professionals.

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