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A linear discrete time system has the characteristics equation, z3 - 0.81 z = 0, the system is
- a)is stable
- b)is marginally stable
- c)is unstable
- d)stability can not be assessed from the given information
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A linear discrete time system has the characteristics equation, z3 - 0...
To determine the stability of a linear discrete-time system, we need to analyze the characteristics equation. The given characteristics equation is z^3 - 0.81z = 0.
We can assess the stability of the system by examining the roots of the characteristics equation. The roots of the equation represent the poles of the system, and their location in the complex plane determines the stability.
Let's solve the characteristics equation to find the roots:
z^3 - 0.81z = 0
Factoring out z, we get:
z(z^2 - 0.81) = 0
This equation has two solutions: z = 0 and z^2 - 0.81 = 0.
Solving the quadratic equation, we have:
z^2 = 0.81
Taking the square root of both sides, we get:
z = ±√0.81
z = ±0.9
Therefore, the roots of the characteristics equation are z = 0, z = 0.9, and z = -0.9.
Now, let's analyze the location of these roots in the complex plane:
1. z = 0: This root lies on the origin of the complex plane. Since it has a magnitude of 0, it is a stable pole.
2. z = 0.9: This root lies outside the unit circle in the complex plane. Since it has a magnitude greater than 1, it represents an unstable pole.
3. z = -0.9: This root also lies outside the unit circle in the complex plane. Similar to the previous root, it has a magnitude greater than 1 and represents an unstable pole.
Based on the analysis of the roots, we can conclude that the given linear discrete-time system is stable (option A). The presence of a stable pole at z = 0 ensures the stability of the system.
We can assess the stability of the system by examining the roots of the characteristics equation. The roots of the equation represent the poles of the system, and their location in the complex plane determines the stability.
Let's solve the characteristics equation to find the roots:
z^3 - 0.81z = 0
Factoring out z, we get:
z(z^2 - 0.81) = 0
This equation has two solutions: z = 0 and z^2 - 0.81 = 0.
Solving the quadratic equation, we have:
z^2 = 0.81
Taking the square root of both sides, we get:
z = ±√0.81
z = ±0.9
Therefore, the roots of the characteristics equation are z = 0, z = 0.9, and z = -0.9.
Now, let's analyze the location of these roots in the complex plane:
1. z = 0: This root lies on the origin of the complex plane. Since it has a magnitude of 0, it is a stable pole.
2. z = 0.9: This root lies outside the unit circle in the complex plane. Since it has a magnitude greater than 1, it represents an unstable pole.
3. z = -0.9: This root also lies outside the unit circle in the complex plane. Similar to the previous root, it has a magnitude greater than 1 and represents an unstable pole.
Based on the analysis of the roots, we can conclude that the given linear discrete-time system is stable (option A). The presence of a stable pole at z = 0 ensures the stability of the system.
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Community Answer
A linear discrete time system has the characteristics equation, z3 - 0...
What about its ROC the system should have common roc which passes through unit circle that's stable but here by just pole how can I say it's stable ROC should be mentioned.
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A linear discrete time system has the characteristics equation, z3 - 0.81 z = 0, the system isa)is stableb)is marginally stablec)is unstabled)stability can not be assessed from the given informationCorrect answer is option 'A'. Can you explain this answer?
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