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We define y[n] = nx[n] – (n-1)x[n]. Now, z[n] = z[n-1] + y[n], is z[n] stable?
- a)Yes
- b)No
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
We define y[n] = nx[n] – (n-1)x[n]. Now, z[n] = z[n-1] + y[n], i...
As we take the sum of y[n], terms cancel out and deem z[n] to be BIBO stable.
Most Upvoted Answer
We define y[n] = nx[n] – (n-1)x[n]. Now, z[n] = z[n-1] + y[n], i...
Explanation:
Given:
y[n] = nx[n] – (n-1)x[n]
z[n] = z[n-1] + y[n]
Stability Analysis:
To determine the stability of z[n], we need to analyze the behavior of the system. One way to do this is by examining the impulse response of the system. An impulse at n=0 will result in z[0] = y[0], z[1] = z[0] + y[1], z[2] = z[1] + y[2], and so on.
Impulse Response:
When an impulse is applied at n=0, y[0] = 0 - (-1)x[0] = x[0]
So, z[0] = x[0]
When n=1, y[1] = 1*x[1] - (1-1)x[1] = 0
So, z[1] = z[0] + y[1] = x[0]
Similarly, for n=2, y[2] = 2*x[2] - (2-1)x[2] = x[2]
So, z[2] = z[1] + y[2] = x[0] + x[2]
Observations:
From the above calculations, we can see that z[n] is a sum of past inputs x[n] with varying coefficients. The system is accumulating the input values x[n] over time.
Stability Conclusion:
Since the sum of past inputs is not growing indefinitely with time (as the coefficients are not increasing without bound), the system is stable. Therefore, the system z[n] = z[n-1] + y[n] is stable.
Therefore, the correct answer is option 'A' - Yes.
Given:
y[n] = nx[n] – (n-1)x[n]
z[n] = z[n-1] + y[n]
Stability Analysis:
To determine the stability of z[n], we need to analyze the behavior of the system. One way to do this is by examining the impulse response of the system. An impulse at n=0 will result in z[0] = y[0], z[1] = z[0] + y[1], z[2] = z[1] + y[2], and so on.
Impulse Response:
When an impulse is applied at n=0, y[0] = 0 - (-1)x[0] = x[0]
So, z[0] = x[0]
When n=1, y[1] = 1*x[1] - (1-1)x[1] = 0
So, z[1] = z[0] + y[1] = x[0]
Similarly, for n=2, y[2] = 2*x[2] - (2-1)x[2] = x[2]
So, z[2] = z[1] + y[2] = x[0] + x[2]
Observations:
From the above calculations, we can see that z[n] is a sum of past inputs x[n] with varying coefficients. The system is accumulating the input values x[n] over time.
Stability Conclusion:
Since the sum of past inputs is not growing indefinitely with time (as the coefficients are not increasing without bound), the system is stable. Therefore, the system z[n] = z[n-1] + y[n] is stable.
Therefore, the correct answer is option 'A' - Yes.
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