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Consider the system with following input-output relation y[n] = (1 + ( -1)n ) x [ n ]where, x[n] is the input and y[n] is the output. The system is
- a)invertible and time invariant
- b)invertible and time varying
- c)non-invertible and time invariant
- d)non-invertible and time varying
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Consider the system with following input-output relation y[n] =(1 + ( ...
Given y [ n ] = (1 + ( -1)n ) x [ n ]
For time invariance
Since Since (1) is not equal to (2) system is time variant
For inverse system
For each unique x [n ] , there should be unique y [n ]
If x[n] = δ[n-1]
y[n] = [ 1 + (-1)n ] δ[n-1]
⇒ y[1] = 0
if x[n] = 2δ[n-1]
y[n] = [ 1+(-1)n ] 2δ (n-1)
y(1) = 0
For two different inputs we have same output. Thus one to one mapping is not possible. Hence the systems is non invertible
Most Upvoted Answer
Consider the system with following input-output relation y[n] =(1 + ( ...
Answer:
Introduction
In this question, we are given the input-output relation of a system, and we need to determine whether the system is invertible and time invariant, invertible and time varying, non-invertible and time invariant, or non-invertible and time varying.
Input-Output Relation
The given input-output relation is:
y[n] = (1 - (-1)^n) * x[n]
Time Invariance
A system is time invariant if a time shift in the input signal results in a corresponding time shift in the output signal. Let's check if the given system is time invariant.
Let's consider two time-shifted input signals: x1[n] = x[n - k] and x2[n] = x[n - k + 1], where k is a positive integer.
The output corresponding to x1[n] is:
y1[n] = (1 - (-1)^n) * x1[n]
= (1 - (-1)^n) * x[n - k]
The output corresponding to x2[n] is:
y2[n] = (1 - (-1)^n) * x2[n]
= (1 - (-1)^n) * x[n - k + 1]
If the system is time invariant, then y1[n] should be equal to y2[n]. Let's check this condition.
Comparing y1[n] and y2[n]:
y1[n] = (1 - (-1)^n) * x[n - k]
y2[n] = (1 - (-1)^n) * x[n - k + 1]
Since k is a positive integer, x[n - k] and x[n - k + 1] are two consecutive samples of the input signal. In the given relation, we see that the output depends on whether n is even or odd, which means it depends on the specific sample of the input signal. Therefore, the system is time varying.
Invertibility
A system is invertible if different input signals produce different output signals. Let's check if the given system is invertible.
Consider two different input signals: x1[n] and x2[n].
The output corresponding to x1[n] is:
y1[n] = (1 - (-1)^n) * x1[n]
The output corresponding to x2[n] is:
y2[n] = (1 - (-1)^n) * x2[n]
If the system is invertible, then y1[n] should be different from y2[n] for different input signals. However, in the given relation, we see that the output depends on whether n is even or odd, which means it depends on the specific sample of the input signal. Therefore, the system is non-invertible.
Conclusion
Based on the analysis, we can conclude that the given system is non-invertible and time varying. Therefore, the correct answer is option D.
Introduction
In this question, we are given the input-output relation of a system, and we need to determine whether the system is invertible and time invariant, invertible and time varying, non-invertible and time invariant, or non-invertible and time varying.
Input-Output Relation
The given input-output relation is:
y[n] = (1 - (-1)^n) * x[n]
Time Invariance
A system is time invariant if a time shift in the input signal results in a corresponding time shift in the output signal. Let's check if the given system is time invariant.
Let's consider two time-shifted input signals: x1[n] = x[n - k] and x2[n] = x[n - k + 1], where k is a positive integer.
The output corresponding to x1[n] is:
y1[n] = (1 - (-1)^n) * x1[n]
= (1 - (-1)^n) * x[n - k]
The output corresponding to x2[n] is:
y2[n] = (1 - (-1)^n) * x2[n]
= (1 - (-1)^n) * x[n - k + 1]
If the system is time invariant, then y1[n] should be equal to y2[n]. Let's check this condition.
Comparing y1[n] and y2[n]:
y1[n] = (1 - (-1)^n) * x[n - k]
y2[n] = (1 - (-1)^n) * x[n - k + 1]
Since k is a positive integer, x[n - k] and x[n - k + 1] are two consecutive samples of the input signal. In the given relation, we see that the output depends on whether n is even or odd, which means it depends on the specific sample of the input signal. Therefore, the system is time varying.
Invertibility
A system is invertible if different input signals produce different output signals. Let's check if the given system is invertible.
Consider two different input signals: x1[n] and x2[n].
The output corresponding to x1[n] is:
y1[n] = (1 - (-1)^n) * x1[n]
The output corresponding to x2[n] is:
y2[n] = (1 - (-1)^n) * x2[n]
If the system is invertible, then y1[n] should be different from y2[n] for different input signals. However, in the given relation, we see that the output depends on whether n is even or odd, which means it depends on the specific sample of the input signal. Therefore, the system is non-invertible.
Conclusion
Based on the analysis, we can conclude that the given system is non-invertible and time varying. Therefore, the correct answer is option D.
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Consider the system with following input-output relation y[n] =(1 + ( -1)n ) x [ n ]where, x[n] is the input and y[n] is the output. The system isa)invertible and time invariantb)invertible and time varyingc)non-invertible and time invariantd)non-invertible and time varyingCorrect answer is option 'D'. Can you explain this answer?
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