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Comment on the time invariance of the following discrete system: y[n] = x[2n+4].
- a)Time invariant
- b)Time variant
- c)Both Time variant and Time invariant
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
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Comment on the time invariance of the following discrete system: y[n] ...
A time shift in the input scale gives double the time shift in the output scale, and hence is time variant.
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Comment on the time invariance of the following discrete system: y[n] ...
Time Invariance of a Discrete System:
A discrete system is said to be time-invariant if its output depends only on the present and past values of the input signal, and not on the time at which the input is applied. In other words, if a time shift is applied to the input signal, the output signal should also undergo the same time shift.
In the given system, y[n] = x[2n+4], we can see that the output depends on the input signal at a shifted time instant.
Explanation:
Let us consider two input signals, x1[n] and x2[n], such that x1[n] = x2[n] for all n.
Now, let us apply a time shift of k samples to both input signals, such that x1[n-k] and x2[n-k] are the new input signals.
For the given system to be time-invariant, the output signals y1[n] and y2[n] corresponding to x1[n-k] and x2[n-k] should be identical to the output signals y1[n-k] and y2[n-k] corresponding to the original input signals x1[n] and x2[n].
y1[n-k] = x1[2(n-k)+4] = x1[2n-2k+4]
y2[n-k] = x2[2(n-k)+4] = x2[2n-2k+4]
y1[n] = x1[2n+4]
y2[n] = x2[2n+4]
Now, if we compare y1[n-k] and y2[n-k] with y1[n] and y2[n], we can see that they are not identical. This is because the output signal depends on the value of n, and a time shift of k samples in the input signal results in a different value of n.
Therefore, the given system is time-variant.
Conclusion:
The given discrete system, y[n] = x[2n+4], is time-variant as the output signal depends on the time at which the input is applied.
A discrete system is said to be time-invariant if its output depends only on the present and past values of the input signal, and not on the time at which the input is applied. In other words, if a time shift is applied to the input signal, the output signal should also undergo the same time shift.
In the given system, y[n] = x[2n+4], we can see that the output depends on the input signal at a shifted time instant.
Explanation:
Let us consider two input signals, x1[n] and x2[n], such that x1[n] = x2[n] for all n.
Now, let us apply a time shift of k samples to both input signals, such that x1[n-k] and x2[n-k] are the new input signals.
For the given system to be time-invariant, the output signals y1[n] and y2[n] corresponding to x1[n-k] and x2[n-k] should be identical to the output signals y1[n-k] and y2[n-k] corresponding to the original input signals x1[n] and x2[n].
y1[n-k] = x1[2(n-k)+4] = x1[2n-2k+4]
y2[n-k] = x2[2(n-k)+4] = x2[2n-2k+4]
y1[n] = x1[2n+4]
y2[n] = x2[2n+4]
Now, if we compare y1[n-k] and y2[n-k] with y1[n] and y2[n], we can see that they are not identical. This is because the output signal depends on the value of n, and a time shift of k samples in the input signal results in a different value of n.
Therefore, the given system is time-variant.
Conclusion:
The given discrete system, y[n] = x[2n+4], is time-variant as the output signal depends on the time at which the input is applied.
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Comment on the time invariance of the following discrete system: y[n] = x[2n+4].a)Time invariantb)Time variantc)Both Time variant and Time invariantd)None of the mentionedCorrect answer is option 'B'. Can you explain this answer?
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