Which one of the following systems described by the following input-output relations is time invariant ? 

Time Invariance in Systems

Definition: A system is considered time-invariant if its behavior and characteristics do not change over time. In other words, if an input signal is delayed by a certain amount of time, the output signal will be equally delayed by the same amount. Mathematically, a system is time-invariant if for any input signal x[n] and any time delay n0, the following condition holds true:

Y[n] = T{x[n]} implies Y[n - n₀] = T{x[n - n₀]}

Here, T represents the system's transformation, Y[n] is the output, and x[n] is the input.

Analysis of the Given Options:

Let's analyze each option to determine if the system is time-invariant:

Option 1: Y[n] = nx[n]

To check for time invariance, we need to examine if a time shift in the input signal results in an equivalent time shift in the output signal:

If the input is x[n], the output is Y[n] = nx[n].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n0] = (n - n₀)x[n - n₀]

Since the output is not simply a time-shifted version of the original output (nx[n]), this system is not time-invariant.

Option 2: Y[n] = x[n] - x[n - 1]

For this option, we again check if a time shift in the input results in an equivalent time shift in the output:

If the input is x[n], the output is Y[n] = x[n] - x[n - 1].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n₀] = x[n - n₀] - x[n - n₀ - 1]

This is indeed the time-shifted version of the original output. Therefore, this system is time-invariant.

Option 3: Y[n] = x[-n]

Here, we check the time invariance by examining the time shift:

If the input is x[n], the output is Y[n] = x[-n].

If the input is shifted by n₀, the new input is x[n - n₀], and the output should be:

Y[n - n₀] = x[-(n - n₀)] = x[-n + n₀]

This is not simply a time-shifted version of the original output. Therefore, this system is not time-invariant.

Option 4: Y[n] = x[n] cos(2πf₀n)

For this option, we check the time invariance by examining the time shift:

If the input is x[n], the output is Y[n] = x[n] cos(2πf₀n).

If the input is shifted by n₀, the new input is x[n - n_0], and the output should be:

Y[n - n₀] = x[n - n₀] cos(2πf₀(n - n₀))

This is not simply a time-shifted version of the original output. Therefore, this system is not time-invariant.

Conclusion:

After analyzing all the options, it is clear that:

Option 2: Y[n] = x[n] - x[n - 1] is the correct answer as it satisfies the condition for time invariance.