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I have this system: $$y[k] = (1 + (-1)^k)x[k]$$ I would like to know if it is time-invariant or not. I've done some work and came out with this: $$y_1[k] = (1 + (-1)^k)x[k - k_0] = y_2[k] = (1 + (-1)^{k-k_0})x[k - k_0]$$ Which simplifies to this: $$x[k - k_0] = (-1)^{k_0}x[k - k_0]$$ Which means that the system is time-invariant but only when $k_0$ is a multiple of $2$. Does the logic make sense here?

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  • $\begingroup$ You, and those who have answered, are working too hard. Suppose $x$ is the unit pulse or discrete impulse and so $x[k]$ is $0$ for all $k\neq 0$ while $x[0]=1$. Then, $y[k]=0$ for all $k$. Delay $x$ by $1$ to get $\hat{x}$ so that it is $\hat{x}[1]$ that has value $1$ now while all other $\hat{x}[k]=0$. Now, $\hat{y}[k]=0$ for all $k$ except $k=1$ when $\hat{y}[1]$ has value $1$. Clearly, delaying the input by $1$ does not delay the output by $1$; $\hat{y}$ is not a delayed version of $y$. Ergo, not a time-invariant system. $\endgroup$ Oct 7 at 3:11
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What you have done is absolutely correct. Since the property is not true for every $k$0, the system is not time invariant

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Which means that the system is time-invariant but only when k0 is a multiple of 2.

If the system is only time invariant at certain starting times, then it's not time-invariant; it's time-varying.

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