Can time-invariance be determined from a given a transfer function?

Question

I've the following function.

$$ G(z) = 2 + \frac{-1+5z^{-1}}{(1-0.5z^{-1})(1-z^{-1})}$$

Calculating it's inverse using $\mathcal Z$-Transform, I get the following function:

$$g[n] = 2\delta[n] + 8u[n] - 9(0.5)^{n}u[n]$$

where $u[n]$ is the unit step and $\delta[n]$ is the impulse.

I tried to do the usual approach to find if a system is time-invariant or not.

That is shifting the input and compute the output $y_1$ and shifting the output $y_2$, compare both; if equal then it's time-invariant, else it isn't time-invariant. However, I don't know what could be the input in this case.

So, I'm supposing that $g[n]$ is the impulse response of the system. But I couldn't find a way to tell if the system is time-invariant or not based on it's impulse response. Maybe I'm missing some key idea here.

Answer 1

The question can only be answered if it is clarified what it is that the function $G(z)$ describes. If $G(z)$ is the system's transfer function then we're done immediately, because only linear time-invariant (LTI) systems can be characterized by a transfer function of that form.

If $G(z)$ is the transfer function, the output sequence $y[n]$ is given by the convolution of the input sequence $x[n]$ and the impulse response $g[n]$, which is the inverse $\mathcal{Z}$-transform of $G(z)$. Note that in order to compute the inverse $\mathcal{Z}$-transform you must know the region of convergence (ROC), or you must be given the additional information that the system is causal. Having computed the impulse response $g[n]$, the output is given by the following convolution sum:

$$y[n]=\sum_{k=-\infty}^{\infty}g[k]x[n-k]\tag{1}$$

It's trivial to show that the response to a shifted input $x[n-n_0]$ is given by $y[n-n_0]$. Hence, any system whose input-output relation is described by a convolution $(1)$ is time-invariant (and, obviously, also linear).