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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.

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  • $\begingroup$ I didn't confirm your math but the inverse z-transfrom (g[n]) is the input response. This is the output you would get with an impulse at the input. In general for an arbitrary input, what is the relationship between the input, impulse response and output? $\endgroup$ – Dan Boschen Jul 2 at 4:45
  • $\begingroup$ As far as I'm concerned, the output is equal to the input convolved with the impulse response. $\endgroup$ – FY Gamer Jul 2 at 12:44
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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).

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  • $\begingroup$ I understand it perfectly. Thank you. Unfortunately, I don't have the information what $G (z) $ is exactly, I guess I'm suppose to make these deductions and imply that it's the system's transfer function. Anyway, I've a question: why should we have to know the system is causal? Couldn't we be able to solve it without this information or knowing that it isn't causal? $\endgroup$ – FY Gamer Jul 2 at 12:54
  • $\begingroup$ @FYGamer: The expression for $G(z)$ alone is not enough to compute $g[n]$. E.g., the inverse Z-transform of $G(z)=1/(1-z^{-1})$ could be $u[n]$ or $-u[-n-1]$ depending on the ROC. $\endgroup$ – Matt L. Jul 2 at 18:00

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