Assuming the system $h[n]$ is LTI (and has an associated $H(z)$ transform), is the whole system below LTI?

I found the impulse response of the system and I got that it is $$h_{0}[n]=\alpha ^{-n}\cdot h[n]$$ where $h_{0}[n]$ is $y[n]$ when $x[n]=\delta [n]$.

That doesn't give us much information, so I thought of testing the system with the input signal being a complex exponential (i.e. $x[n]=a^{n}$). Because complex exponentials are eigenfunctions of LTI systems, if the output is not something like $y[n]=z^{n}\cdot H(z)$ then the system would not be LTI.

However, the output with $x[n]=a^{n}$ turned out to be $$y[n]=a^{n}\cdot H(a\cdot \alpha)$$

So, apparently, complex exponentials are eigenfunctions of the whole system. Also, I think that $H_{0}(z)=H(z\cdot \alpha)$. Is that right?

Is this enough to affirm that the whole system is LTI? Or does it just prove that it could be LTI?

Thank you for your time!


1 Answer 1


You are right that determining the response to an impulse will generally not lead to any useful description of system behavior, unless the system is LTI. Your reasoning using eigenfunctions is correct. However, I would approach the problem as follows. If (and only if) the input/output relation can be formulated as a convolution of the input signal with a sequence that is independent of the input signal (the impulse response), then the system is LTI. And this is indeed possible:


where $\tilde{h}[n]=\alpha^{-n}h[n]$ is the impulse response of the total system, as you've already found out by yourself.

The $\mathcal{Z}$-transform of $\tilde{h}[n]$ is indeed easily expressed in terms of $H(z)$:

$$\tilde{H}(z)=\sum_{n=-\infty}^{\infty}\tilde{h}[n]z^{-n}=\sum_{n=-\infty}^{\infty}h[n]\alpha^{-n}z^{-n}=H(\alpha z)$$

  • $\begingroup$ I'd just like to add that this is not true if $\alpha=0$. There's also trouble if $\alpha$ is negative and $n$ is rational, and I suspect it's even worse if they're allowed to be complex. $\endgroup$
    – MBaz
    Commented Jan 26, 2016 at 20:27
  • $\begingroup$ Thank you Matt, brilliant as usual. And nice observation by MBaz $\endgroup$
    – Tendero
    Commented Jan 27, 2016 at 1:21
  • $\begingroup$ Sorry to come back here, but I was checking this again and I don't really get your answer. Where did the $y[n]=\alpha^{-n}\sum_{k=-\infty}^{\infty}x[k]\alpha^kh[n-k]$ come from? $\endgroup$
    – Tendero
    Commented Feb 10, 2016 at 1:26
  • 1
    $\begingroup$ @M.S.: This is just the mathematical representation of the system in the figure: $x[n]$ is multiplied by $\alpha^n$, then convolved with $h[n]$, and finally multiplied by $\alpha^{-n}$. $\endgroup$
    – Matt L.
    Commented Feb 10, 2016 at 8:50

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