0
$\begingroup$

What is the transfer function of the system described by the following affine input ($x$)-output ($y$) relationship: $$ y[n] = \alpha x[n] + \beta. $$

Using the Z-transform we find: $$ Y[z] = \alpha X[z]+ \beta $$ and $$ H[z] = \frac{Y[z]}{X[z]} = \alpha + \frac{\beta}{X[z]} $$

But I don't know how to take it from here to compute the transfer function! I'm not sure if this can be done. Any help would be appreciated.

$\endgroup$
3
  • 1
    $\begingroup$ You have a static system. No memory. No state. Why do you need its Z-transform? $\endgroup$ Mar 27 '21 at 9:45
  • $\begingroup$ If a system is LTI, then $Y (z) = H (z) X (z)$. Thus, if $X (z) = 0$, then $Y (z) = 0$. Is that the case for your static affine system? $\endgroup$ Mar 27 '21 at 9:50
  • $\begingroup$ @RodrigodeAzevedo, I have a system with an output that I know can be described by this affine relationship and for which I do not know the input. It is when I was thinking of estimating the parameters $\alpha$, $\beta$ using a blind system identification approach and I started thinking about identifiability questions that I thought about getting its transfer function, then got stuck! $\endgroup$
    – Likely
    Mar 27 '21 at 15:00
3
$\begingroup$

That system cannot be described by a transfer function because it is not linear. It neither satisfies the condition of additivity nor does it satisfy the condition of homogeneity.

If $y_1[n]$ is the response to an input $x_1[n]$, and $y_2[n]$ is the response to an input $x_2[n]$, then the response to $x_1[n]+x_2[n]$ doesn't equal $y_1[n]+y_2[n]$ (additivity is violated). Furthermore, the response to $\alpha x_1[n]$ doesn't equal $\alpha y_1[n]$ (homogeneity is violated).

This is unrelated to existence questions concerning the $\mathcal{Z}$-transform. The $\mathcal{Z}$-transform of a constant $\beta$ doesn't equal $\beta$, but it simply doesn't exist.

$\endgroup$
1
$\begingroup$

It does not have a Z-transform.

The constant sequence $f[n] = \beta$, $ \forall n, \beta \neq 0$ does not have a Z-transform, as the sum $$F(z) = \sum_{n=-\infty}^{\infty} \beta z^{-n}$$ diverges for any $z$; i.e., there's no set of $z$ values (region of convergence - ROC) for which the sum is finite (convergent).

Hence your affine function $$y[n] = \alpha x[n] + \beta$$ does not have a Z-transform.

May be you want to restrict the domain of input to $n\geq 0$ in which case you will have a mapping of $$y[n] = (\alpha x[n] + \beta) u[n] $$ which may have a Z-transform, provided taht $x[n]u[n]$ does have a Z-transform.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.