Is it possible to explicit $y(n)$ of this mathematical succession in recursive form using z inverse transform?:

$ y(0) = 1 \\ y(n+1) = 2y(n) + 3 $

I can't write this as a linear system in z (linear difference equation) because there is the $3$ that doesn't depends on input or output.

I know some methods to solve this but I'm trying to use z inverse transform with no success.

  • $\begingroup$ Is your function implicitly defined for non-negative n only? If so, then you may be able to attach the u[n] step function to the constant. In that case, it indeed does have a Z-transform. $\endgroup$ – Envidia May 17 '18 at 20:44
  • $\begingroup$ Thank you @Envidia it solves, if you write an aswer I will choose it as solution $\endgroup$ – Andrea May 18 '18 at 21:30

If you are allowed to define your system to be causal so that it is non-zero only for $n \ge 0$, then you can use the unit step function $u[n]$ so that your system becomes:

$$y[0] = 1$$ $$y[n+1] = 2y[n] + 3u[n]$$

Which does have a Z-transform representation.

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