First of all, thank you all for your answers. I know the z transform for

$$ x(n)=3^n \space ; \space n\geqslant 3 $$

or rather

$$ x(n)= 3^n u(n-3) $$$$\begin{align}X(z)&=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&=\sum_{n=-\infty}^\infty3^nu(n-3)z^{-n}\\& \text{change of variables: } n'=n-3 \\& =\sum_{n=-\infty}^\infty 3^{(n'+3)} u(n') z^{(-n'-3)}\\& \text{recall: } n'-> n \\&=3^3z^{-3}\sum_{n=0}^{\infty}(3 z^{-1})^{n}\\& X(z) =3^3 z^{-3}\frac{1}{1-3{z^{-1}}},\quad |z|>3 \tag{1}\end{align} $$\begin{align} \\& \text{instead, for the sequence: } x(n)=3^{n} \ ; n\leqslant -3 \\&=\sum_{n=-\infty}^\infty3^nu(-n-3)z^{-n}\\& \text{change of variables: } n'=-n-3 \\& =\sum_{n=-\infty}^\infty 3^{(-n'-3)} u(n') z^{-(-n'-3)}\\& \text{recall: } n'-> n \\&=3^{-3} z^{3}\sum_{n=0}^{\infty}(3^{-1} z)^{n}\\& X(Z)= 3^{-3}z^{3}\frac{1}{1- \frac{z}{3}} \quad |z|<3\tag{2}\end{align} Is it correct? And for @Matt L. I don't understand your mathematical step initial when you write: $$\begin {align} \sum_{n=-\infty}^3 3^nz^{-n} =\sum_{n=-\infty}^0 3^{(n+3)} u(n') z^{-(n+3)} \end {align} $$

Can you write me the generic formula?

  • $\begingroup$ Welcome to DSP.SE! Your "or rather" equation is not equivalent to your first equation. Your $z$-transform expression is missing the ROC (region of convergence). What is the region of convergence? $\endgroup$
    – Peter K.
    Nov 18, 2015 at 1:38
  • $\begingroup$ I've edited your question according to Peter's comments, i.e. I corrected the argument of the step function, and I added the ROC to the expression for the $\mathcal{Z}$-transform. $\endgroup$
    – Matt L.
    Nov 18, 2015 at 8:20
  • $\begingroup$ @MattL. Sorry, i forgot to write the ROC for my Z Transform, thank you Matt for your added, it's okay! The exercise required the calculation of the Z transform and the ROC for the sequence. I proceed in this way : first I calculate the Z transformed , then I have to make considerations for the ROC. I have just one more question that I write in response post, watch it ;) thank you so much, from Italy . $\endgroup$
    – P_B
    Nov 18, 2015 at 11:38
  • $\begingroup$ @MattL. my next edited is because i develop for n<= -3 , for the second sequence.. ok ?. We arrive at the same result , I believe that my method is also correct, No? $\endgroup$
    – P_B
    Nov 18, 2015 at 14:54

1 Answer 1


The signal $x[n]=3^n$ for $n\le -3$ (and zero otherwise, I assume) can be written as


To compute its $\mathcal{Z}$-transform you simply need to use the formula:

$$\begin{align}X(z)&=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&=\sum_{n=-\infty}^{-3}3^nz^{-n}\\&=\sum_{n=-\infty}^03^{(n-3)}z^{-(n-3)}\\&=3^{-3}z^{3}\sum_{n=0}^{\infty}\left(\frac{z}{3}\right)^n\\&=3^{-3}z^{3}\frac{1}{1-\frac13 z},\quad |z|<3\tag{2}\end{align}$$

It's important to specify the region of convergence ($|z|<3$), otherwise the $\mathcal{Z}$-transform does not uniquely identify the corresponding sequence $x[n]$.

  • $\begingroup$ Ok , if I understand , for the second sequence 3 ^ n for n > = - 3 : if i take indices from -∞ to 0 instead of from -∞ to -3 , i have decrease to the index in the formula by 3 to obtain the same result $\endgroup$
    – P_B
    Nov 18, 2015 at 15:08
  • $\begingroup$ I beg you to excuse me , but yesterday I was wrong and I did not put " - " in front of the 3 for the second sequence . I understand that I am creating some confusion and is not what I want . I would just like to be sure of what I did and I thank you so much for helping me $\endgroup$
    – P_B
    Nov 18, 2015 at 15:16
  • $\begingroup$ @P_B: OK, no problem, I've updated my answer accordingly, and we obtain the same result. $\endgroup$
    – Matt L.
    Nov 18, 2015 at 16:55

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.