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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?

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  • $\begingroup$ 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 '15 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 '15 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 '15 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 '15 at 14:54
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The signal $x[n]=3^n$ for $n\le -3$ (and zero otherwise, I assume) can be written as

$$x[n]=3^nu[-n-3]\tag{1}$$

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 $x[n]$.

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  • $\begingroup$ 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 '15 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 '15 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 '15 at 16:55

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