Practice Problem on Z-transform computation

Compute the z-transform (including the ROC) of the following DT signal:

$ x[n]=3^n u[-n+3] \ $

Then use your answer to obtain the Fourier transform of the signal. (Write enough intermediate steps to fully justify your answer.)

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Answer 1

x[n] = 3ⁿu[-n + 3]

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

$ X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n} $

Let k = -n+3, n = -k+3

$ X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{-k+3} $

$ X(z) = (\frac{3}{z})^{3} \sum_{k=0}^{+\infty} (\frac{z}{3})^{k} $

$ X(z) = (\frac{27}{z^3}) \sum_{k=0}^{+\infty} (\frac{z}{3})^{k} $

By geometric series formula,

$ X(z) = (\frac{27}{z^3}) (\frac{1}{1-(\frac{z}{3})}) $ ,for |z| < 3

X(z) = diverges, else

So,

$ X(z) = (\frac{3}{3-z}) $ with ROC, |z| < 3

Answer 2

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n} $

Let k=-n+3, n=3-k, then

$ X(z) = \sum_{k=-\infty}^{+\infty} (3)^{n-k}u[k](z)^{-3+k} $

$ X(z) = (\frac{3}{z})^{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^{k} $

$ X(z) = \left\{ \begin{array}{l l} (\frac{3}{z})^3 \frac{1}{1-\frac{z}{3}} &, if \quad |z| < 3\\ \text{diverges} &, \quad \text{otherwise} \end{array} \right. $

$ \mathcal{F}(x[n]r^{-n}) = X(3e^{jw}) = \mathcal{X}(w) = \frac{\frac{3}{3e^{jw}}}{1-e^{jw}} $

Answer 3

Kyungjun Kim

$ X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n} $

Let l=-n+3, n=3-l, then

$ X(z) = \sum_{l=-\infty}^{+\infty} (3)^{n-l}u[k]z^{-3+l} $

$ X(z) = (\frac{3}{z})^{3}\sum_{l=0}^{+\infty} (\frac{z}{3})^{l} $

$ X(z) = (\frac{3}{z})^3 \frac{1}{1-\frac{z}{3}} $ if |z| < 3

Answer 4

$ X[Z] = \sum_{n=-\infty}^{+\infty} 3^{n}u[n+3] Z^{-n} $

$ X[Z] = \sum_{n=-3}^{+\infty} 3^{n}Z^{-n} $

$ X[Z] = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} $

$ X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} $

$ for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, no effect, because this converges everywhere on plane. $

$ for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n}) = \frac{1}{1-\frac{3}{z}}, if |\frac{3}{z}|<1, |z|>3 $

or diverges else.

for the DTFT for this signal,

$ for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \frac{1}{1-\frac{3}{z}}, |z|>3, so it is impossible to have e^{j\omega}, because ROC is bigger at 3 $

$ for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, the DTFT is follow: $

$ \sum_{n=-3}^{n=-1} (\frac{3}{e^{j\omega}})^{n} $

for all, this signal can't have DTFT.

Answer 5

x[n] = 3ⁿu[-n + 3]

$ X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} $

$ X(z) = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n} $

$ X(z) = \sum_{-\infty}^{3} 3^n z^{-n} $

Let k = -n,

$ X(z) = \sum_{k=-3}^{+\infty} 3^{-k} z^k $

$ X(z) = \sum_{k=-3}^{+\infty} (\frac{z}{3})^{k} $

For |z| < 3, we have, by geometric series, that:

$ X(z) = (\frac{27 z^{-3}}{1+(\frac{z}{3})}) $

By simplification,

$ X(z) = (\frac{-81 z^{-4}}{1-3 z^{-1}}) $, ROC |z| < 3.

Since the ROC contains the unit circle, there exists a Fourier transform representation for the signal. Therefore, to find the signal`s Fourier representation, we just need to replace z be $ e^{j w} $

So,

$ X(\omega) = -(\frac{81 e^{-j 4 w}}{1-3 e^{-j w}}) $

Answer 6

$ x[n] = 3^n u[-n+3] $

$ X(z) = \sum_{n = -\infty}^{\infty} x[n] z^{-n} $

$ = \sum_{n=-\infty}^{\infty} 3^n u[-n+3] z^{-n} $

let k = -n+3 => n = -k+3

$ X(z) = \sum_{k=0}^{\infty} 3^{-k+3} z^{k-3} $

$ = \frac{3^3}{z^3} \sum_{k=0}^{\infty} {\frac{z}{3}}^k $

$ = \frac{3^3}{z^3} \frac{1}{1-\frac{z}{3}}, \left |\frac{z}{3} \right | < 1 $

$ X(z) = \frac{3^3}{z^3} \frac{1}{1-\frac{z}{3}}, \left | z \right | < 3 $

diverges , else

$ F{{x[n]r^{-n}}} = X(3e^{jw}) = X(\omega) = \frac{e^{-j \omega 3}}{1-e^{jw}} $

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