Revision as of 17:31, 19 September 2013

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.)

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

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}} $

Back to ECE438 Fall 2013 Prof. Boutin