Revision as of 19:21, 12 September 2013 by Wang701 (Talk | contribs)

$ <math>Insert formula here $</math>

Practice Problem on Z-transform computation

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

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

(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! No need to write your name: we can find out who wrote what by checking the history of the page.


Answer 1

alec green

Green26 ece438 hmwrk3 power series.png

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

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

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

Let k = n+3:

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

Using the geometric series property:

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

Answer 2

Muhammad Syafeeq Safaruddin

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

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

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

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

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

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

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

By geometric series formula,

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

X(z) = diverges, else

So,

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



Answer 3

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

$ X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{3}{z})^{-3} + (\frac{3}{z})^{-2} + (\frac{3}{z})^{-1} $

$ X(z) = \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3}) $

if |3/z|<1,i.e z<-3 or z>3,

$ (z) = (\frac{1}{1-(\frac{3}{z})}) + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3}) $

if -3<z<3,

X(z) diverges
















Answer 4

$ x[n] = 3^{n}u[n+3] $

$ 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.







Answer 5

Yixiang Liu

$ x[n] = 3^{n} u[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

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

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

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

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

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

using geometric series formula

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

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

Back to ECE438 Fall 2013 Prof. Boutin


Answer 6

Xi Wang

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

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

if z > 3

  $ X[z] = (\frac{1}{1-(\frac{3}{z})}) $

if z < 3

  $ Diverges $

Answer 7

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

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

With that we obtain,

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

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

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

Thus, ROC is |z| > 3 because it is restricted by geometric series.

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman