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<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math> | <math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math> | ||
− | <math>X(z) = \sum_{n=-3}^{+\infty} (3 | + | <math>X(z) = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math> |
Let k = n+3, n = k-3 | Let k = n+3, n = k-3 | ||
− | <math>X(z) = \sum_{k=0}^{+\infty} (3 | + | <math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math> |
− | <math>X(z) = (z | + | <math>X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> |
− | <math>X(z) = (z^3 | + | <math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> |
− | <math>X(z) = (z^3 | + | <math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math> |
− | <math>X(z) = (z^3 | + | By geometric series formula, |
+ | |||
+ | <math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math> ,for |z| < 3 | ||
X(z) = diverges, else | X(z) = diverges, else | ||
+ | |||
+ | So, | ||
+ | |||
+ | <math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| < 3 | ||
===Answer 3=== | ===Answer 3=== |
Revision as of 16:47, 12 September 2013
Contents
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.)
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
$ 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
Write it here.
Answer 4
Write it here.