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<math> X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} </math> | <math> X[Z] = \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n} + \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} </math> | ||
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+ | <math> for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, no effect, because this converges everywhere on plane. </math> | ||
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+ | <math> for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \{ | ||
Revision as of 17:40, 12 September 2013
$ <math>Insert formula here $</math>
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
$ 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) = (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3}) $
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} = \{ ---- ===Answer 5=== Yixiang Liu <math>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} $