$ <math>Insert formula here $</math>
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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.)
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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) = \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.
Answer 2
Cary Wood
$ 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) = 3^{n} z^{-n} for all n > -3 $
and
$ X(z) = 0, else $
Thus, we re-write X(z) as...
$ = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n} $
By the geometric series formula,
$ X(z) = (\frac{1}{1-(\frac{3}{z})}) $ , for |3/z| < 1
X(z) = diverges, elsewhere
ROC, |z| > 3