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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} $
$ = \sum_{n=-3}^{+\infty} (\(frac{3}{z})^{n} <math> <math>= \sum_{n=0}^{+\infty} (\(frac{3}{z})^{n} + \sum_{n=-3}^{-1} (\(frac{3}{z})^{n}<math> <math>= \sum_{n=0}^{+\infty} (\(frac{3}{z})^{n} + (\frac{3}{z})^{-3} + (\frac{3}{z})^{-2} + (\frac{3}{z})^{-1} <math>= \sum_{n=0}^{+\infty} (\(frac{3}{z})^{n} + (\frac{z^{3}}{27}) + (\frac{z^{2}}{9}) + (\frac{z}{3}) ===Answer 4=== <math> x[n] = 3^{n}u[n+3] $
$ X[Z] = \sum_{n=-\infty}^{+\infty} 3^{n}u[n+3] Z^{-n} $
$ = \sum_{n=-3}^{+\infty} 3^{n}Z^{-n} $
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
