Revision as of 17:23, 12 September 2013

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

Write it here.

Answer 4

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

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

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