Revision as of 19:21, 12 September 2013

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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

$X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}$

$X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}$

$X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}$

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,

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