Line 53: Line 53:
 
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math>
 
<math>X(z) = \sum_{n=-3}^{+\infty} 3^n z^{-n}</math>
  
<math>X(z) = \sum_{n=-3}^{+\infty} (3/z)^{n}</math>
+
<math>X(z) = \sum_{n=-3}^{+\infty} (\frac{3}{z})^{n}</math>
  
 
Let k = n+3, n = k-3
 
Let k = n+3, n = k-3
  
<math>X(z) = \sum_{k=0}^{+\infty} (3/z)^{k-3}</math>
+
<math>X(z) = \sum_{k=0}^{+\infty} (\frac{3}{z})^{k-3}</math>
  
<math>X(z) = (z/3)^{3} \sum_{k=0}^{+\infty} (3/z)^{k}</math>
+
<math>X(z) = (\frac{z}{3})^{3} \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>
  
<math>X(z) = (z^3/27) \sum_{k=0}^{+\infty} (3/z)^{k}</math>
+
<math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>
  
<math>X(z) = (z^3/27) \sum_{k=0}^{+\infty} (3/z)^{k}</math>
+
<math>X(z) = (\frac{z^3}{27}) \sum_{k=0}^{+\infty} (\frac{3}{z})^{k}</math>
  
<math>X(z) = (z^3/27) 1/(1-3/z) </math> ,for |z| < 3
+
By geometric series formula,
 +
 
 +
<math>X(z) = (\frac{z^3}{27}) (\frac{1}{1-(\frac{3}{z})}) </math>   ,for |z| < 3
  
 
X(z) = diverges, else
 
X(z) = diverges, else
 +
 +
So,
 +
 +
<math>X(z) = (\frac{z}{z-3}) </math> with ROC, |z| < 3
  
 
===Answer 3===
 
===Answer 3===

Revision as of 16:47, 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.)


Share your answers below

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

Green26 ece438 hmwrk3 power series.png

$ 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

Write it here.


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