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<math> F(x[n]r^{-n}) = X(3^{e^{jw}}) = X(w) = (\frac{(\frac{3}{3e^{jw}})}{1-e^{jw}})
+
<math> F(x[n]r^{-n}) = X(3^{e^{jw}}) = X(w) = (\frac{(\frac{3}{3e^{jw}})}{1-e^{jw}}) </math>
  
 
===Answer 3===
 
===Answer 3===

Revision as of 11:45, 19 September 2013


Practice Problem on Z-transform computation

Compute the z-transform (including the ROC) of the following DT signal:

$ x[n]=3^n u[-n+3] \ $

Then use your answer to obtain the Fourier transform of the signal. (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!


Answer 1

x[n] = 3nu[-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, n = -k+3

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

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

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

By geometric series formula,

$ X(z) = (\frac{27}{z^3}) (\frac{1}{1-(\frac{z}{3})}) $ ,for |z| < 3

X(z) = diverges, else

So,

$ X(z) = (\frac{3}{3-z}) $ with ROC, |z| < 3

Answer 2

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

Let k=-n+3, n=3-k, then

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

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

$ X(z) = \left\{ \begin{array}{l l} (\frac{3}{z})^3 \frac{1}{1-\frac{z}{3}} &, if \quad |z| < 3\\ \text{diverges} &, \quad \text{otherwise} \end{array} \right. $

$ F(x[n]r^{-n}) = X(3^{e^{jw}}) = X(w) = (\frac{(\frac{3}{3e^{jw}})}{1-e^{jw}}) $

Answer 3

Write it here.

Answer 4

Write it here.


Back to ECE438 Fall 2013 Prof. Boutin

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn