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=== Answer 2=== | === Answer 2=== | ||
| − | + | <math>X(z) = \sum_{n=-\infty}^{+\infty} x[n] z^{-n} = \sum_{n=-\infty}^{+\infty} 3^n u[-n+3] z^{-n}</math> | |
| + | |||
| + | Let k=-n+3, n=3-k, then | ||
| + | |||
| + | <math>X(z) = \sum_{k=-\infty}^{+\infty} (3)^{n-k}u[k](z)^{-3+k}</math> | ||
| + | |||
| + | <math>X(z) = (\frac{3}{z})^{3}\sum_{k=0}^{+\infty} (\frac{z}{3})^{k}</math> | ||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
Revision as of 10:38, 19 September 2013
Contents
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.)
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} $
Answer 3
Write it here.
Answer 4
Write it here.
