Revision as of 16:35, 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.)

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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. $

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

Answer 3

Kyungjun Kim

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

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

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

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

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

Answer 4

$ 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.

for the DTFT for this signal,

$ for \sum_{n=0}^{+\infty} (\frac{3}{z})^{n} = \frac{1}{1-\frac{3}{z}}, |z|>3, so it is impossible to have e^{j\omega}, because ROC is bigger at 3 $

$ for \sum_{n=-3}^{n=-1} (\frac{3}{z})^{n}, the DTFT is follow: $

$ \sum_{n=-3}^{n=-1} (\frac{3}{e^{j\omega}})^{n} $

for all, this signal can't have DTFT.

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