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  <math>X(z)=X(re^{j\omega})</math>
 
  <math>X(z)=X(re^{j\omega})</math>
  Then <math>X(z) = F(x[n]r^-n)</math>
+
 
+
  Then <math>X(z) = F(x[n]r^{-n})</math>
 +
 
  <math>X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} = \sum_{n = -\infty}^\infty x[n](re^{j\omega})^{-n} = \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n}</math>
 
  <math>X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} = \sum_{n = -\infty}^\infty x[n](re^{j\omega})^{-n} = \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n}</math>

Revision as of 14:08, 30 November 2008

Z Transform

Discrete analog of Laplace Transform

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

    Where z is a complex variable.

Relationship between Z-Transform and F.T.

$ X(\omega) = X(e^{j\omega}) $
$ X(z)=X(re^{j\omega}) $

Then $ X(z) = F(x[n]r^{-n}) $

$ X(z) = \sum_{n = -\infty}^\infty x[n]z^{-n} = \sum_{n = -\infty}^\infty x[n](re^{j\omega})^{-n} = \sum_{n = -\infty}^\infty x[n]r^{-n}e^{-j\omega n} $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010