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Relationship between DTFT and z-transform

Recall that

  • The Discrete-time Fourier transform (DTFT) is $ {\mathcal X}(\omega) = {\mathcal F} \left( x[n] \right) = \sum_{n=-\infty}^\infty x[n]e^{-j\omega n} $.
  • The z-transform is $ X(z)= {\mathcal Z} \left( x[n] \right)= \sum_{n=-\infty}^\infty x[n] z^{-n} $

1. One can obtain the DTFT from the z-transform X(z) by as follows:

$ \left. X(z)\right|_{z=e^{jw}} = {\mathcal X}(\omega) $

In other words, if you restrict the z-transoform to the unit circle in the complex plane, then you get the Fourier transform (DTFT).

2. One can also obtain the Z-Transform from the DTFT.

Write the z-transform $ X(z)=X(re^{jw}) $ using polar coordinates for the complex number z. Then

$ \begin{align} X(z)&= \sum_{-\infty}^\infty x[n]z^{-n}\\ & = \sum_{-\infty}^\infty x[n](re^{jw})^{-n} \\ & = \sum_{-\infty}^\infty x[n]r^{-n}e^{-jwn} \\ & = {\mathcal F} \left( x[n]r^{-n} \right) \end{align} $

So the z-transform is like a DTFT after multiplying the signal by the signal $ y[n]=r^{-n} $.

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang