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=Relationship between DTFT and [[Info_z-transform|z-transform]]=
 
=Relationship between DTFT and [[Info_z-transform|z-transform]]=
<math>X(w) = F{x[n]} = \sum_{n=-\infty}^\infty x[n]e^{-jwn}</math>
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----
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Recall that
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*The Discrete-time Fourier transform (CTFT) is <math>{\mathcal X}(\omega) = {\mathcal F} \left( x[n] \right) = \sum_{n=-\infty}^\infty x[n]e^{-j\omega n}</math>.
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*The z-transform is <math>X(z)= {\mathcal Z} \left( x[n] \right)= \sum_{n=-\infty}^\infty x[n] z^{-n}</math>
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----
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'''1. One can obtain the  DTFT from the z-transform X(z) by as follows:'''
  
<math>X(z)|_{z=e^{jw}} = X(e^{jw})</math>
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<center><math>\left. X(z)\right|_{z=e^{jw}} = X(e^{jw}) </math></center>
  
Can compute Z-Transform as a DTFT
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In other words, if you restrict the z-transoform to the unit circle in the complex plane, then you get the Fourier transform (DTFT).
write <math>X(z)=X(re^{jw})</math>
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then <math>X(z)= \sum_{-\infty}^\infty x[n]z^{-n}</math>
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'''2. One can also obtain the Z-Transform from the DTFT'''.
  
<math>X(z)= \sum_{-\infty}^\infty x[n](re^{jw})^{-n}</math>
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Write the z-transform <math>X(z)=X(re^{jw})</math> using polar coordinates for the complex number z. Then
  
<math>X(z)= \sum_{-\infty}^\infty x[n]r^{-n}e^{-jwn}</math>
 
  
<math> = F{x[n]r^{-n}}</math>
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<math>\begin{align} X(z)&= \sum_{-\infty}^\infty x[n]z^{-n}\\
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& = \sum_{-\infty}^\infty x[n](re^{jw})^{-n} \\
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& = \sum_{-\infty}^\infty x[n]r^{-n}e^{-jwn} \\
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& = {\mathcal F} \left( x[n]r^{-n} \right)
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\end{align}</math>
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So the z-transform is like a DTFT after multiplying the signal by the signal <math>y[n]=r^{-n}</math>.
 
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Revision as of 14:02, 30 April 2015


Relationship between DTFT and z-transform


Recall that

  • The Discrete-time Fourier transform (CTFT) 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}} = X(e^{jw}) $

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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