(New page: <math>X(w) = F{x[n]} = \sum_{n=-\infty}^\infty x[n]e^{-jwn}</math> <math>X(z)|_{z=e^{jw}} = X(e^{jw})</math> Can compute Z-Transform as a DTFT write <math>X(z)=X(re^{jw})</math> then <m...) |
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| − | + | [[Category:discrete-time Fourier transform]] | |
| + | [[Category:z-transform]] | ||
| − | <math>X( | + | =Relationship between [[Discrete-time_Fourier_transform_info|DTFT]] and [[Info_z-transform|z-transform]]= |
| + | ---- | ||
| + | Recall that | ||
| + | *The Discrete-time Fourier transform (DTFT) is <math>{\mathcal X}(\omega) = {\mathcal F} \left( x[n] \right) = \sum_{n=-\infty}^\infty x[n]e^{-j\omega n}</math>. | ||
| + | *The z-transform is <math>X(z)= {\mathcal Z} \left( x[n] \right)= \sum_{n=-\infty}^\infty x[n] z^{-n}</math> | ||
| + | ---- | ||
| + | '''1. One can obtain the DTFT from the z-transform X(z) by as follows:''' | ||
| − | + | <center><math>\left. X(z)\right|_{z=e^{jw}} = {\mathcal X}(\omega) </math></center> | |
| − | + | ||
| − | then | + | 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'''. | |
| − | <math>X(z)= | + | Write the z-transform <math>X(z)=X(re^{jw})</math> using polar coordinates for the complex number z. Then |
| − | <math> = | + | |
| + | <math>\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}</math> | ||
| + | |||
| + | So the z-transform is like a DTFT after multiplying the signal by the signal <math>y[n]=r^{-n}</math>. | ||
| + | ---- | ||
Latest revision as of 14:49, 1 May 2015
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:
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} $.
