Haar Example
Before we leave the case of amplitude-complementary, two-channel,
critically sampled, perfect reconstruction filter banks, let's see
what happens when
is the simplest possible lowpass filter
having unity dc gain, i.e.,
![]() |
(12.29) |
This case is obtained above by setting
![$ E_0(z^2)=1/2$](http://www.dsprelated.com/josimages_new/sasp2/img2033.png)
![$ o=1$](http://www.dsprelated.com/josimages_new/sasp2/img2034.png)
![$ h_0(1)=1/2$](http://www.dsprelated.com/josimages_new/sasp2/img2035.png)
![$ H_0(z)$](http://www.dsprelated.com/josimages_new/sasp2/img1982.png)
![]() |
(12.30) |
Choosing
![$ H_1(z)=1-H_0(z)$](http://www.dsprelated.com/josimages_new/sasp2/img2037.png)
![$ F_0(z)$](http://www.dsprelated.com/josimages_new/sasp2/img1984.png)
![$ F_1(z)$](http://www.dsprelated.com/josimages_new/sasp2/img1985.png)
![\begin{eqnarray*}
H_0(z) &=& \frac{1}{2} + \frac{1}{2}z^{-1} \eqsp E_0(z^2)+z^{-1}E_1(z^2)\\ [5pt]
H_1(z) &=& 1-H_0(z) \eqsp \frac{1}{2} - \frac{1}{2}z^{-1} \eqsp E_0(z^2)-z^{-1}E_1(z^2)\\ [5pt]
F_0(z) &=& \;\;\, H_1(-z) \eqsp \frac{1}{2} + \frac{1}{2}z^{-1} \eqsp \;\;\,H_0(z)\\ [5pt]
F_1(z) &=& -H_0(-z) \eqsp -\frac{1}{2} + \frac{1}{2}z^{-1} \eqsp -H_1(z).
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/sasp2/img2038.png)
Thus, both the analysis and reconstruction filter banks are scalings
of the familiar Haar filters (``sum and difference'' filters
). The frequency responses are
![\begin{eqnarray*}
H_0(e^{j\omega}) &=&\;\;\,F_0(e^{j\omega}) \eqsp \frac{1}{2} + \frac{1}{2}e^{-j\omega}\eqsp e^{-j\frac{\omega}{2}} \cos\left(\frac{\omega}{2}\right)\\ [5pt]
H_1(e^{j\omega}) &=& -F_0(e^{j\omega}) \eqsp \frac{1}{2} - \frac{1}{2}e^{-j\omega}\eqsp j e^{-j\frac{\omega}{2}} \sin\left(\frac{\omega}{2}\right)
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/sasp2/img2040.png)
which are plotted in Fig.11.16.
Next Section:
Polyphase Decomposition of Haar Example
Previous Section:
Amplitude-Complementary 2-Channel Filter Bank