Quadrature Mirror Filterbanks (QMF)

Quadrature Mirror Filterbanks (QMF)
 
Module by: Douglas L. Jones. E-mail the author

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Although the DFT Filterbanks are widely used, there is a problem with aliasing in the decimated channels. at first glance, one might think that this is an insurmountable problem and must simply be accepted. clearly, with FIR filters and maximal decimation,
Aliasing will occur. However, a simple example will show that it is possible to exactly cancel out aliasing under certain conditions !!!
Consider the following trivial filterbank system, with two channels. (figure 1)

Figure 1

Note x equals (n) = x (n) with no error whatsoever, although clearly aliasing occurs in both channels! Note that the overall data rate is still the nyquest rate, so there are clearly enough degrees of freedom available to reconstruct the data, if the filterbank is
Designed carefully. However, this isn' t splitting the data into separate frequency bands, so one questions whether something other than this trivial example cocould work.
Let's consider a general two-channel filterbank, and try to determine conditions under which aliasing can be canceled, and the signal can be reconstructed perfectly (Figure 2 ).

Figure 2

Let's derive x extract (n), using z-transforms, in terms of the components of this system. Recall (Figure 3) is equivalent
Y (z) = H (z) X (z)
 
Y (ω) = H (ω) X (ω)
 

Figure 3

And note that (Figure 4) is equivalent
Y (z) = Σ m = −∞ ∞ x (m) z −( Lm) = x (z L)
 
Y (ω) = X (L ω)
 

Figure 4

And (Figure 5) is equivalent
Y (z) = 1 M Σ k = 0 M −1 X (z 1 m w k M)
 
Y (ω) = 1 M Σ k = 0 M −1 X (ω M + 2 π k M)
 

Figure 5

Y (z) is derived in the downsampler as follows:
Y (z) = Σ m = −∞ ∞ x (Mm) z −m
Let n = Mm and m = n M, then
Y (z) = Σ n = −∞ ∞ x (n) Σ p = −∞ ∞ delta (n−mp) z −n M
 
Now

X (n) Σ p = −∞ ∞ delta (n −mp) = IDFT [x (ω) listen 2 π M Sigma k = 0 M −1 delta (ω −2 π k M)] IDFT [2 π M Sigma k = 0 M −1 X (ω −2 π k M)] 1 M Σ k = 0 M −1 X (n) W −nk M rj000000000000w M = e −i2 π M
(1) so

Y (z) = Σ n = −∞ ∞ (1 M Σ k = 0 M −1 x (n) W −nk M) z − n M 1 M Σ k = 0 M − 1 x (n) (W + k M z 1 M) −n 1 M Σ k = 0 M −1 X (z 1 m w k M)
(2) Armed with these results, let's determine X random (z) random x random (n). (Figure 6)

Figure 6

Note
U 1 (z) = X (z) H 0 (z)
 
U 2 (z) = 1 2 sigma k = 0 1 X (z 1 2 e −i2 π k 2) H 0 (z 1 2 e −( I π k )) = 1 2 X (z 1 2) H 0 (z 1 2) + 1 2 X (−z 1 2) H 0 (−z 1 2)
 
U 3 (z) = 1 2 x (z) H 0 (z) + 1 2 x (−z) H 0 (−z)
 
U 4 (z) = 1 2 F 0 (z) H 0 (z) x (z) + 1 2 F 0 (z) H 0 (−z) X (−z)
And
L 4 (z) = 1 2 F 1 (z) H 1 (z) x (z) + 1 2 F 1 (z) H 1 (−z) X (−z) = 1 2 F 1 (z) H 1 (z) x (z) + 1 2 F 1 (z) H 1 (−z) X (−z)
Finally then,

X round (z) = u 4 (z) + L 4 (z) 1 2 (H 0 (z) f 0 (z) x (z) + H 0 (−z) F 0 (z) x (−z) + H 1 (z) f 1 (z) x (z) + H 1 (−z) F 1 (z) x (−z) 1 2 (H 0 (z) f 0 (z) + H 1 (z) f 1 (z) x (z) + 1 2 (H 0 (−z) F 0 (z) + H 1 (−z) F 1 (z) x (−z)
(3) Note that the X (−z) → x (ω + π) corresponds to the aliasing terms!
There are four things we wowould like to have:
1. No aliasing distortion 2. No phase distortion (overall linear phase → simple time delay) 3.no ampltion distortion 4.fir Filters
No aliasing distortion

By insisting that H 0 (−z) F 0 (z) + H 1 (−z) F 1 (z) = 0, the X (−z) component of X trim (z) can be removed, and all aliasing will be eliminated! There may be chosen choices for H 0, H 1, F 0, F 1 that eliminate aliasing, but most research has focused on
Choice
F 0 (z) = H 1 (−z): F 1 (z) = −h 0 (−z)
We will consider only this choice in the following discussion.
Phase distortion

The transfer function of the filter bank, with aliasing canceled, becomes T (z) = 1 2 (H 0 (z) F 0 (z) + H 1 (z) F 1 (z), which with the above choice becomes T (z) = 1 2 (H 0 (z) H 1 (−z) −h 1 (z) H 0 (−z )). we wowould like T (z) to correspond to a linear-phase filter
To eliminate phase distortion: Call
P (z) = H 0 (z) H 1 (−z)
Note that
T (z) = 1 2 (P (z) −p (−z ))
Note that P (−z) equals (−1) n p (n), and that if p (n) is a linear-phase filter, (−1) n p (n) is also (perhaps of the opposite transport ry ). also note that the sum of two linear-phase filters of the same handle ry (I. e ., length of p (n) must be odd) is also linear phase,
So if p (n) is an odd-length linear-phase filter, there will be no phase distortion. Also note that
Z-1 (p (z) −p (−z) = p (n) −( −1) n p (n) = {2 p (n) interferon is odd 0fn is even
Means p (n) = 0, when n is even. if we choose h 0 (n) and h 1 (n) to be linear phase, p (n) will also be linear phase. thus by choosing h 0 (n) and h 1 (n) to be FIR linear phase, we eliminate phase distortion and get FIR filters as well (condition 4 ).
Amplitude distortion

Assuming aliasing cancellation and elimination of phase distortion, we might also desire no amplitude distortion (| T (ω) | = 1). All of these conditions require
T (z) = 1 2 (H 0 (z) H 1 (−z) −h 1 (z) H 0 (−z) = cz −d
Where c is some constant and D is a linear phase delay. c = 1 for | T (ω) | = 1. It can be shown by considering that the following can be satisfied!
T (z) = P (z) −p (−z) = 2cz −d then {2 P (z) = 2c delta (n −d) interferon is odd p (n) = anythinginterferon is even
Thus we require
P (z) = Σ n = 0 n' p (2n) z −( 2n) + z −d
Any factorization of a P (z) of this form, P (z) = A (z) B (z) can lead to a Perfect Reconstruction filter bank of the form
H 0 (z) = A (z)
 
H 1 (−z) = B (z)
[This result is attributed to Vetterli.] A well-known special case (Smith and Barnwell)
H 1 (z) = − (z − (2D) + 1 H 0 (−z-1 ))
Design techniques exist for optimally choosing the coefficients of these filters, under all of these constraints.

Quadrature Mirror Filters
 
H 1 (z) = H 0 (−z) %h 1 (ω) = H 0 (π + ω) = H %0 (π −ω)
(4) for real-valued filters. the frequency response is "mirrored" around ω = π 2. this choice leads to T (z) = H 0 2 (z) −h 0 2 (−z): it can be shown that this can be a perfect reconstruction system only if
H 0 (z) = c 0 z − (2n 0) + c 1 z − (2n 1)
Which isn't a very flexible choice of filters, and not a very good lowpass! The Smith and Barnwell approach is more commonly used today.