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)

Note x ^ ⁢n=x⁢n x ^ n x n with no error whatsoever, although clearly aliasing occurs in both channels! Note that the overall data rate is still the Nyquist 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 could work.

Let's consider a general two-channel filterbank, and try to determine conditions under which aliasing can be cancelled, and the signal can be reconstructed perfectly (Figure 2).

Let's derive x ^ ⁢n x ^ n , using z-transforms, in terms of the components of this system. Recall (Figure 3) is equivalent to Y⁢z=H⁢z⁢X⁢z Y z H z X z Y⁢ω=H⁢ω⁢X⁢ω Y ω H ω X ω and note that (Figure 4) is equivalent to Y⁢z=∑m=−∞∞x⁢m⁢z−(L⁢m)=x⁢zL Y z m x m z L m x z L Y⁢ω=X⁢L⁢ω Y ω X L ω and (Figure 5) is equivalent to Y⁢z=1M⁢∑k=0M−1X⁢z1M⁢ W M k Y z 1 M k 0 M 1 X z 1 M W M k Y⁢ω=1M⁢∑k=0M−1X⁢ωM+2⁢π⁢kM Y ω 1 M k 0 M 1 X ω M 2 k M Y⁢z Y z is derived in the downsampler as follows: Y⁢z=∑m=−∞∞x⁢M⁢m⁢z−m Y z m x M m z m Let n=M⁢m n M m and m=nM m n M , then Y⁢z=∑n=−∞∞x⁢n⁢∑p=−∞∞δ⁢n−M⁢p⁢z−nM Y z n x n p δ n M p z n M

Now

so Armed with these results, let's determine X ^ ⁢z⇔ x ^ ⁢n ⇔ X ^ z x ^ n . (Figure 6) Note U 1 ⁢z=X⁢z⁢ H 0 ⁢z U 1 z X z H 0 z U 2 ⁢z=12⁢∑k=01X⁢z12⁢e−j⁢2⁢π⁢k2⁢ H 0 ⁢z12⁢e−(j⁢π⁢k)=12⁢X⁢z12⁢ H 0 ⁢z12+12⁢X⁢−z12⁢ H 0 ⁢−z12 U 2 z 1 2 k 0 1 X z 1 2 2 k 2 H 0 z 1 2 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=12⁢X⁢z⁢ H 0 ⁢z+12⁢X⁢−z⁢ H 0 ⁢−z U 3 z 1 2 X z H 0 z 1 2 X z H 0 z U 4 ⁢z=12⁢ F 0 ⁢z⁢ H 0 ⁢z⁢X⁢z+12⁢ F 0 ⁢z⁢ H 0 ⁢−z⁢X⁢−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=12⁢ F 1 ⁢z⁢ H 1 ⁢z⁢X⁢z+12⁢ F 1 ⁢z⁢ H 1 ⁢−z⁢X⁢−z=12⁢ F 1 ⁢z⁢ H 1 ⁢z⁢X⁢z+12⁢ F 1 ⁢z⁢ H 1 ⁢−z⁢X⁢−z 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, Note that the X⁢−z→X⁢ω+π → X z X ω corresponds to the aliasing terms!

There are four things we would like to have:

No aliasing distortion

By insisting that H 0 ⁢−z⁢ F 0 ⁢z+ H 1 ⁢−z⁢ F 1 ⁢z=0 H 0 z F 0 z H 1 z F 1 z 0 , the X⁢−z X z component of X ^ ⁢z X ^ z can be removed, and all aliasing will be eliminated! There may be many choices for H 0 H 0 , H 1 H 1 , F 0 F 0 , F 1 F 1 that eliminate aliasing, but most research has focused on the choice F 0 ⁢z= H 1 ⁢−z : F 1 ⁢z=− H 0 ⁢−z 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 cancelled, becomes T⁢z=12⁢( H 0 ⁢z⁢ F 0 ⁢z+ H 1 ⁢z⁢ F 1 ⁢z) T z 1 2 H 0 z F 0 z H 1 z F 1 z , which with the above choice becomes T⁢z=12⁢( H 0 ⁢z⁢ H 1 ⁢−z− H 1 ⁢z⁢ H 0 ⁢−z) T z 1 2 H 0 z H 1 z H 1 z H 0 z . We would like T⁢z T z to correspond to a linear-phase filter to eliminate phase distortion: Call P⁢z= H 0 ⁢z⁢ H 1 ⁢−z P z H 0 z H 1 z Note that T⁢z=12⁢(P⁢z−P⁢−z) T z 1 2 P z P z Note that P⁢−z⇔−1n⁢p⁢n ⇔ P z 1 n p n , and that if p⁢n p n is a linear-phase filter, −1n⁢p⁢n 1 n p n is also (perhaps of the opposite symmetry). Also note that the sum of two linear-phase filters of the same symmetry (i.e., length of p⁢n p n must be odd) is also linear phase, so if p⁢n p n is an odd-length linear-phase filter, there will be no phase distortion. Also note that Z-1p⁢z−p⁢−z=p⁢n−−1n⁢p⁢n={2⁢p⁢n  if  n is odd0  if  n is even Z p z p z p n 1 n p n 2 p n n is odd 0 n is even means p⁢n=0 p n 0 , when nn is even. If we choose h 0 ⁢n h 0 n and h 1 ⁢n h 1 n to be linear phase, p⁢n p n will also be linear phase. Thus by choosing h 0 ⁢n h 0 n and h 1 ⁢n 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 T ω 1 ). All of these conditions require T⁢z=12⁢( H 0 ⁢z⁢ H 1 ⁢−z− H 1 ⁢z⁢ H 0 ⁢−z)=c⁢z−D T z 1 2 H 0 z H 1 z H 1 z H 0 z c z D where cc is some constant and DD is a linear phase delay. c=1 c 1 for |T⁢ω|=1 T ω 1 . It can be shown by considering that the following can be satisfied! T⁢z=P⁢z−P⁢−z=2⁢c⁢z−D⇔{2⁢p⁢z=2⁢c⁢δ⁢n−D  if  n is oddp⁢n=anything  if  n is even ⇔ T z P z P z 2 c z D 2 p z 2 c δ n D n is odd p n anything n is even Thus we require P⁢z=∑n=0 N ′ p⁢2⁢n⁢z−(2⁢n)+z−D P z n 0 N ′ p 2 n z 2 n z D Any factorization of a P⁢z P z of this form, P⁢z=A⁢z⁢B⁢z P z A z B z can lead to a Perfect Reconstruction filter bank of the form H 0 ⁢z=A⁢z H 0 z A z H 1 ⁢−z=B⁢z H 1 z B z [This result is attributed to Vetterli.] A well-known special case (Smith and Barnwell) H 1 ⁢z=−(z−(2⁢D)+1⁢ H 0 ⁢−z-1) H 1 z z 2 D 1 H 0 z Design techniques exist for optimally choosing the coefficients of these filters, under all of these constraints.

for real-valued filters. The frequency response is "mirrored" around ω=π2 ω 2 . This choice leads to T⁢z= H 0 2z− H 0 2−z T z H 0 z 2 H 0 z 2 : it can be shown that this can be a perfect reconstruction system only if H 0 ⁢z= c 0 ⁢z−(2⁢ n 0 )+ c 1 ⁢z−(2⁢ n 1 ) H 0 z c 0 z 2 n 0 c 1 z 2 n 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.

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This work is licensed by Douglas L. Jones under a Creative Commons Attribution License (CC-BY 2.0), and is an Open Educational Resource.

Last edited by Connexions on Jun 3, 2009 3:44 pm -0500.