Does anyone know or is anyone able to explain why a 4th order digital butterworth bandpass/bandstop filter can't be represented/decomposed into a parallel connection of two real allpass filters? A 2nd order and a 6th order butterworth bandpass filter can be decomposed into a parallel connection of two real allpass filters by why not a 4th order filter? Sorry, if this question is lame/stupid.
Allpass decomp
Started by ●April 11, 2012
Reply by ●April 11, 20122012-04-11
On 4/11/12 8:31 AM, niarn wrote:> Does anyone know or is anyone able to explain why a 4th order digital > butterworth bandpass/bandstop filter can't be represented/decomposed into a > parallel connection of two real allpass filters? > > A 2nd order and a 6th order butterworth bandpass filter can be decomposed > into a parallel connection of two real allpass filters by why not a 4th > order filter? > > Sorry, if this question is lame/stupid.no, it's a very sophisticated question assuming the premise is true. i surely don't have an answer for you (yet) since i haven't thunk about it. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●April 11, 20122012-04-11
On 4/11/12 10:30 AM, robert bristow-johnson wrote:> On 4/11/12 8:31 AM, niarn wrote: >> Does anyone know or is anyone able to explain why a 4th order digital >> butterworth bandpass/bandstop filter can't be represented/decomposed >> into a >> parallel connection of two real allpass filters? >> >> A 2nd order and a 6th order butterworth bandpass filter can be decomposed >> into a parallel connection of two real allpass filters by why not a 4th >> order filter? >> >> Sorry, if this question is lame/stupid. > > no, it's a very sophisticated question assuming the premise is true. i > surely don't have an answer for you (yet) since i haven't thunk about it. >now i'm suspicious of the premise of the question. can you show me an example of s 2nd-order BPF (with complex-conjugate poles) being decomposed into 2 APFs with real coefficients? is one of those APFs a wire? or a wire with gain? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●April 12, 20122012-04-12
> >now i'm suspicious of the premise of the question. can you show me an >example of s 2nd-order BPF (with complex-conjugate poles) being >decomposed into 2 APFs with real coefficients? is one of those APFs a >wire? or a wire with gain? > >-- > >r b-j rbj@audioimagination.com > >"Imagination is more important than knowledge." >You're right. The 2nd-order case is special in that one of the APFs is a wire. Here is Matlab/Octave code that shows an example. % second order BPF [B_bpf,A_bpf] = butter(1,[0.3 0.5]); % cutoff freqs are arbitrary A_apf1 = A_bpf; B_apf1 = fliplr(A_apf1); % in this special case where one of the APFs is a wire we get A_bpf2 = A_apf1; B_bpf2 = 0.5*(A_apf1 - B_apf1); You can then confirm that B_bpf2 == B_bpf. 'Of course' A_bpf2 == A_bpf for this example. My own explanation that the decomp is not valid for 4th order butterworth BPF is the following. 1. For Butterworth LPF the decomp is valid for uneven order. Although I can't proof this in any way or make any serious math that backs up this fact then intuitively I can see this because at nyquist the phase difference between the output of the two APFs must be pi or some uneven number times pi. This can only be possible (for real APFs) when the order of the LPF is uneven. 2. The BPF is created by a LP2BP transformation that increases the order by a factor of two. So because the decomp is valid for orders 1, 3, 5,... for LPF, then the decomp is valid for orders 2, 6, 10,... for BPF. 3. Hence, the decomp is not valid for 4th order BPF. If the above explanation is valid, then it is not really an explanation I'm satisfied with. What do you think?
