Confused about the noble identity?
I'm studying polyphase filtering and decimation and one of the things that gets mentioned occasionally are the noble identities. They are rendered here in figure 24.2: https://www.dsprelated.com/freebooks/sasp/Multirate_Noble_Identities.html.
Going by that figure, you'd think that you could simply throw away N-1 of N samples and use the remaining samples on the transfer function, but if I understand things correctly, that's not the case: you still need all samples, but they just get grouped and then applied to N smaller transfer functions.
In other words, the noble identity seems like a simple rearranging and grouping of terms? If so, what is the big deal about it? To me, the representation of the noble identity in 24.2 doesn't clarify anything, it adds confusion?
I’ll answer my own question after spending a lot more time on this:
The key is the H(z^M) part. I didn’t understand that notation.
When H(z) = h0 + h1 z^-1 + h2 z^-2
then H(z^3) = h0 + h3 z^-3 + h6 z^-6
My mistake was thinking that the noble identity works for any random transfe function that doesn’t have interstitial zeros in the coefficients.
I had reduced a decimating filter into a polyphase by moving the decimation and then rearranging terms to reduce resources. Doing so, I implicitly used the noble identity without knowing what it was.
The noble identities state that downsampling by a factor N followed by a filter with transfer function H(z) is equivalent to first filtering with H(z^N) and then downsampling by the factor N. The latter structure saves computations.
To apply the noble identities, H(z) must be decomposed into polyphase components, which are downsampled and phase-shifted versions of the original filter. For example, if there are 12 coefficients and N=4, the filter can be split into 4 polyphase subfilters. Each subfilter contains 3 coefficients (downsampled by 4 and phase-shifted), and together they represent the same overall filter H(z).
The advantage of this approach is that downsampling occurs before filtering, so no unnecessary computations are performed on samples that would otherwise be discarded.
Example topologies can be found here:
Yes, I understand that you can save a lot of operations (either lower clocks or lower number of MACs) by splitting into polyphase and applying the noble identities. In fact, for a non-DSP expert HW engineers like me and my friend, these steps are second nature. We were confused by how the noble identies are formulated, but the benefits of polyphase sub-filters were always obvious.
I’m not sure what you’re looking for. Are you looking for a proof of the Noble identities, or have you already found the answer?
I already had my answer. ;-)
Please note that in your post, the coefficients are changing, but it only inserts zeros:
```
When H(z) = h0 + h1 z^-1 + h2 z^-2
then H(z^3) = h0 + h1 z^-3 + h2 z^-6






