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As part of an exercise, I have been assigned to find an alternative method of executing the additions of a transposed FIR filter. There is a hint stating that the alternative method should be in a form resembling a binary tree. I figured that the leaves of the tree should be the product of x(n)*h(i), i=0,..,k but I am having trouble finding the actual solution. What am I missing?

Edit:

After further research, this paper was found. It explains the tree-like structure of the adders of an FIR Filter.

Peter K.
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samiu
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  • That is indeed an odd exercise . An FIR filter is easy enough to transpose but the result doesn't look anything like a binary tree (to me). Do you have any constraint on the number of coefficients? Is it a power of 2? – Hilmar Jul 07 '22 at 18:23
  • @Hilmar The transposed FIR structure is given and the exercise requests to change only the part where additions and delays are made into a binary tree. There is not any constrain on the number of coefficients, it is just referred as k. – samiu Jul 07 '22 at 18:40

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Try looking at this paper from Xilinx.

In particular, Figure 2 seems to do something like what you want:

Xilinx Figure 2

The cascaded adders on the right look like they'd eventually form a binary tree if you had a long enough filter.

This paper has more of the transpose information, but the connection might not be as clear.

Transpose version.

Peter K.
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  • Thanks for the answer. I have seen this particular paper and fig2 is the structure of a non transposed fir filter. I am looking for the binary version of the transposed one. – samiu Jul 07 '22 at 19:57
  • @samiu In that case, you should expand upon your question and add references to what you've looked at that doesn't help. Please edit your question to add links to these references. Also, please add the "working" (thinking) that you've done to date. – Peter K. Jul 07 '22 at 20:30