How is the output size of UPFIRDN determined?

2 ビュー (過去 30 日間)
Harry
Harry 2014 年 10 月 26 日
コメント済み: Bradford Watson 2021 年 1 月 29 日
I want to understand resampling better, so I am trying to replicate the functionality of Matlab's built-in upfirdn() function. To clarify, I'm not looking for a fast implementation (at first), I just want to see how to get the exact same result. As such, I understand the basic steps (conceptually) are:
(1) Upsampling by P (zero insertion).
(2) FIR filtering.
(3) Downsampling by Q (throwing away samples).
This is fine. However, what is bugging me is that (as stated by Matlab help) the final output size is:
Ly = ceil( ((Lx-1)*P + Lh)/Q ) where Lx = length(X) and Lh = length(H)
and I don't understand why. With reference to the 3 stages listed above, a naive implementation might give stage-by-stage output sizes:
(1) P*Lx
(2) P*Lx
(3) floor(P*Lx/Q)
Therefore, I really want to know exactly what upfirdn is doing. Can anyone help?
  1 件のコメント
Bradford Watson
Bradford Watson 2021 年 1 月 29 日
Whenever two sequences are convolved together, the length of the resulting sequence is Lx + Ly - 1. The inner part of this equation is expressing the length of the convolution of the upsampled sequence (Lx-1)*P and the filter length (Lh), though neglecting the -1 term. This is then divided by the decimation, and the ceiling function is taken to get rid of any fraction.

サインインしてコメントする。

回答 (0 件)

カテゴリ

Help Center および File ExchangeMultirate Signal Processing についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by