How to resolve the algorithm Apply a digital filter (direct form II transposed) step by step in the Nim programming language

Published on 12 May 2024 09:40 PM
#Nim

How to resolve the algorithm Apply a digital filter (direct form II transposed) step by step in the Nim programming language

Table of Contents

Problem Statement

Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the "direct form II transposed" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [1] Filter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667] The signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589] [Wikipedia on Butterworth filters]

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Apply a digital filter (direct form II transposed) step by step in the Nim programming language

Source code in the nim programming language

import strformat

func filter(a, b, signal: openArray[float]): seq[float] =

  result.setLen(signal.len)

  for i in 0..signal.high:
    var tmp = 0.0
    for j in 0..min(i, b.high):
      tmp += b[j] * signal[i - j]
    for j in 1..min(i, a.high):
      tmp -= a[j] * result[i - j]
    tmp /= a[0]
    result[i] = tmp

#———————————————————————————————————————————————————————————————————————————————————————————————————

let a = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17]
let b = [0.16666667, 0.5, 0.5, 0.16666667]

let signal = [-0.917843918645,  0.141984778794, 1.20536903482,   0.190286794412,
              -0.662370894973, -1.00700480494, -0.404707073677,  0.800482325044,
               0.743500089861,  1.01090520172,  0.741527555207,  0.277841675195,
               0.400833448236, -0.2085993586,  -0.172842103641, -0.134316096293,
               0.0259303398477, 0.490105989562, 0.549391221511,  0.9047198589]

let result = filter(a, b, signal)
for i in 0..result.high:
  stdout.write fmt"{result[i]: .8f}"
  stdout.write if (i + 1) mod 5 != 0: ", " else: "\n"

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