How to resolve the algorithm Apply a digital filter (direct form II transposed) step by step in the Lua programming language
How to resolve the algorithm Apply a digital filter (direct form II transposed) step by step in the Lua 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 Lua programming language
Source code in the lua programming language
function filter(b,a,input)
local out = {}
for i=1,table.getn(input) do
local tmp = 0
local j = 0
out[i] = 0
for j=1,table.getn(b) do
if i - j < 0 then
--continue
else
tmp = tmp + b[j] * input[i - j + 1]
end
end
for j=2,table.getn(a) do
if i - j < 0 then
--continue
else
tmp = tmp - a[j] * out[i - j + 1]
end
end
tmp = tmp / a[1]
out[i] = tmp
end
return out
end
function main()
local sig = {
-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
}
--Constants for a Butterworth filter (order 3, low pass)
local a = {1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17}
local b = {0.16666667, 0.5, 0.5, 0.16666667}
local result = filter(b,a,sig)
for i=1,table.getn(result) do
io.write(result[i] .. ", ")
end
print()
return nil
end
main()
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