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]

Task

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]

with javascript_semantics
function direct_form_II_transposed_filter(sequence a, b, signal)
    sequence result = repeat(0,length(signal))
    for i=1 to length(signal) do
        atom tmp = 0
        for j=1 to min(i,length(b)) do tmp += b[j]*signal[i-j+1] end for
        for j=2 to min(i,length(a)) do tmp -= a[j]*result[i-j+1] end for
        result[i] = tmp/a[1]
    end for
    return result
end function
 
constant acoef = {1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17},
         bcoef = {0.16666667, 0.5, 0.5, 0.16666667},
         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}
 
pp(direct_form_II_transposed_filter(acoef, bcoef, signal),{pp_FltFmt,"%9.6f",pp_Maxlen,110})
