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Biquad normalizes feedforward but not feedback by a0 · case 05

With a0 = 2 the poles are twice as strong as specified and the filter can blow up.

Member previewVariant 5 · 3 implementations · 7 checks per implementation

Case contract

Input [[b0, b1, b2, a0, a1, a2], samples]; coefficients are rational strings, normalized by a0 ("a0-zero" if a0 == 0). y[n] = b0 x[n] + b1 x[n-1] + b2 x[n-2] - a1 y[n-1] - a2 y[n-2] with zero initial state; return outputs as exact fraction strings.

Why this case matters

The biquad is the workhorse IIR section; sign, state or normalization slips produce unstable or wrong-gain filters.

One recorded failure

Sample boundary fixture

This sample comes from the broken implementation of a controlled reproducer.

Boundary fixtureActualExpectedOutcome
regression: random biquad 17["-4/3", "5/3", "-17/12"]["-4/3", "10/3", "-7"]Failed

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