Revert of Compute tail time from Biquad coefficients

third_party/WebKit/Source/platform/audio/Biquad.cpp

Index: third_party/WebKit/Source/platform/audio/Biquad.cpp
diff --git a/third_party/WebKit/Source/platform/audio/Biquad.cpp b/third_party/WebKit/Source/platform/audio/Biquad.cpp
index 4876bf26283ea3a39e88a37c70c7a3b540196090..51944e38075bc7937dade8c185be5132509e24f5 100644
--- a/third_party/WebKit/Source/platform/audio/Biquad.cpp
+++ b/third_party/WebKit/Source/platform/audio/Biquad.cpp
@@ -585,298 +585,4 @@
   }
 }
 
-static double RepeatedRootResponse(double n,
-                                   double c1,
-                                   double c2,
-                                   double r,
-                                   double log_eps) {
-  // The response is h(n) = r^(n-2)*[c1*(n+1)*r^2+c2]. We're looking
-  // for n such that |h(n)| = eps.  Equivalently, we want a root
-  // of the equation log(|h(n)|) - log(eps) = 0 or
-  //
-  //   (n-2)*log(r) + log(|c1*(n+1)*r^2+c2|) - log(eps)
-  //
-  // This helps with finding a nuemrical solution because this
-  // approximately linearizes the response for large n.
-
-  return (n - 2) * log(r) + log(fabs(c1 * (n + 1) * r * r + c2)) - log_eps;
-}
-
-// Regula Falsi root finder, Illinois variant
-// (https://en.wikipedia.org/wiki/False_position_method#The_Illinois_algorithm).
-//
-// This finds a root of the repeated root response where the root is
-// assumed to lie between |low| and |high|.  The response is given by
-// |c1|, |c2|, and |r| as determined by |RepeatedRootResponse|.
-// |log_eps| is the log the the maximum allowed amplitude in the
-// response.
-static double RootFinder(double low,
-                         double high,
-                         double log_eps,
-                         double c1,
-                         double c2,
-                         double r) {
-  // Desired accuray of the root (in frames).  This doesn't need to be
-  // super-accurate, so half frame is good enough, and should be less
-  // than 1 because the algorithm may prematurely terminate.
-  const double kAccuracyThreshold = 0.5;
-  // Max number of iterations to do.  If we haven't converged by now,
-  // just return whatever we've found.
-  const int kMaxIterations = 10;
-
-  int side = 0;
-  double root = 0;
-  double f_low = RepeatedRootResponse(low, c1, c2, r, log_eps);
-  double f_high = RepeatedRootResponse(high, c1, c2, r, log_eps);
-
-  // The function values must be finite and have opposite signs!
-  DCHECK(std::isfinite(f_low));
-  DCHECK(std::isfinite(f_high));
-  DCHECK_LE(f_low * f_high, 0);
-
-  int iteration;
-  for (iteration = 0; iteration < kMaxIterations; ++iteration) {
-    root = (f_low * high - f_high * low) / (f_low - f_high);
-    if (fabs(high - low) < kAccuracyThreshold * fabs(high + low))
-      break;
-    double fr = RepeatedRootResponse(root, c1, c2, r, log_eps);
-
-    DCHECK(std::isfinite(fr));
-
-    if (fr * f_high > 0) {
-      // fr and f_high have same sign.  Copy root to f_high
-      high = root;
-      f_high = fr;
-      side = -1;
-    } else if (f_low * fr > 0) {
-      // fr and f_low have same sign. Copy root to f_low
-      low = root;
-      f_low = fr;
-      if (side == 1)
-        f_high /= 2;
-      side = 1;
-    } else {
-      // f_low * fr looks like zero, so assume we've converged.
-      break;
-    }
-  }
-
-  // Want to know if the max number of iterations is ever exceeded so
-  // we can understand why that happened.
-  DCHECK_LT(iteration, kMaxIterations);
-
-  return root;
-}
-
-double Biquad::TailFrame(int coef_index, double max_frame) {
-  // The Biquad filter is given by
-  //
-  //   H(z) = (b0 + b1/z + b2/z^2)/(1 + a1/z + a2/z^2).
-  //
-  // To compute the tail time, compute the impulse response, h(n), of
-  // H(z), which we can do analytically.  From this impulse response,
-  // find the value n0 where |h(n)| <= eps for n >= n0.
-  //
-  // Assume first that the two poles of H(z) are not repeated, say r1
-  // and r2.  Then, we can compute a partial fraction expansion of
-  // H(z):
-  //
-  //   H(z) = (b0+b1/z+b2/z^2)/[(1-r1/z)*(1-r2/z)]
-  //        = b0 + C2/(z-r2) - C1/(z-r1)
-  //
-  //  where
-  //    C2 = (b0*r2^2+b1*r2+b2)/(r2-r1)
-  //    C1 = (b0*r1^2+b1*r1+b2)/(r2-r1)
-  //
-  // Expand H(z) then this in powers of 1/z gives:
-  //
-  //   H(z) = b0 -(C2/r2+C1/r1) + sum(C2*r2^(i-1)/z^i + C1*r1^(i-1)/z^i)
-  //
-  // Thus, for n > 1 (we don't care about small n),
-  //
-  //   h(n) = C2*r2^(n-1) + C1*r1^(n-1)
-  //
-  // We need to find n0 such that |h(n)| < eps for n > n0.
-  //
-  // Case 1: r1 and r2 are real and distinct, with |r1|>=|r2|.
-  //
-  // Then
-  //
-  //   h(n) = C1*r1^(n-1)*(1 + C2/C1*(r2/r1)^(n-1))
-  //
-  // so
-  //
-  //   |h(n)| = |C1|*|r|^(n-1)*|1+C2/C1*(r2/r1)^(n-1)|
-  //          <= |C1|*|r|^(n-1)*[1 + |C2/C1|*|r2/r1|^(n-1)]
-  //          <= |C1|*|r|^(n-1)*[1 + |C2/C1|]
-  //
-  // by using the triangle inequality and the fact that |r2|<=|r1|.
-  // And we want |h(n)|<=eps which is true if
-  //
-  //   |C1|*|r|^(n-1)*[1 + |C2/C1|] <= eps
-  //
-  // or
-  //
-  //   n >= 1 + log(eps/C)/log(|r1|)
-  //
-  // where C = |C1|*[1+|C2/C1|] = |C1| + |C2|.
-  //
-  // Case 2: r1 and r2 are complex
-  //
-  // Thne we can write r1=r*exp(i*p) and r2=r*exp(-i*p).  So,
-  //
-  //   |h(n)| = |C2*r^(n-1)*exp(-i*p*(n-1)) + C1*r^(n-1)*exp(i*p*(n-1))|
-  //          = |C1|*r^(n-1)*|1 + C2/C1*exp(-i*p*(n-1))/exp(i*n*(n-1))|
-  //          <= |C1|*r^(n-1)*[1 + |C2/C1|]
-  //
-  // Again, this is easily solved to give
-  //
-  //   n >= 1 + log(eps/C)/log(r)
-  //
-  // where C = |C1|*[1+|C2/C1|] = |C1| + |C2|.
-  //
-  // Case 3: Repeated roots, r1=r2=r.
-  //
-  // In this case,
-  //
-  //   H(z) = (b0+b1/z+b2/z^2)/[(1-r/z)^2
-  //
-  // Expanding this in powers of 1/z gives:
-  //
-  //   H(z) = C1*sum((i+1)*r^i/z^i) - C2 * sum(r^(i-2)/z^i) + b2/r^2
-  //        = b2/r^2 + sum([C1*(i+1)*r^i + C2*r^(i-2)]/z^i)
-  // where
-  //   C1 = (b0*r^2+b1*r+b2)/r^2
-  //   C2 = b1*r+2*b2
-  //
-  // Thus, the impulse response is
-  //
-  //   h(n) = C1*(n+1)*r^n + C2*r^(n-2)
-  //        = r^(n-2)*[C1*(n+1)*r^2+C2]
-  //
-  // So
-  //
-  //   |h(n)| = |r|^(n-2)*|C1*(n+1)*r^2+C2|
-  //
-  // To find n such that |h(n)| < eps, we need a numerical method in
-  // general, so there's no real reason to simplify this or use other
-  // approximations.  Just solve |h(n)|=eps directly.
-  //
-  // Thus, for an set of filter coefficients, we can compute the tail
-  // time.
-  //
-
-  // If the maximum amplitude of the impulse response is less than
-  // this, we assume that we've reached the tail of the response.
-  // Currently, this means that the impulse is less than 1 bit of a
-  // 16-bit PCM value.
-  const double kMaxTailAmplitude = 1 / 32768.0;
-
-  // Find the roots of 1+a1/z+a2/z^2 = 0.  Or equivalently,
-  // z^2+a1*z+a2 = 0.  From the quadratic formula the roots are
-  // (-a1+/-sqrt(a1^2-4*a2))/2.
-
-  double a1 = a1_[coef_index];
-  double a2 = a2_[coef_index];
-  double b0 = b0_[coef_index];
-  double b1 = b1_[coef_index];
-  double b2 = b2_[coef_index];
-
-  double tail_frame = 0;
-  double discrim = a1 * a1 - 4 * a2;
-
-  if (discrim > 0) {
-    // Compute the real roots so that r1 has the largest magnitude.
-    double r1;
-    double r2;
-    if (a1 < 0) {
-      r1 = (-a1 + sqrt(discrim)) / 2;
-    } else {
-      r1 = (-a1 - sqrt(discrim)) / 2;
-    }
-    r2 = a2 / r1;
-
-    double c1 = (b0 * r1 * r1 + b1 * r1 + b2) / (r2 - r1);
-    double c2 = (b0 * r2 * r2 + b1 * r2 + b2) / (r2 - r1);
-
-    DCHECK(std::isfinite(r1));
-    DCHECK(std::isfinite(r2));
-    DCHECK(std::isfinite(c1));
-    DCHECK(std::isfinite(c2));
-
-    // It's possible for kMaxTailAmplitude to be greater than c1 + c2.
-    // This may produce a negative tail frame.  Just clamp the tail
-    // frame to 0.
-    tail_frame = clampTo(
-        1 + log(kMaxTailAmplitude / (fabs(c1) + fabs(c2))) / log(r1), 0);
-
-    DCHECK(std::isfinite(tail_frame));
-  } else if (discrim < 0) {
-    // Two complex roots.
-    // One root is -a1/2 + i*sqrt(-discrim)/2.
-    double x = -a1 / 2;
-    double y = sqrt(-discrim) / 2;
-    std::complex<double> r1(x, y);
-    std::complex<double> r2(x, -y);
-    double r = hypot(x, y);
-
-    DCHECK(std::isfinite(r));
-
-    // It's possible for r to be 1. (LPF with Q very large can cause this.)
-    if (r == 1) {
-      tail_frame = max_frame;
-    } else {
-      double c1 = abs((b0 * r1 * r1 + b1 * r1 + b2) / (r2 - r1));
-      double c2 = abs((b0 * r2 * r2 + b1 * r2 + b2) / (r2 - r1));
-
-      DCHECK(std::isfinite(c1));
-      DCHECK(std::isfinite(c2));
-
-      tail_frame = 1 + log(kMaxTailAmplitude / (c1 + c2)) / log(r);
-      DCHECK(std::isfinite(tail_frame));
-    }
-  } else {
-    // Repeated roots.  This should be pretty rare because all the
-    // coefficients need to be just the right values to get a
-    // discriminant of exactly zero.
-    double r = -a1 / 2;
-
-    if (r == 0) {
-      // Double pole at 0.  This just delays the signal by 2 frames,
-      // so set the tail frame to 2.
-      tail_frame = 2;
-    } else {
-      double c1 = (b0 * r * r + b1 * r + b2) / (r * r);
-      double c2 = b1 * r + 2 * b2;
-
-      DCHECK(std::isfinite(c1));
-      DCHECK(std::isfinite(c2));
-
-      // It can happen that c1=c2=0.  This basically means that H(z) =
-      // constant, which is the limiting case for several of the
-      // biquad filters.
-      if (c1 == 0 && c2 == 0) {
-        tail_frame = 0;
-      } else {
-        // The function c*(n+1)*r^n is not monotonic, but it's easy to
-        // find the max point since the derivative is
-        // c*r^n*(1+(n+1)*log(r)).  This has a root at
-        // -(1+log(r))/log(r). so we can start our search from that
-        // point to max_frames.
-
-        double low = clampTo(-(1 + log(r)) / log(r), 1.0,
-                             static_cast<double>(max_frame - 1));
-        double high = max_frame;
-
-        DCHECK(std::isfinite(low));
-        DCHECK(std::isfinite(high));
-
-        tail_frame = RootFinder(low, high, log(kMaxTailAmplitude), c1, c2, r);
-      }
-    }
-  }
-
-  return tail_frame;
-}
-
 }  // namespace blink