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 64d6b26c99985233b5871340dd47557e49e9bd0b..8505bda61b733fda333b2df268566257f13cedbe 100644 |
--- a/third_party/WebKit/Source/platform/audio/Biquad.cpp |
+++ b/third_party/WebKit/Source/platform/audio/Biquad.cpp |
@@ -591,4 +591,298 @@ void Biquad::GetFrequencyResponse(int n_frequencies, |
} |
} |
+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 |