Optimally truncating head-related impulse response by dynamic programming with its applications

Abstract

We propose a method to optimally truncate the head-related impulse responses (HRIRs) in this paper. The truncated HRIR consists of a portion of the original HRIR and a flat line. An algorithm based on dynamic programming is used to optimally select the portions of the original HRIRs and the constants of the flat lines to minimize the modeling errors. The truncated HRIRs can be used to reproduce multi-channel sound for headphones with a significantly lower computational cost. The proposed method is compared with another approximation method, the CAPZ (Common-Acoustical-Pole and Zero) approach. The experimental results show that the proposed method yields lower composition as well as modeling errors for the same amount of computation. Compared with the direct implementation, the proposed approach requires about 35 % of the computational cost while maintaining acceptable composition errors.

Appendix A. Computational complexity of exhaustive search

This appendix briefly discusses the lower-bound time complexity of using an exhaustive search to find optimal section IIs, i.e., the optimal starting points and lengths (N _i ^II) of section IIs. Let the number of impulse responses be F, the number of coefficients in a response be N, $M = \sum\nolimits_{{i = 1}}^F {N_i^{{II}}}$, and, for simplicity, N = M (M is proportional to N in practice). An allocation of M is an assignment of the values of N ₁ ^II … N _F ^II satisfying the constraint that $M = \sum\nolimits_{{i = 1}}^F {N_i^{{II}}}$. For example, if F = 5 and M = 100, a possible allocation is N ₁ ^II = 1, N ₂ ^II = 1, N ₂ ^II = 1, N ₃ ^II = 1, N ₄ ^II = 96. In this case, since N ₁ ^II = 1, the section II of the first impulse response has a total of N possible starting points. An exhaustive search must compute the modeling errors of all possible distinct allocations and starting points, and record the one with the smallest error. We will show that there exists at least Ω(M ^F−1) distinct allocations, and each of the distinct allocation has at least Ω(M ^F−1) distinct combinations of starting points. Therefore, an exhaustive search must compute $\varOmega ({M^{{F - 1}}} \times {M^{{F - 1}}}) = \varOmega ({M^{{2F - 2}}})$ distinct modeling errors. Since computing a particular modeling error requires Ω(N) = Ω(M) time (Eq. 10 or Eq. 11). The complexity of an exhaustive search is Ω(M ^2F−1).

Let’s consider the number of distinct allocations first. Since we are concerned with the lower bound, we do not need to calculate all possible distinct allocations. Instead, we consider a subset of all possible distinct allocations, one with the restriction that $1 \leqslant N_1^{{II}} \leqslant \frac{M}{F}$, …, $1 \leqslant N_{{F - 1}}^{{II}} \leqslant \frac{M}{F}$. Under this restriction, N ₁ ^II have $\frac{M}{F}$ different possible values; so are N ₂ ^II …N _F−1 ^II . Note that this restriction requires $F - 1 \leqslant \sum\limits_{{i = 1}}^{{F - 1}} {N_i^{{II}}} \leqslant \frac{{(F - 1)}}{F}M$, and therefore there must exist a value of N _F ^II that satisfies $M = \sum\nolimits_{{i = 1}}^F {N_i^{{II}}}$. In other words, all combinations of possible values of N ₁ ^II…N _F−1 ^II are legal and are distinct allocations. Therefore, there exist $\varOmega \left( {\frac{{{M^{{F - 1}}}}}{{{F^{{F - 1}}}}}} \right)$ distinct allocations. Since F is typically a constant (e.g., 5), F ^F−1 is also a constant (e.g., 5 ⁴ = 625). The number of distinct allocations can be simplified as Ω(M ^F−1).

We now calculate the number of possible starting points for each distinct allocation. Normally, for the i ^th impulse response, depending on the value of N _i ^II, the number of distinct starting points can be as small as 1 (when N _i ^II = N), and as large as N (when N _i ^II = 1). However, for the first (F−1) impulse responses, we have made the restriction that $1 \leqslant N_i^{{II}} \leqslant \frac{M}{F}$. When N _i ^II = 1 and $N_i^{{II}} = \frac{M}{F}$, the number of distinct starting points are N and $N - \frac{M}{F}$, respectively. Thus, each of the first (F−1) impulse response has at least Ω(M) distinct starting points. Therefore, the first (F−1) impulse responses alone have Ω(M ^F−1) possible combinations of starting points. Note that we did not count the starting points of the F ^th impulse response. This is safe because we are calculating a lower bound.

Since there are Ω(M ^F−1) distinct allocations and each allocation has Ω(M ^F−1) distinct combination of starting points, an exhaustive search must compute $\varOmega ({M^{{F - 1}}} \times {M^{{F - 1}}}) = \varOmega ({M^{{2F - 2}}})$ different possible modeling errors. The modeling error of each allocation can be calculated in Ω(N) = Ω(M) time according to Eq. 10 or Eq. 11. Therefore, the complexity of an exhaustive search becomes $\varOmega ({M^{{2F - 2}}} \times M) = \varOmega ({M^{{2F - 1}}})$. If F = 5 (five-channel audio) and M = 220, the number of computations is at least proportional to ${M^9} = 1.2 \times {10^{{21}}}$. For a computer that can compute one square and one addition in Eq. 10 in 10⁻⁸ s, it would take 3.8 × 10⁵ years to find an optimal solution, which is clearly impractical.