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Optimal Bit Allocation

Module by: Phil Schniter. E-mail the author

Summary: For transform coding with a fixed total bit rate, the optimal (SNR-maximizing) allocation of bit rates is derived.

  • Motivating Question: Assuming that T is an N×NN×N orthogonal matrix, what is the MSE-optimal bitrate allocation strategy assuming independent uniform quantization of the transform outputs? In other words, what {R}{R} minimize average reconstruction error for fixed average rate 1NR1NR?
  • Say that the kthkth element of the transformed output vector y(n)y(n) has variance {σyk2}{σyk2}. With uniform quantization, example 1 from Background and Motivation showed that the kthkth quantizer error power is
    σqk2=γykσyk22-2Rk,σqk2=γykσyk22-2Rk,
    (1)
    where Rk is the bit rate allocated to the kthkth quantizer output and where γykγyk depends on the distribution of yk. From this point on we make the simplifying assumption that γykγyk is independent of k. As shown in example 1 from Background and Motivation, orthogonal matrices imply that the mean squared reconstruction error equals the mean squared quantization error, so that
    σr2=1Nk=0N-1σqk2=γyNk=0N-1σyk22-2Rk.σr2=1Nk=0N-1σqk2=γyNk=0N-1σyk22-2Rk.
    (2)
    Thus we have the constrained optimization problem
    min {Rk}k=0N-1σyk22-2Rks.t.R=1Nk=0N-1Rk. min {Rk}k=0N-1σyk22-2Rks.t.R=1Nk=0N-1Rk.
    (3)
    Using the Lagrange technique, we first set
    Rkσyk22-2Rk-λ1NkRk=0.Rkσyk22-2Rk-λ1NkRk=0.
    (4)
    Since 2-2Rk=(eln2)-2Rk=e-2Rkln22-2Rk=(eln2)-2Rk=e-2Rkln2, the zero derivative implies
    0=-2ln2·2-2R·σy2-λNR=-12log2λ-2Nσy2ln2.0=-2ln2·2-2R·σy2-λNR=-12log2λ-2Nσy2ln2.
    (5)
    Hence
    R=1NkRk=-12Nklog2λ-2Nσyk2ln2=-12log2λ-2Nln2+12Nklog2σyk2R=1NkRk=-12Nklog2λ-2Nσyk2ln2=-12log2λ-2Nln2+12Nklog2σyk2
    (6)
    so that
    -12log2λ-2Nln2=R-12log2kσyk21/N.-12log2λ-2Nln2=R-12log2kσyk21/N.
    (7)
    Rewriting Equation 5 and plugging in the expression above,
    R=-12log2λ-2Nln2+12log2σy2=R-12log2kσyk21/N+12log2σy2=R+12log2σy2k=0N-1σyk21/N.R=-12log2λ-2Nln2+12log2σy2=R-12log2kσyk21/N+12log2σy2=R+12log2σy2k=0N-1σyk21/N.
    (8)
  • The optimal bitrate allocation expression Equation 8 (lower equation) is meaningful only when R0R0, and practical only for integer numbers of quantization levels 2-2R2-2R (or practical values of R for a particular coding scheme). Practical strategies typically
    • set R=0R=0 to when Equation 8 (lower equation) suggests that the optimal R is negative,
    • round positive R to practical values, and
    • iteratively re-optimize {R}{R} using these rules until all R have practical values.
  • Plugging Equation 8 (lower equation) into Equation 1, we find that optimal bit allocation implies
    σq2=γy2-2Rk=0N-1σyk21/N,σq2=γy2-2Rk=0N-1σyk21/N,
    (9)
    which means that, with optimal bit allocation, each coefficient contributes equally to reconstruction error. (Recall a similar property of the Lloyd-Max quantizer.)

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